From fe96dc60a0aecde9b3b01b756674a8aae849591c Mon Sep 17 00:00:00 2001 From: Billy Quarles <4674360+saturnaxis@users.noreply.github.com> Date: Mon, 20 May 2024 20:02:08 -0400 Subject: [PATCH] updated Ch 8 --- Chapter_8/statistical-physics.ipynb | 86 ++++++- _config.yml | 3 +- docs/Chapter_1/birth-of-modern-physics.html | 50 ++-- docs/Chapter_10/semiconductors.html | 2 +- docs/Chapter_11/nuclear-physics.html | 2 +- docs/Chapter_12/particle-physics.html | 2 +- .../special-theory-of-relativity.html | 162 ++++++------- .../experimental-quantum-physics.html | 71 +++--- docs/Chapter_4/structure-of-the-atom.html | 66 +++--- docs/Chapter_5/quantum-mechanics-part1.html | 86 +++---- docs/Chapter_6/quantum-mechanics-part2.html | 194 +++++++-------- docs/Chapter_7/hydrogen-atom.html | 130 +++++------ docs/Chapter_8/statistical-physics.html | 220 +++++++++++------- docs/Chapter_9/molecules-and-lasers.html | 2 +- docs/LICENSE.html | 2 +- docs/Preamble/HW_template.html | 2 +- docs/Preamble/Markdown-basics.html | 2 +- docs/Preamble/Python-basics.html | 2 +- docs/Preamble/who-for.html | 2 +- docs/README.html | 2 +- .../Chapter_8/statistical-physics.ipynb | 86 ++++++- docs/genindex.html | 2 +- docs/home.html | 2 +- docs/search.html | 2 +- docs/searchindex.js | 2 +- 25 files changed, 703 insertions(+), 479 deletions(-) diff --git a/Chapter_8/statistical-physics.ipynb b/Chapter_8/statistical-physics.ipynb index b648175..e00f007 100644 --- a/Chapter_8/statistical-physics.ipynb +++ b/Chapter_8/statistical-physics.ipynb @@ -1000,9 +1000,10 @@ "\n", "Let's retain the \"free electron\" assumption of the Drude model and use the results of the 3-D infinite square-well potential because it corresponds physically to a cubic lattice of ions. The allowed energies are\n", "\n", - "\\begin{align*}\n", + "```{math}\n", + ":label: box_energy\n", "E = \\frac{h^2}{8mL^2}\\left(n_1^2 + n_2^2 + n_3^2 \\right),\n", - "\\end{align*}\n", + "```\n", "\n", "where $L$ is the side-length of a cube and $n_i$ are the integer quantum numbers. \n", "\n", @@ -1268,6 +1269,87 @@ "## Bose-Einstein Statistics\n", "\n", "### Blackbody Radiation\n", + "Recall the definition of an *ideal* blackboddy as a nearly perfect absorbing cavity that emits a spectrum of electromagnetic radiation (see Section [3.5](https://saturnaxis.github.io/ModernPhysics/Chapter_3/experimental-quantum-physics.html#blackbody-radiation)). The problem is to find the intensity of the emitted radiation as a function of temperature and wavelength:\n", + "\n", + "```{math}\n", + ":label: spectral_density\n", + "\n", + "\\mathcal{I}(\\lambda,T) = \\frac{2\\pi c^2 h}{\\lambda^5} \\frac{1}{e^{hc/(\\lambda kT)}-1}. \n", + "```\n", + "\n", + "In quantum theory, we begin with the assumption that the electromagnetic radiation is a collection of photons with energy $hc/\\lambda$. Photons are *bosons* with spin $1$. We use the Bose-Einstein distribution to find how the photons are distributed by energy, and then convert the energy distribution into a function of wavlength via $E= hc/\\lambda.$\n", + "\n", + "The key to the problem is the density of states $g(E)$. We model the photon gas just as we did for the electron gas: a collection of free particles within a 3-D infinite potential well. We cannot use Eq. {eq}`box_energy` for the energy states because the photons are masseless. \n", + "\n", + "We recast the solution to the particle-in-a-box problem in terms of mementum states rather than energy states. For a free particle of mass $m$, the energy $E=p^2/(2m)$. We rewrite Eq. {eq}`box_energy` as\n", + "\n", + "\\begin{align}\n", + "p &= \\sqrt{p_x^2 + p_y^2 + p_z^2}, \\\\\n", + "&= \\frac{h}{2L}\\sqrt{n_1^2 + n_2^2 + n_3^2}.\n", + "\\end{align}\n", + "\n", + "The energy of a photon is $E=pc$ so that\n", + "\n", + "\\begin{align}\n", + "E = \\frac{hc}{2L}\\sqrt{n_1^2 + n_2^2 + n_3^2}.\n", + "\\end{align}\n", + "\n", + "Again by thinking of $n_i$ as the coordinates of a number space and defining $r^2 = n_1^2 + n_2^2 + n_3^2$, we find the number of allowed energy states within \"radius\" $r$ is\n", + "\n", + "$$ N_r = 2\\left(\\frac{1}{8}\\right) \\left( \\frac{4}{3} \\pi r^3\\right). $$\n", + "\n", + "This time the factor of $2$ comes from the two possible *photon polarizations*. Also, the energy is proportional to $r$ by\n", + "\n", + "\\begin{align}\n", + "E &= \\frac{hc}{2L} r.\n", + "\\end{align}\n", + "\n", + "We can then rewrite $N_r$ in terms of $E$ as\n", + "\n", + "\\begin{align}\n", + "N_r = \\frac{1}{3} \\pi r^3 = \\frac{8\\pi L^3}{3h^3 c^3} E^3.\n", + "\\end{align}\n", + "\n", + "The density of states $g(E)$ is\n", + "\n", + "\\begin{align}\n", + "g(E) = \\frac{dN_r}{dE} = \\frac{8\\pi L^3}{h^3 c^3}E^2.\n", + "\\end{align}\n", + "\n", + "The energy distribution is the product of the density of states and the a statistical factor. In this case, the Bose-Einstein factor:\n", + "\n", + "\\begin{align}\n", + "n(E) &= g(E)F_{\\rm BE} = \\frac{8\\pi L^3}{h^3 c^3} E^2 \\frac{1}{e^{E/(kT)}-1}.\n", + "\\end{align}\n", + "\n", + "The normalization factor $B_{\\rm BE} = 1$ because we have a non-normalized collection of photons. As photons are absorbed and emitted by the walls of the cavity, the number of photons is not constant.\n", + "\n", + "The next step is to convert from a number distribution to an energy density distribution $u(E)$. Instead of number density $N/V$, we use a factor $E/L^3$ (energy per unit volume) as a multiplicative factor:\n", + "\n", + "\\begin{align*}\n", + "u(E) = n(E)\\frac{E}{L^3} = \\frac{8\\pi}{h^3 c^3}E^3 \\frac{1}{e^{E/(kT)}-1}.\n", + "\\end{align*}\n", + "\n", + "For all photons between $E$ and $E+dE$,\n", + "\n", + "\\begin{align}\n", + "u(E)\\ dE = \\frac{8\\pi}{h^3 c^3} \\frac{E^3\\ dE}{e^{E/(kT)}-1}.\n", + "\\end{align}\n", + "\n", + "Using $E=hc/\\lambda$ and $|dE| = \\left(hc/\\lambda^2\\right) d\\lambda$., we find\n", + "\n", + "\\begin{align}\n", + "u(\\lambda, T)d\\lambda = \\frac{8\\pi hc}{\\lambda^5} \\frac{ dE}{e^{E/(kT)}-1}.\n", + "\\end{align}\n", + "\n", + "In the SI system, multiplying by a factor $c/4$ is required to change energy density $u(\\lambda, T)$ to a spectral density $\\mathcal{I}(\\lambda, T)$, or Eq. {eq}`spectral_density`.\n", + "\n", + "A few notes:\n", + "\n", + "- Planck **did not** use the Bose-Einstein distribution to derive his radiation law. However, it shows the power of the statistical approach.\n", + "- This problem was first solved by [Satyendra Nath Bose](https://en.wikipedia.org/wiki/Satyendra_Nath_Bose) in 1924 before the concept of spin in quantum theory.\n", + "- Einstein's name was added because he helped Bose publish his work in the West and later applied the distribution to other problems. \n", + "\n", "\n", "### Liquid Helium\n", "\n", diff --git a/_config.yml b/_config.yml index f977ec7..42ce8e2 100644 --- a/_config.yml +++ b/_config.yml @@ -7,7 +7,7 @@ # Book settings title : PHYS2700 Modern Physics # The title of the book. Will be placed in the left navbar. author : Billy Quarles # The author of the book -copyright : "2022" # Copyright year to be placed in the footer +copyright : "2024" # Copyright year to be placed in the footer logo : "bohr_gif.gif" # A path to the book logo # Patterns to skip when building the book. Can be glob-style (e.g. "*skip.ipynb") exclude_patterns : [_build, Thumbs.db, .DS_Store, "**.ipynb_checkpoints","*.png"] @@ -35,6 +35,7 @@ parse: - dollarmath - html_image - html_admonition + - colon_fence ####################################################################################### # LaTeX-specific settings diff --git a/docs/Chapter_1/birth-of-modern-physics.html b/docs/Chapter_1/birth-of-modern-physics.html index ce8176c..369874c 100644 --- a/docs/Chapter_1/birth-of-modern-physics.html +++ b/docs/Chapter_1/birth-of-modern-physics.html @@ -552,15 +552,15 @@

1.1.1. Mechanics

Newton’s first law: An object in motion with a constant velocity will continue in motion unless acted upon by some net external force. This is a restatement of Galileo’s inertia experiments with inclined planes. Newton’s first law is also called the law of inertia and is used to describe inertial reference frames.

  • Newtons second law: The acceleration \(\vec{a}\) of a body is proportional to the net external force \(\vec{F}\) and inversely proportional to the mass \(m\) of the body. Mathematically, it is stated as

    • -
    -(1.1)#\[\begin{align} +
    +(1.1)#\[\begin{align} \vec{F} = m\vec{a} = \frac{d\vec{p}}{dt} \end{align}\]
    -
    -(1.2)#\[\begin{align} +
    +(1.2)#\[\begin{align} \vec{F}_{21} = - \vec{F}_{12}. \end{align}\]

    In this form, it is often called the law of action and reaction.

    @@ -580,8 +580,8 @@

    1.1.2. ElectromagnetismHeinrich Hertz.

    Each of these contributions were held separately until Maxwell showed that electricity and magnetism were intimately connected. He also showed that electricity and magnetism were related through a change in the inertial frame of reference. Maxwell’s work led to the understanding of electromagnetic radiation, which describes the behavior of light. Maxwell combined the ideas of the time into four equations that form electromagnetism as

    -
    -(1.3)#\[\begin{align} +
    +(1.3)#\[\begin{align} \text{Gauss' law for electricity} \quad \oint \vec{E} \cdot d\vec{A} &= \frac{q}{\epsilon_o} ,\\ \text{Gauss' law for magnetism} \quad \oint \vec{B} \cdot d\vec{A} &= 0, \\ \text{Faraday's law for induction} \quad \oint \vec{E} \cdot d\vec{s} &= - \frac{d\Phi_B}{dt}, \\ @@ -613,8 +613,8 @@

    1.1.3. Thermodynamics

  • First law of thermodynamics: The change in the internal energy \(\Delta U\) of a system is equal to the heat \(Q\) added to the system plus the work \(W\) done on the system. The first law of thermodynamics generalized the conservation of energy by including heat. Mathematically it is written as

    • -
    -(1.5)#\[\begin{align} +
    +(1.5)#\[\begin{align} \Delta U = Q + W \end{align}\]
    -
    -(1.13)#\[\begin{align} +
    +(1.13)#\[\begin{align} \vec{F}_g = - \frac{Gm_1m_2}{r^2}\hat{r}. \end{align}\]

    • The electromagnetic force (or force between charges) depends on two charges (\(q_1\) and \(q_2\)), the permittivity of free space \(\epsilon_o\), and the distance between the charges \(r\). Similar to the gravitational force, it is a vector. It follows an inverse square law for the case of electrostatics and includes the cross-product for electrodynamics (see Eqn. (1.4)). The electromagnetic force is responsible for all chemical and biological processes (or practically all nongravitational forces that we experience). The electrostatic, or Coulomb, force is given by

    -

    If we rotate the apparatus by \(90^\circ\) so that the ether passes along \(d_1\). The time difference \(\Delta t^\prime\) is

    -
    -(2.3)#\[\begin{align} +
    +(2.3)#\[\begin{align} \Delta t^\prime = \frac{2}{c}\left(\frac{d_2}{1-v^2/c^2} - \frac{d_1}{\sqrt{1-v^2/c^2}} \right). \end{align}\]

    Michelson looked for a shift in the interference pattern and found the time difference as

    @@ -621,8 +621,8 @@

    2.3. Einstein’s Postulates -(2.6)#\[\begin{align} +
    +(2.6)#\[\begin{align} \beta &= \frac{v}{c} \\ & \text{and} \\ \gamma &= \frac{1}{\sqrt{1-v^2/c^2}}, @@ -633,8 +633,8 @@

    2.3. Einstein’s Postulates2.4. Lorentz Transformations#

    Ultimately, we want to know which transformation is necessary so that all inertial frames of reference are valid for the laws of physics (i.e., both Newton’s mechanics and Maxwell’s equations).

    Consider two inertial reference frames (\(K\) and \(K^\prime\)) that are moving relative to each other along their \(x\) and \(x^\prime\) axes with a uniform velocity \(v\), as shown in Figure 2.1. A flash lamp goes off at \(t=t^\prime=0\), where the speed of light will be \(c\) in both systems according to postulate 2. The wavefronts observed in both systems must be spherical and described by

    -
    -(2.7)#\[\begin{align} +
    +(2.7)#\[\begin{align} x^2 + y^2 + z^2 &= c^2 t^2, \\ x^{\prime 2} + y^{\prime 2} + z^{\prime 2} &= c^2 t^{\prime 2}. \end{align}\]
    @@ -650,27 +650,27 @@

    2.4. Lorentz Transformations\(\gamma\) and solve for \(t^\prime\) in terms of un-primed coordinates.

    To solve for \(\gamma\), we transform the above equations into

    -
    -(2.10)#\[\begin{align} +
    +(2.10)#\[\begin{align} t^\prime &= \gamma t\left(1-\frac{v}{c}\right), \\ t &= \gamma t^\prime \left(1+\frac{v}{c}\right), \end{align}\]

    and through direct substitution, we get

    -
    -(2.11)#\[\begin{align} +
    +(2.11)#\[\begin{align} t^\prime &= \gamma t^\prime \left(1-\frac{v}{c}\right) \left(1+\frac{v}{c}\right). \end{align}\]

    Eliminating \(t^\prime\) and solving for \(\gamma\) produces

    (2.12)#\[\gamma = \frac{1}{\sqrt{1-v^2/c^2}}.\]

    Now we can solve for \(t^\prime\) in un-primed coordinates directly by

    -
    -(2.13)#\[\begin{align} +
    +(2.13)#\[\begin{align} t^\prime = \frac{t\left(1-v/c\right)}{\sqrt{1-v^2/c^2}}, \end{align}\]

    and using the substitution \(t = x/c\) in the \(vt/c\) term to get

    -
    -(2.14)#\[\begin{align} +
    +(2.14)#\[\begin{align} t^\prime = \frac{t-vx/c^2}{\sqrt{1-v^2/c^2}}. \end{align}\]

    Now we have a the complete Lorentz transformations or Eqn. (2.5). The inverse transformation equation are obtained by replacing \(v\) by \(-v\) and by exchanging the primed and unprimed quantities (i.e., \(x\rightarrow x^\prime\)).

    @@ -1064,14 +1064,14 @@

    2.9. Spacetime

    Invariant quantities have the same value in all inertial frames, where they serve a special role in physics because their values do not change from one system to another (e.g., the speed of light is invariant). In Euclidean geometry, we define the distance \(d\) through \(d^2 = x^2 + y^2 + z^2\) and can obtain the same result for \(d^2\) in any inertial frame of reference. In spacetime, we can define a distance using a similar form for two systems \(K\) and \(K^\prime\) as

    -
    -(2.19)#\[\begin{align} +
    +(2.19)#\[\begin{align} s^2 &= x^2 -(ct)^2, \\ s^{\prime 2} &= x^{\prime 2} -(ct^\prime)^2. \end{align}\]

    If we use the Lorentz transformation for \(x\) and \(t\), we find that \(s^2 = s^{\prime 2}\), which makes \(\mathbf{s^2}\) an invariant quantity. This relationship can be extended to include the two other spatial coordinates, \(y\) and \(z\), so that

    -
    -(2.20)#\[\begin{align} +
    +(2.20)#\[\begin{align} s^2 &= x^2 + y^2 + z^2 -(ct)^2. \end{align}\]

    For simplicity, the distance \(\Delta s^2\) between two events can be represented by only a single spatial coordinate \(x\) as

    @@ -1085,8 +1085,8 @@

    2.9. Spacetime

    A 3-vector \(\vec{R}\) can be defined using Cartesian coordinates \(x,\ y,\ z\) in Euclidean space. There are two geometries in Newtonian spacetime, where one is the 3D Euclidean geometry (\(d\ell^2 = dx^2 + dy^2 + dz^2\)) and the other is a 1D time interval \(dt\). Minkowski noticed that both space and time will not independently suffice under a Lorentz transformation, and only a union of both will be useful.

    A 4-vector has four components \(x,\ y,\ z,\ ict\) and the equivalent of Eqn. (2.21) becomes

    -
    -(2.22)#\[\begin{align} +
    +(2.22)#\[\begin{align} ds^2 = dx^2 + dy^2 + dz^2 -c^2dt^2. \end{align}\]

    The spacetime distance \(ds^2\) can be positive, negative, or zero. It is also now invariant under the Lorentz transformation.

    @@ -1105,24 +1105,24 @@

    2.10. Doppler Effect (2.23)#\[f = \frac{cn}{cT-vT}.\]

    In the source’s rest frame, \(n\) waves of frequency \(f_o\) are emitted during the proper time \(T_o^\prime\), or

    -
    -(2.24)#\[\begin{align} +
    +(2.24)#\[\begin{align} n = f_o T_o^\prime. \end{align}\]

    The proper time interval \(T_o^\prime\) measured on the clock at rest in the moving system is related to the time interval \(T\) measured on a clock fixed by the receiver in system \(K\) by

    -
    -(2.25)#\[\begin{align} +
    +(2.25)#\[\begin{align} T_o^\prime = \frac{T}{\gamma}. \end{align}\]

    The clock moving with the source measures the proper time because it is present with both the beginning and end of the wave.

    To calculate the doppler frequency relativistically, we use the number of waves \(n\) relative to the time interval \(T\) as

    -
    -(2.26)#\[\begin{align} +
    +(2.26)#\[\begin{align} n = \frac{f_o T}{\gamma}, \end{align}\]

    and substitute into Eqn. (2.23) to determine the frequency as

    -
    -(2.27)#\[\begin{align} +
    +(2.27)#\[\begin{align} f = \frac{f_o cT/\gamma}{cT-vT} = \frac{1}{1-v/c}\frac{f_o}{\gamma} = \frac{\sqrt{1-v^2/c^2}}{1-v/c}f_o. \end{align}\]

    If we use \(\beta = v/c\), then we can write more compactly the doppler frequency as

    @@ -1130,8 +1130,8 @@

    2.10. Doppler Effect(2.28)#\[f = \frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}f_o.\]

    Equation (2.28) is also valid when the source is fixed and the receiver approaches it with a velocity \(v\).

    When the source and receiver are both receding from each other with a velocity \(v\), the distance between the beginning and end of the wave train becomes

    -
    -(2.29)#\[\begin{align} +
    +(2.29)#\[\begin{align} \text{Length of wave train} = cT + vT. \end{align}\]

    The change in sign propagates through the entire derivation wth the final result as

    @@ -1142,8 +1142,8 @@

    2.10. Doppler Effect(2.31)#\[f = \frac{\sqrt{1+\beta}}{\sqrt{1-\beta}}f_o.\]

    Elements absorb and emit characteristic frequencies of light due to the existence of particular atomic energy levels. Scientists have observe these characteristic frequencies in starlight and observed shifts in the frequencies. One reason for the shifts is the doppler effect, and the shifts are used to determine the speed of the emitting object with respect to us. Objects moving away from us shift the light to shorter frequencies (i.e., longer wavelengths) and are called redshifts. The measurement of redshifts for distant galaxies led to a linear relationship, where the farther away the galaxy; the higher the redshift. An implication of this discovery is that the universe is expanding, which was first suggested by Harlow Shapely and Edwin Hubble.

    The Doppler effect measures the radial motion (toward or away), which is maximized if the source and receiver are directly approaching or receding. It is also possible for the source and receiver to be moving at an angle with respect to one another. The angles \(\theta\) and \(\theta^\prime\) are the angles that the light signal makes with the \(x\) axes in the \(K\) and \(K^\prime\) systems. They are related mathematically by,

    -
    -(2.32)#\[\begin{align} +
    +(2.32)#\[\begin{align} f \cos \theta &= f_o \gamma (\cos \theta^\prime + \beta) = \frac{f_o(\cos \theta^\prime + \beta)}{\sqrt{1-\beta^2}}, \\ f \sin \theta &= f_o \sin \theta^\prime. \end{align}\]
    @@ -1171,38 +1171,38 @@

    2.11. Relativistic Momentum\(\vec{F} = d\vec{p}/dt\)) keeps its same form under a Galilean transformation, where it might be different under a Lorentz transformation. From Newton’s second law, an acceleration of a particle already moving at very high speeds could lead to a speed \(v>c\). That would be in conflict with the Lorentz transformation, and a modification may be necessary at high speed.

    Consider a collision that has no external forces, where a ball of mass \(m\) is held at rest in the fixed frame \(K\) and a similar ball in a moving frame \(K^\prime\) is moving in the \(x\) direction with a velocity \(v\) with respect to the fixed frame. The ball in the fixed frame is launched along its positive \(y\) axis, while the ball in the moving frame is launched with the same speed along its negative \(y\) axis. The two balls collide in a perfectly elastic collision and return to their starting position along their respective \(y\) axis.

    According to an observer in the \(K\) frame, the initial velocity of the ball has components

    -
    -(2.34)#\[\begin{align} +
    +(2.34)#\[\begin{align} u_x &= 0, \\ u_y &= u_o, \end{align}\]

    where \(u_o\) is the launch speed. Using the definition of momentum (\(\vec{p}=m\vec{v}\)), the momentum of the ball is entirely in the \(y\) direction, or \(p_y = mu_o\). Because the collision is perfectly elastic, the ball returns with a speed \(u_o\) along the \(-y\) axis, where the change in momentum in system \(K\) is

    -
    -(2.35)#\[\begin{align} +
    +(2.35)#\[\begin{align} \Delta p^{\text{FF}} = \text{After }p - \text{Before }p = -mu_o - (+mu_o) = -2mu_o, \end{align}\]

    where the superscript \(\text{FF}\) refers to a measurement of the fixed frame relative to the fixed observer. To confirm the conservation of linear momentum, we need to determine the change in momentum of the ball in the moving frame \(K^\prime\) from the perspective of an observer in the fixed frame \(K\).

    In the moving frame, the initial velocity of the ball has components: \(u_x^\prime = 0\) and \(u_y^\prime = -u_o\) because the ball is launched along its \(-y^\prime\) axis. To determine the velocity of the ball as measured by an observer in the \(K\) frame, we need to use Eqn. (2.18). If we insert the appropriate values for the speeds, we obtain

    -
    -(2.36)#\[\begin{gather} +
    +(2.36)#\[\begin{gather} u_x &=& \frac{0 + v}{1 + (v/c^2)(0)} &=& v, \\ u_y &=& \frac{-u_o}{\gamma (1+(v/c^2)(0))} &=& -u_o \sqrt{1-v^2/c^2}. \end{gather}\]

    Before the collision, the momentum of the ball in the moving frame \(K^\prime\) as measured by an observer in the \(K\) frame is

    -
    -(2.37)#\[\begin{align} +
    +(2.37)#\[\begin{align} \text{Before } & p_x = mv, \\ \text{Before } & p_y = -mu_o \sqrt{1-v^2/c^2}. \end{align}\]

    For a perfectly elastic collision, the momentum after the collision is

    -
    -(2.38)#\[\begin{align} +
    +(2.38)#\[\begin{align} \text{After } & p_x = mv, \\ \text{After } & p_y = +mu_o \sqrt{1-v^2/c^2}. \end{align}\]

    Then, the change in momentum of the ball in the moving frame \(K^\prime\) according to the fixed frame \(K\) is

    -
    -(2.39)#\[\begin{align} +
    +(2.39)#\[\begin{align} \Delta p^{\text{MF}} = \Delta p_y = \text{After }p_y - \text{Before }p_y = 2mu_o \sqrt{1-v^2/c^2}, \end{align}\]

    where the superscript \(\text{MF}\) refers to a measurement of the moving frame relative to the fixed observer. The conservation of linear momentum requires the total change in the momentum of the collision \(\Delta p^{\text{FF}} + \Delta p^{\text{MF}} = 0\). Clearly, \(2mu_o \neq 2mu_o/\gamma\). There is not a problem with the \(x\) direction, but there is a problem with the \(y\) direction (i.e., the direction which the ball is launched in each system).

    @@ -1219,8 +1219,8 @@

    2.11. Relativistic Momentum\(\Gamma(u)\) in Eqn. (2.40) is the same as the \(\gamma\) in the Lorentz transformation. However this \(\Gamma(u)\) is different because it contains the speed of the particle \(u\), whereas the Lorentz transformation contains the relative speed \(v\) between two inertial reference frames. Among physicists it is common to replace \(\Gamma(u)\) with \(\gamma\), but we must remember the distinction.

    A plausible determination for the correct form of the momentum requires the proper time \(\tau\), where the momentum becomes

    -
    -(2.41)#\[\begin{align} +
    +(2.41)#\[\begin{align} \vec{p} = m\frac{d\vec{r}}{d\tau} = m\frac{d\vec{r}}{dt}\frac{dt}{d\tau}. \end{align}\]

    The classical velocity is \(\vec{u} = d\vec{r}/dt\), where \(\vec{r}\) is the position vector. All observers do not agree to the value of \(\vec{u}\), but they do agree to the value of \(d\vec{r}/d\tau\), where \(d\tau\) is the proper time measured in the moving system \(K^\prime\). From Eqn. (2.15), the value of \(dt/d\tau = \gamma\), where the speed \(u\) is used.

    @@ -1265,13 +1265,13 @@

    2.12. Relativistic Energy (2.43)#\[\vec{F} = \frac{d\vec{p}}{dt} = \frac{d}{dt}\left(\gamma m\vec{u} \right) = \frac{d}{dt}\left(\frac{m\vec{u}}{\sqrt{1-u^2/c^2}}\right).\]

    Kinetic energy can be expressed as the work done on a particle by a net force. The work \(W_{12}\) done by a force \(\vec{F}\) to move a particle from position 1 to position 2 along a path \(\vec{s}\) is

    -
    -(2.44)#\[\begin{align} +
    +(2.44)#\[\begin{align} W_{12} = \int_1^2 \vec{F}\cdot d\vec{s} = K_2 - K_1, \end{align}\]

    where \(K_1\) is defined as the kinetic energy of the particle at position 1. For simplicity, let the particle start from rest under the influence of the force \(\vec{F}\). The work \(W\) and kinetic energy \(K\) are

    -
    -(2.45)#\[\begin{align} +
    +(2.45)#\[\begin{align} W = K = \int \frac{d}{dt}\left(\gamma m\vec{u} \right) \cdot \vec{u}\ dt, \end{align}\]

    where the integral is performed over the differential path \(d\vec{s} = \vec{u}\ dt\). The mass is invariant, but the relativistic factor \(\gamma\) depends on \(u\) and must remain in the integral. The kinetic energy becomes

    @@ -1289,8 +1289,8 @@

    2.12. Relativistic Energy\((u/c)^4\) and greater) are neglected because \(u\ll c\). The relativistic kinetic energy at low speeds is

    -
    -(2.47)#\[\begin{align} +
    +(2.47)#\[\begin{align} K = mc^2 + \frac{1}{2}mu^2 - mc^2 = \frac{1}{2}mu^2, \end{align}\]

    which is the expected classical result. The relativistic and classical kinetic energies diverge considerably for \(u/c > 0.6\).

    @@ -1387,13 +1387,13 @@

    2.12. Relativistic Energy

    2.12.1. Total Energy and Rest Energy#

    The relativistic kinetic energy can be rewritten as

    -
    -(2.49)#\[\begin{align} +
    +(2.49)#\[\begin{align} \gamma mc^2 = \frac{mc^2}{\sqrt{1-u^2/c^2}} = K + mc^2. \end{align}\]

    The rest energy \(E_o\) is the term mc^2, or

    -
    -(2.50)#\[\begin{align} +
    +(2.50)#\[\begin{align} E_o = mc^2. \end{align}\]

    The sum of the kinetic energy and rest energy is interpreted as the total energy \(E\) of the particle, which is

    @@ -1416,23 +1416,23 @@

    2.12.2. Equivalence of Mass and Energy

    Let’s examine the conservation of mass-energy. The energy before the collision is

    -
    -(2.52)#\[\begin{align} +
    +(2.52)#\[\begin{align} E_{\rm before} = \text{(rest mass-energy)} + \text{kinetic energy} = 2mc^2 + 2K, \end{align}\]

    and the energy after the collision is

    -
    -(2.53)#\[\begin{align} +
    +(2.53)#\[\begin{align} E_{\rm after} = \text{(rest mass-energy)} = Mc^2, \end{align}\]

    where the (rest) mass of the system is \(M\). Through energy conservation, \(2(mc^2+K) = Mc^2\), and the new mass \(M\) is greater than the individual masses \(2m\). The kinetic energy went into compressing the spring, so the spring has increased potential energy. Kinetic energy has been converted into mass. The difference in mass \(\Delta M\) is determined by

    -
    -(2.54)#\[\begin{align} +
    +(2.54)#\[\begin{align} \Delta M = M - 2m = \frac{2K}{c^2}, \end{align}\]

    and linear momentum is conserved in this head-on collision. The fractional mass \((f_r = \Delta M/2m)\) is quite small, which is

    -
    -(2.55)#\[\begin{align} +
    +(2.55)#\[\begin{align} f_r = \frac{M-2m}{2m} = \frac{K}{mc^2}. \end{align}\]

    For typical masses and kinetic energies of wood blocks, this fraction increase in mass is too small to measure. For example, if mass of a wood block \(m = 0.1\ {\rm kg}\) and the speed of the wood block \(v = 10\ {\rm m/s}\), then the fractional increase in mass is

    @@ -1488,8 +1488,8 @@

    2.12.3. Relationship of Energy and Momen

    2.12.4. Massless Particles#

    Equation (2.56) is valid for the total energy of massless (zero mass) particles. For example, the total energy of a photon is

    -
    -(2.57)#\[\begin{align} +
    +(2.57)#\[\begin{align} E = pc. \quad \text{(Photon)} \end{align}\]

    The energy is completely due to its motion (or momentum), where it has no rest energy. Through the relativistic equations, we can show that the speed of a photon must be the speed of light \(c\). Using the relativistic equations for total energy and kinetic energy

    @@ -1499,8 +1499,8 @@

    2.12.4. Massless Particles \[ \gamma mc^2 = \gamma muc, \]

    and

    -
    -(2.58)#\[\begin{align} +
    +(2.58)#\[\begin{align} u = c. \quad \text{(Massless particle)} \end{align}\]
    @@ -1511,8 +1511,8 @@

    2.13. Computations in Modern Physics \[ W = (1\ e)(1\ {\rm V}) = 1\ {\rm eV}. \]

    The electron volt (eV) is a unit of energy and it is related to the SI unit joule through the charge of an electron, so that

    -
    -(2.59)#\[\begin{align} +
    +(2.59)#\[\begin{align} 1\ {\rm eV} = 1.602 \times 10^{-19}\ {\rm J}. \end{align}\]

    The eV is more often used in modern physics than the SI unit J. The ev can also carry the SI prefixes where applicable (e.g., \(10^6\ {\rm eV} = 1\ {\rm MeV}\); mega-electron-volt). Since work is related to kinetic energy, we speak of a particle in terms of its kinetic energy, where a \(6\ {\rm GeV}\) proton would have \(7\ {\rm GeV}\) of total energy (i.e., kinetic + rest energy).

    @@ -1534,15 +1534,15 @@

    2.13. Computations in Modern PhysicsNotice the difference in font from \(u\) (velocity) and \({\rm u}\) (amu).

    The conversion between kg and amu are determined by comparing the mass of one \(^{12}{\rm C}\) atom:

    -
    -(2.60)#\[\begin{align} +
    +(2.60)#\[\begin{align} \text{Mass}(^{12}{\rm C}\text{ atom}) &= \frac{\text{molar mass of carbon-12}}{\text{Avogadro's number}}, \\ &= \frac{12\ {\rm g/mol}}{6.02\times 10^{23}\ {\rm atoms/mol}} = 1.99 \times 10^{-23}\ {\rm g/atom}, \\ \text{Mass}(^{12}{\rm C}\text{ atom}) &= 1.99 \times 10^{-26}\ {\rm kg} = 12\ {\rm u/atom}. \end{align}\]

    Therefore, the conversions are (up to 6 significant figures):

    -
    -(2.61)#\[\begin{align} +
    +(2.61)#\[\begin{align} 1\ {\rm u} &= 1.66054 \times 10^{-27}\ {\rm kg}\\ 1\ {\rm u} &= 931.494\ {\rm MeV/c^2}. \end{align}\]
    @@ -1577,8 +1577,8 @@

    2.13. Computations in Modern Physics

    The speed of a 2 GeV proton is \(0.95c\) or \(2.8 \times 10^8\ {\rm m/s}\).

    (b) When the two protons collide head-on, the protons could behave similarly to the wood blocks, but the time for the two protons to interact is very short \((<10^{-29}\ {s})\). If the tow protons did momentarily stop at rest, then the two proton system would have its mass increased to \(4\ {\rm GeV/c^2}\). This would be highly excited system, where several outcomes are possible. The two protons could remain or disappear, where new particles are created to conserved mass-energy, angular momentum, and charge. Two of the possibilities are:

    -
    -(2.62)#\[\begin{align} +
    +(2.62)#\[\begin{align} p + p &\rightarrow p + p + p + \overline{p}, \\ p + p &\rightarrow \pi^+ + d, \end{align}\]
    @@ -1623,16 +1623,16 @@

    2.13.1. Binding Energy\(E_B\). The binding energy is the work required to pull the particles out of the bound system into separate, free particles at rest. The conservation of energy is written as

    -
    -(2.63)#\[\begin{align} +
    +(2.63)#\[\begin{align} {\rm M_{bound}}c^2 + E_B = \sum_i m_i c^2, \end{align}\]

    which depends on the mass of the bound system \({\rm M_{bound}}\) and the masses of the free particles \(m_i\). The binding energy is the difference between the rest energy of the individual particles and the rest energy of the combined, bound system, or

    (2.64)#\[E_B = \sum_i m_i c^2 - {\rm M_{bound}}c^2.\]

    For the simple case of two final particles with a particle 1 mass \(m_1\), particle 2 mass \(m_e\), and the bound mass \({\rm M_{bound}}\), we have

    -
    -(2.65)#\[\begin{align} +
    +(2.65)#\[\begin{align} E_B = (m_1 + m_2 - {\rm M_{bound}})c^2 = \Delta Mc^2. \end{align}\]

    To bind a proton and a neutron together into a deuteron, part of the rest energy of the individual particles is lost and makes up the binding energy of the system. The rest energy of the combined system must be the reduced by this amount. The rest energies of the particles are:

    @@ -1891,7 +1891,7 @@

    2.15. Homework Problems

    - © Copyright 2022. + © Copyright 2024.

    diff --git a/docs/Chapter_3/experimental-quantum-physics.html b/docs/Chapter_3/experimental-quantum-physics.html index 1688385..db80690 100644 --- a/docs/Chapter_3/experimental-quantum-physics.html +++ b/docs/Chapter_3/experimental-quantum-physics.html @@ -522,8 +522,8 @@

    3.1. Discovery of the X-Ray and the Elec

    Hertz performed a similar experiment, but observed an effect on the cathode rays due to the deflecting voltage. In Hertz’s experiment, the poorer vacuum allowed the cathode rays to interact with and ionize residual gas. Thomson found the same result at first, but improved the vacuum to obtain his final result.

    Thomson’s method of measuring the ratio of the electron’s charge to mass \(e/m\) is now a stand technique. With the magnetic field turned off, the electron entering the region between the plates is accelerated upward by the electric field as

    -
    -(3.1)#\[\begin{align} +
    +(3.1)#\[\begin{align} F_y = ma_y = qE, \end{align}\]

    where the ratio of the electron charge \(q\) and mass \(m\) are determined by measuring the acceleration \(a_y\). The time \(t\) for the electron to traverse the deflecting plates of length \(\ell\) is \(t \approx \ell/v_o\). The exit angle \(\theta\) of the electron is then given by

    @@ -537,8 +537,8 @@

    3.1. Discovery of the X-Ray and the Elec

    The ratio \(q/m\) can be determined if the initial velocity \(v_o\) is known. The initial velocity can be determined by turning on the magnetic field and adjusting the strength of \(\vec{B}\) until no deflection occurs. The condition for zero deflection is the case of equilibrium or

    -
    -(3.3)#\[\begin{align} +
    +(3.3)#\[\begin{align} \vec{F} = q\vec{E} + q\vec{v} \times \vec{B} = 0, \end{align}\]

    and

    @@ -548,8 +548,8 @@

    3.1. Discovery of the X-Ray and the Elec
    (3.4)#\[v_x = v_o = E/B.\]

    We can extract the ratio \(q/m\) through substitution by

    -
    -(3.5)#\[\begin{align} +
    +(3.5)#\[\begin{align} \frac{q}{m} = \frac{v_o^2 \tan \theta}{E\ell} = \frac{E\tan \theta}{B^2 \ell}. \end{align}\]
    @@ -608,16 +608,16 @@

    3.2. Determination of Electron Charge(3.6)#\[\vec{F}_j = -b\vec{v}.\]

    This drag force always opposes the velocity, hence the negative sign. The constant \(b\) is determined via Stokes’ law and is proportional to the oil drop’s radius. Millikan showed that Stokes’ law was incorrect for small-diameter spheres due to the atomic nature of the medium, where an appropriate correction was required. For a first-order calculation, we neglect the buoyancy of the air that produces an upward force on the drop.

    To suspend the oil drop at rest between the plates, the electric and gravitational forces must be balanced. This makes the frictional force equal to zero because the oil drop’s velocity is zero. The force balance is

    -
    -(3.7)#\[\begin{align} +
    +(3.7)#\[\begin{align} \vec{F}_E = q\vec{E}=-m\vec{g}, \quad (\text{when } v=0). \end{align}\]

    The magnitude of the electric field is \(E=V/d\), which depends on the voltage \(V\) across large, flat plates separated by a small distance \(d\). The magnitude of the electron charge is determined as

    (3.8)#\[q = \frac{mgd}{V}.\]

    To calculate \(q\), we need to know the mass \(m\) of the oil drops. Millikan found he could determine \(m\) by measuring the terminal velocity of the oil drop while the electric field was off. From Stoke’s law (or Eqn. (3.6)), the radius of the oil drop is related to the terminal velocity. Using the radius \(r\) and the density \(\rho\) of the oil, the mass of the oil can be determined by

    -
    -(3.9)#\[\begin{align} +
    +(3.9)#\[\begin{align} m = \frac{4}{3}\pi r^3 \rho. \end{align}\]

    The oil drop can be moved up and down in the apparatus at will by adjusting the voltage magnitude and polarity. millikan reported that he was able to observe an oil drop for up to six hours, where the drop changed its charge several times. Millikan made 1000s of measurements using different oils and showed that there is a basic quantized electron charge. Millikan’s value was very close to the modern value \(e=1.602 \times 10^{-19}\ {\rm C}\).

    @@ -862,8 +862,8 @@

    3.3. Line SpectraWavelengths were formerly listed in units of angstroms (Å) which are named after Anders Ångstrom. Ångstrom was one of the first people to observe and measure the wavelengths of the four visible lines of hydrogen. \(1\ {\rm \unicode{x212B}} = 0.1\ {\rm nm} = 10^{-10}\ {\rm m}\).

    Equation (3.11) can be rewritten as its reciprocal into the following form

    -
    -(3.12)#\[\begin{align} +
    +(3.12)#\[\begin{align} \frac{1}{\lambda} = \frac{1}{365.56\ {\rm nm}}\left(\frac{k^2 -4}{k^2}\right) = \frac{4}{365.56\ {\rm nm}}\left( \frac{1}{2^2}-\frac{1}{k^2}\right) = R_{\rm H} \left( \frac{1}{2^2}-\frac{1}{k^2}\right), \end{align}\]

    where \(R_{\rm H}\) represents the Rydberg constant (for hydrogen) and \(k\) is an integer. The ratio (\(4/364.56\ {\rm nm}\)) is \(1.0972 \times 10^7\ {\rm m^{-1}}\), where the more accurate value is \(1.096776 \times 10^7\ {\rm m^{-1}}\). Johannes Rydberg and Walther Ritz determined a more general empirical equation for calculating the wavelengths known as the Rydberg equation, given by

    @@ -1453,18 +1453,18 @@

    3.6.3. Einstein’s theory (3.19)#\[\lambda f = c.\]

    Einstein proposed that light should behave with its well-known wave-like aspect and light should also be considered to have a particle-like aspect. The quantum of light delivers it entire energy \(hf\) to a single electron in the material. To leave the material, the struck electron must give up an amount of energy \(\phi\) to overcome its binding in the material. The electron may lose some additional energy by electron interactions on its way to the surface. Whatever remaining energy will appear as the electron’s kinetic energy as it leaves the emitter. The conservation of energy requires that

    -
    -(3.20)#\[\begin{align} +
    +(3.20)#\[\begin{align} hf = \phi + \text{K.E. (electron)}. \end{align}\]

    We are safe in using the nonrelativistic form of the electron’s kinetic energy \((\frac{1}{2}mv^2)\) because the energies involved are on the order of \({\rm eV}\). The electron’s energy will degrade as it passes through the emitter material. Experimentally, we detect the maximum value of the kinetic energy as

    -
    -(3.21)#\[\begin{align} +
    +(3.21)#\[\begin{align} hf = \phi + \frac{1}{2}mv^2_{\rm max}. \end{align}\]

    The retarding potentials are opposing potentials needed to stop (i.e., work against) the most energetic electrons, or

    -
    -(3.22)#\[\begin{align} +
    +(3.22)#\[\begin{align} W = eV_o = \frac{1}{2}mv^2_{\rm max}. \end{align}\]
    @@ -1515,9 +1515,8 @@

    3.6.4. Quantum Interpretation 
    The value of h from Millikan is 6.65e-34 J s.
-
    -
    -
    The error in Millikan's result is 0.4 percent.
+
+The error in Millikan's result is 0.4 percent.
    ../_images/787e59ebf74337bc8fb03eb1de212cf0528b5ef7f9b6b254eb044a2c50bdfeec.png @@ -1640,8 +1639,8 @@

    3.7. X-Ray Production -(3.26)#\[\begin{align} +
    +(3.26)#\[\begin{align} E_f = E_i - hf. \end{align}\]

    The nucleus absorbs very little energy because linear momentum must be conserved. One or more photons may be created in this way as electrons pass through matter.

    @@ -1759,21 +1758,21 @@

    3.8. Compton Effect \[ E_e^2 = (m_ec^2)^2 + p_e^2c^2. \]

    The conservation laws for energy and momentum are:

    -
    -(3.28)#\[\begin{align} +
    +(3.28)#\[\begin{align} \text{Energy:} & \quad & hf + m_ec^2 = hf^\prime + E_e, \\ p_x: & \quad & \frac{h}{\lambda} = \frac{h}{\lambda^\prime}\cos \theta + p_e \cos \phi, \\ p_y: & \quad & 0 = \frac{h}{\lambda^\prime}\sin \theta - p_e\sin \phi. \end{align}\]

    The recoil angle \(\phi\) can be eliminated by finding the momentum squared of each component,

    -
    -(3.29)#\[\begin{align} +
    +(3.29)#\[\begin{align} p_x^2: & \quad & \left(\frac{h}{\lambda}\right)^2 +\left(\frac{h}{\lambda^\prime}\right)^2\cos^2 \theta - 2\left(\frac{h}{\lambda}\right) \left(\frac{h}{\lambda^\prime}\right)\cos \theta = p_e^2 \cos^2 \phi , \\ p_y^2: & \quad & \left(\frac{h}{\lambda^\prime}\right)^2\sin^2 \theta = p_e^2 \sin^2 \phi, \end{align}\]

    and adding them together (i.e., \(\cos^2 \phi + \sin^2 \phi = 1\)) to get

    -
    -(3.30)#\[\begin{align} +
    +(3.30)#\[\begin{align} p_e^2 = \left(\frac{h}{\lambda}\right)^2 + \left(\frac{h}{\lambda^\prime} \right)^2 - 2\left(\frac{h}{\lambda}\right) \left(\frac{h}{\lambda^\prime}\right)\cos \theta. \end{align}\]

    Now, we can substitute values into the conservation energy equation (using \(\lambda = c/f\)) to get,

    @@ -1929,13 +1928,13 @@

    3.9. Pair Production and Annihilation

    Consider the conversion of a photon into an electron and a positron that takes place inside an atom where the electric field of the nucleus is large. The nucleus recoils and takes away a negligible amount of energy, but a considerable amount of momentum. The conservation of energy becomes

    -
    -(3.36)#\[\begin{align} +
    +(3.36)#\[\begin{align} hf = E_{-} + E_{+} + \text{K.E. (nucleus)}. \end{align}\]

    The photon energy must be equal to \(2m_ec^2\) in order to create the masses of the electron and positron, or

    -
    -(3.37)#\[\begin{align} +
    +(3.37)#\[\begin{align} hf > 2 m_ec^2 = 1.022\ {\rm MeV}. \qquad \text{(for pair production)} \end{align}\]

    The probability of pair production increases dramatically with higher: photon energy and atomic number \(Z\) fo the atom’s nucleus (due to the correspondingly higher electric field).

    @@ -1944,8 +1943,8 @@

    3.9. Pair Production and Annihilation \[ e^+ + e^- \rightarrow \gamma + \gamma. \]

    Consider a positronium “atom” at rest in free space. It must emit at least two photons to conserve energy and momentum. If the positronium annihilation takes place near a nucleus, it is possible that only one photon will be created because the missing momentum can be supplied by the nucleus recoil. The conservation laws for the above process will be

    -
    -(3.38)#\[\begin{align} +
    +(3.38)#\[\begin{align} \text{Energy: } & \quad & 2m_e c^2 = hf_1 + hf_2, \\ \text{Momentum: } & \quad & 0 = \frac{hf_1}{c} - \frac{hf_2}{c}. \end{align}\]
    @@ -2129,7 +2128,7 @@

    3.10. Homework Problems

    - © Copyright 2022. + © Copyright 2024.

    diff --git a/docs/Chapter_4/structure-of-the-atom.html b/docs/Chapter_4/structure-of-the-atom.html index 558c7c7..9a51590 100644 --- a/docs/Chapter_4/structure-of-the-atom.html +++ b/docs/Chapter_4/structure-of-the-atom.html @@ -613,8 +613,8 @@

    4. Structure of the Atom (4.3)#\[\Delta \vec{p} = \vec{p}_f - \vec{p}_i,\]

    where the subscripts \(f\) and \(i\) indicate the final and initial momentum, respectively. Since \(p_f \approx p_i\), the vector triangle in Fig. 4.3 is isosceles and the angle \(\theta\) can be bisected. The magnitude of \(\Delta p\) is now

    -
    -(4.4)#\[\begin{align} +
    +(4.4)#\[\begin{align} \Delta p = 2mv_o \sin \frac{\theta}{2}. \end{align}\]

    The direction of \(\Delta \vec{p}\) is the \(z^\prime\) axis, where we need to determine the \(z^\prime\) component of \(\vec{F}\). The Coulomb force \(\vec{F}\) is defined as

    @@ -651,19 +651,19 @@

    4. Structure of the Atom(4.7)#\[\sigma = \pi b^2.\]

    The cross section \(\sigma\) is related to the probability for a particle being scattered by the nucleus.

    If we have a target foil of thickness \(t\) with \(n\) atoms/volume, the target nuclei per unit area is \(nt\). Because we assumed a thin target area \(A\) and all nuclei are exposed, the number of target nuclei is \(ntA\). The number density \(n\) is related to the density \(\rho\), Avogadro’s number \(N_A\), atoms per molecule \(N_M\), and the molecular weight by

    -
    -(4.8)#\[\begin{align} +
    +(4.8)#\[\begin{align} n &= \rho \left(\frac{\rm g}{\rm cm^3}\right) N_A \left(\frac{\rm molecules}{\rm mol}\right)N_M \left(\frac{\rm atoms}{\rm molecule}\right) \left(M_g ({\rm g/mol})\right)^{-1}, \\ & = \frac{\rho N_A N_M}{M_g} \frac{\rm atoms}{\rm cm^3}. \end{align}\]

    Then the number of scattering nuclei per unit area is

    -
    -(4.9)#\[\begin{align} +
    +(4.9)#\[\begin{align} nt = \frac{\rho N_A N_M t}{M_g} \frac{\rm atoms}{\rm cm^2}, \end{align}\]

    and the number of target nuclei \(N_s\) is

    -
    -(4.10)#\[\begin{align} +
    +(4.10)#\[\begin{align} N_s = ntA = \frac{\rho N_A N_M tA}{M_g}\ {\rm atoms}. \end{align}\]

    The probability of the particle begin scattered is equal number of target nuclei times the cross section divided by the total target area \(A\), or

    @@ -786,18 +786,18 @@

    4.3. The Classical Atomic ModelThomson had previously considered a planetary model resembling the Solar System but rejected it because the planets attract one another while orbiting the Sun, whereas the electrons would repel one another.

    Let’s examine the hydrogen atom, which consists of a single proton and electron. Assuming also that the electron has a circular orbit. The force of attraction due to the proton is

    -
    -(4.13)#\[\begin{align} +
    +(4.13)#\[\begin{align} \vec{F}_e = \frac{-1}{4\pi \epsilon_o}\frac{e^2}{r^2}\hat{e}_r, \end{align}\]

    where the negative sign indicates the force is attractive. The electrostatic force provides the centripetal force need for the electron to move in a circular orbit at constant speed. Its radial acceleration is

    -
    -(4.14)#\[\begin{align} +
    +(4.14)#\[\begin{align} a_r = \frac{v^2}{r}, \end{align}\]

    which depends on the tangential velocity \(v\) of the electron. Newton’s second law gives,

    -
    -(4.15)#\[\begin{align} +
    +(4.15)#\[\begin{align} \frac{1}{4\pi \epsilon_o}\frac{e^2}{r^2} = \frac{m_ev^2}{r}, \end{align}\]

    and

    @@ -840,8 +840,8 @@

    4.3. The Classical Atomic Model

    The kinetic energy of the system due to the electron is \(K = m_ev^2/2\), where we assume that the nucleus is essentially at rest because it is so massive compared with the electron \((m_p/m_e = 1836)\). The potential energy is \(V = -e^2/(4\pi \epsilon_o r)\). The total mechanical energy is

    -
    -(4.17)#\[\begin{align} +
    +(4.17)#\[\begin{align} E = K + V = \frac{1}{2}m_e v^2 - \frac{e^2}{4\pi \epsilon_o r}. \end{align}\]

    If we substitute from Eqn. (4.16), then we have

    @@ -874,8 +874,8 @@

    4.4. The Bohr Model of the Hydrogen Atom
    \[ L = |\vec{r} \times \vec{p}| = m_evr. \]

    Assumption 4 states this should be

    -
    -(4.19)#\[\begin{align} +
    +(4.19)#\[\begin{align} L = m_evr = n\hbar, \end{align}\]

    which is defined by the principal quantum number \(n\) (an integer). Then the equation for the orbital velocity is

    @@ -994,8 +994,8 @@

    4.4.1. The Correspondence Principle—Bohr’s correspondence principle

    Consider the predictions of the two radiation laws. The frequency of the radiation produced by the atomic electrons in the Bohr model should agree with the predictions from classical electrodynamics where the Planck’s constant is unimportant (i.e., for large quantum number \(n\)). Classically, the frequency of emitted radiation is equal to the electron’s orbital frequency \(f_{\rm orb}\) around the nucleus:

    -
    -(4.31)#\[\begin{align} +
    +(4.31)#\[\begin{align} f_{\rm classical} &= f_{\rm orb} = \frac{\omega}{2\pi} = \frac{v}{2\pi r},\\ &= \frac{1}{2\pi}\sqrt{\frac{e^2}{4\pi \epsilon_o m_e r^3}}, \\ & = \frac{1}{n^3}\frac{m_e e^4}{4\epsilon_o^2 h^3}. @@ -1009,8 +1009,8 @@

    4.4.1. The Correspondence Principle \[ f_{\rm Bohr} \approx \frac{E_o}{h}\left[ \frac{2n}{n^4} \right] = \frac{2E_o}{hn^3}. \]

    If we substitute Bohr’s \(E_o\) from Eqn. (4.24), the result is

    -
    -(4.33)#\[\begin{align} +
    +(4.33)#\[\begin{align} f_{\rm Bohr} = \frac{1}{n^3}\frac{m_e e^4}{4\epsilon_o^2 h^3} = f_{\rm classical}. \end{align}\]

    The frequencies of the radiated energy agree between the classical theory and the Bohr model for large values of the principle quantum number \(n\). Bohr’s correspondence principle is verified for large electron orbits, where classical and quantum physics should agree.

    @@ -1021,8 +1021,8 @@

    4.5. Success and Failures of the Bohr Mo

    4.5.1. Reduced Mass Correction#

    The electron and hydrogen nucleus actually revolve about their mutual center of mass, where this is a two-body problem. Instead of just \(r\), we should have used the electron’s distance \(r_e\) and the nucleus’ distance \(r_{\rm nucleus}\) from the center of mass. A straightforward analysis from classical mechanics shows that the two-body problem can be reduced to an equivalent one-body problem in terms of the reduced mass \(\mu_e\) that is moving around the center of mass. The reduced mass depends on the mass of the nucleus \(M\) and the electron mass \(m_e\), or

    -
    -(4.34)#\[\begin{align} +
    +(4.34)#\[\begin{align} \mu_e = \frac{m_eM}{m_e + M} = \frac{m_e}{1+m_e/M}. \end{align}\]

    In the case of the hydrogen atom, the mass of the nucleus \(M\) is the proton mass, and the correction for the hydrogen atom is \(\mu_e = 0.999456\ m_e\). Athough it is small, the difference is still measurable experimentally. The Rydberg constant \(R_\infty\) (see (4.27)) should be replaced by

    @@ -1148,15 +1148,15 @@

    4.6. Characteristic X-Ray Spectra and At

    If a vacancy occurs in the \(K\) shell, then there is still one electron remaining in the \(K\) shell. An electron in the \(L\) shell will feel an effective charge of \((Z-1)e\) due to the \(+Ze\) from the nucleus and \(-e\) from the remaining \(K\) shell electron. The other electrons outside the \(K\) shell hardly affect the \(L\) shell electron. The x-ray produced when a transition occurs from the \(n=2\rightarrow 1\) shell has the wavelength (from Eqn. (4.36)) of

    -
    -(4.38)#\[\begin{align} +
    +(4.38)#\[\begin{align} \frac{1}{\lambda_{K_\alpha}} &= (Z-1)^2 R\left(\frac{2^2 -1^2}{1^22^2}\right), \\ &= \frac{3}{4}R(Z-1)^2, \\ f_{K_\alpha} &= \frac{c}{\lambda_{K_\alpha}} = \frac{3}{4}cR(Z-1)^2. \end{align}\]

    Moseley’s equation describing the \(K_\alpha\) shell x-rays is an application of Bohr’s model. The general form for the \(K\) series of x-ray wavelengths is

    -
    -(4.39)#\[\begin{align} +
    +(4.39)#\[\begin{align} \frac{1}{\lambda_K} &= R(Z-1)^2 \left(\frac{n^2 -1}{n^2}\right). \end{align}\]

    Moseley correctly concluded that the atomic number \(Z\) was the determining factor for the ordering of the periodic table, where the reordering by atomic number was more consistent with chemical properties than one based on atomic weight. He also concluded that the atomic number of an element should be identified with the number of positive charges (protons) in the nucleus. He tabulated all the atomic numbers between aluminum \({\rm Al}\ (Z=13)\) to gold \({\rm Au}\ (Z=79)\) and pointed out that there were still three elements (\(Z=43,\ 61,\ \text{and } 75\)) yet to be discovered! The element promethium (\(Z=61\)) was discovered around 1940.

    @@ -1167,13 +1167,13 @@

    4.6. Characteristic X-Ray Spectra and At

    Moseley found experimentally an equation describing the frequency of the \(L_\alpha\) line as \(f_{L_\alpha} = (5/36)cR(Z-7.4)^2\). How can the Bohr model explain this result? What is the general form for the \(L\) series wavelengths \(\lambda_L\)?

    The \(L_\alpha\) x-ray results from the transition of \(n=3\rightarrow 2\), where the \(L\) series refer to transition to the \(n=2\) state from higher states. For the case of the \(K\) series, the effective charge of the nucleus \(Z_{\rm eff}\) was simply \(Z-1\). Since we don’t yet know the detailed arrangement of electrons, we just let \(Z=Z_{\rm eff}\) and its value will be determined empirically. From Bohr’s model we can proceed to find

    -
    -(4.40)#\[\begin{align} +
    +(4.40)#\[\begin{align} f_{L_\alpha} = cRZ_{\rm eff}^2\left(\frac{3^2-2^2}{2^23^2}\right) = \frac{5}{36}cRZ_{\rm eff}^2. \end{align}\]

    According to Moseley’s data, the effective charge must be \(Z-7.4\). We can rewrite our result by replacing \(3\rightarrow n\) (and changing to wavelength) to get

    -
    -(4.41)#\[\begin{align} +
    +(4.41)#\[\begin{align} \frac{1}{\lambda_L} = RZ_{\rm eff}^2\left(\frac{n^2 -4}{4n^2}\right) = R(Z - 7.4)^2\left(\frac{n^2 -4}{4n^2}\right). \end{align}\]
    @@ -1344,7 +1344,7 @@

    4.8. Homework Problems<

    - © Copyright 2022. + © Copyright 2024.

    diff --git a/docs/Chapter_5/quantum-mechanics-part1.html b/docs/Chapter_5/quantum-mechanics-part1.html index ba593dc..3eaab8c 100644 --- a/docs/Chapter_5/quantum-mechanics-part1.html +++ b/docs/Chapter_5/quantum-mechanics-part1.html @@ -584,13 +584,13 @@

    5.1. X-Ray Scattering

    5.2. De Broglie Waves#

    By 1920 it was clear that x-rays exhibited wave properties, where x-ray crystallography could be used to study the crystalline structure of atoms and molecules. A detailed understanding of the atom was not yet proposed, where a more general theory was necessary to replace the Bohr model of the atom. A first step was made by Prince Louis deBroglie, who was well versed in the work of Planck, Einstein, and Bohr. De Broglie was struck by how photons (EM radiation) had both wave (e.g., crystallography) and particle (e.g., photoelectric effect) properties. If electromagnetic radiation must have both wave and particle properties, they material particles should have both wave and particle properties as well. According to de Broglie, the symmetry of nature encourages such an idea, and no laws of physics prohibit it.

    De Broglie presented his new hypothesis within his doctoral thesis to the University of Paris in 1924, where it aroused both interest and skepticism. De Broglie combined the special theory of relativity with quantum theory to establish the wave properties of particles. He predicted a relationship between the wavelength \(\lambda\) and momentum \(p\) for a particle as

    -
    -(5.2)#\[\begin{align} +
    +(5.2)#\[\begin{align} \lambda = \frac{h}{p}. \end{align}\]

    De Broglie was guided by the concepts of phase and group velocities of waves. Recall that photons have energy through momentum \(E= pc\) and with a frequency \(E=hf\), so that

    -
    -(5.3)#\[\begin{align} +
    +(5.3)#\[\begin{align} hf &= pc = p\lambda f,\\ h &= p\lambda, \\ \lambda &= \frac{h}{p}. @@ -682,8 +682,8 @@

    5.3. Electron Scattering \[ n\lambda = 2d \sin \theta = 2d \cos \alpha. \]

    Bragg’s law is written in terms of the lattice plane spacing \(d\), where we want to know the interatomic distance \(D\) and have the relation \(d= D\sin \alpha\). Applying to Bragg’s law, we find

    -
    -(5.4)#\[\begin{align} +
    +(5.4)#\[\begin{align} n\lambda &= 2d\cos\alpha = 2D \sin \alpha \cos \alpha, \\ &= D\sin 2\alpha = D\sin \phi. \end{align}\]
    @@ -739,8 +739,8 @@

    5.3. Electron Scattering \[ U = \text{K.E.} = \frac{3}{2}kT = \frac{p^2}{2m_N}, \]

    which can be solved to get \(p = \sqrt{3m_NkT}\) using the mass \(m_N\) of a neutron. With the momentum, we can write the de Broglie wavelength as

    -
    -(5.5)#\[\begin{align} +
    +(5.5)#\[\begin{align} \lambda &= \frac{h}{p} = \frac{h}{\sqrt{3m_NkT}}, \\ &= \frac{1}{\sqrt{T}} \frac{6.6261 \times 10^{-34}\ {\rm J\cdot s }}{\sqrt{3(1.675 \times 10^{-27}\ {\rm kg})(1.381 \times 10^{-23}\ {J/K})}}, \\ &= \frac{2.515\ {\rm nm \cdot K^{1/2}}}{\sqrt{T}}. @@ -787,8 +787,8 @@

    5.3. Electron Scattering5.4. Wave Motion#

    It must be possible to formulate a wave description of particle motion because both massless and massive particles exhibit wave behavior. The quantum of theory of physics is based heavily on waves, where we develop a framework for waves before applying it to particles.

    The simplest form of a wave has a sinusoidal form, and at an initial time \(t=0\), its spatial variation (as a function of \(x\)) looks like

    -
    -(5.6)#\[\begin{align} +
    +(5.6)#\[\begin{align} \Psi(x,t=0) = A \sin (kx), \end{align}\]

    in terms of the wave number \(k=2\pi/\lambda\) and maximum displacement \(A\) (or amplitude). The wavelength \(\lambda\) is defined as the distance between two points on the wave with the same value of \(\Psi\). The time required for a wave to travel one wavelength \(\lambda\) is called the period \(T\).

    @@ -864,28 +864,28 @@

    5.4. Wave Motion (5.8)#\[\frac{\partial^2\Psi}{\partial x^2} = \frac{1}{v^2}\frac{\partial^2\Psi}{\partial t^2}.\]

    Using \(\lambda = vT\) , we can Eqn. (5.7) as

    -
    -(5.9)#\[\begin{align} +
    +(5.9)#\[\begin{align} \Psi(x,t) = A\sin \left(kx - \frac{2\pi}{T}t \right), \end{align}\]

    where the angular velocity is \(\omega= 2\pi/T\) and

    -
    -(5.10)#\[\begin{align} +
    +(5.10)#\[\begin{align} \Psi(x,t) = A\sin \left(kx - \omega t \right). \end{align}\]

    This is the mathematical description of a sine curve traveling in the positive \(x\) direction that has a zero-displacement (\(\Psi = 0\)) at \(x=0\) and \(t=0\). A similar wave traveling in the negative \(x\) direction has the form

    -
    -(5.11)#\[\begin{align} +
    +(5.11)#\[\begin{align} \Psi(x,t) = A\sin \left(kx + \omega t \right). \end{align}\]

    The phase velocity \(v_{\rm ph}\) is the velocity for a point on the wave that has a given phase (e.g., the crest), which is

    -
    -(5.12)#\[\begin{align} +
    +(5.12)#\[\begin{align} v_{\rm ph} = \frac{\lambda}{T} = \frac{\omega}{k}. \end{align}\]

    A wave can be shifted by a phase constant \(\phi\), which produces

    -
    -(5.13)#\[\begin{align} +
    +(5.13)#\[\begin{align} \Psi(x,t) = A\sin(kx-\omega t + \phi). \end{align}\]

    When two or more waves traverse the same region, then they act independently of each other. We add the displacements of all the waves present through the principle of superposition. For sound, beating occurs when two waves have nearly equal frequencies. The net displacement depends on the harmonic amplitude, phase, and frequency of each individual wave. Combining waves is accomplished by adding their instantaneous displacements.

    @@ -986,8 +986,8 @@

    5.4. Wave Motion \[ \cos\ \alpha + \cos\ \beta = 2\cos\ \left(\frac{\alpha+\beta}{2}\right) \cos\ \left(\frac{\alpha-\beta}{2}\right), \]

    we can obtain

    -
    -(5.15)#\[\begin{align} +
    +(5.15)#\[\begin{align} \Psi(x,t) &= 2A \cos \left[\frac{k_1+k_2}{2}x - \frac{\omega_1+\omega_2}{2}t \right] \cos \left[\frac{1}{2}(k_1-k_2)x - \frac{1}{2}(\omega_1-\omega_2)t \right], \\ &= 2A \cos\left(k_{\rm av}x - \omega_{\rm av}t\right) \cos\left(\frac{1}{2}\Delta kx - \frac{1}{2}\Delta \omega t\right). \end{align}\]
    @@ -999,13 +999,13 @@

    5.4. Wave MotionA similar reasoning can be applied to determine a time \(\Delta t\) over which the wave is localized and obtain \(\Delta \omega \Delta t = 2\pi\).

    The relation (in Eqn. (5.16)) implies that to determine a precise position \(\Delta x\) of the wave packet (\(\Delta x \rightarrow \text{small}\)), we must have a large range of wave numbers (\(\Delta k \rightarrow \text{large}\)). A similar reasoning can be applied for the relation between angular frequency and time.

    Equation (5.14) can be extended to consider the sum over many waves with possibly different wave numbers, angular frequencies, and amplitudes through a Fourier series, or

    -
    -(5.17)#\[\begin{align} +
    +(5.17)#\[\begin{align} \Psi(x,t) = \sum_i A_i \cos(k_ix -\omega_i t). \end{align}\]

    Gaussian wave packets are often used to represent the position of particles because the associated integrals are relatively easy to evaluate. A Gaussian wave (at \(t = 0\)) can be expressed as

    -
    -(5.18)#\[\begin{align} +
    +(5.18)#\[\begin{align} \Psi(x,0) = \psi(x) = A e^{-\Delta k^2 x^2}\cos(k_ox), \end{align}\]

    where the range of wave numbers \(\Delta k\) are used to form the wave packet. The \(\cos(k_o x)\) term describes an oscillating wave within the (Gaussian) exponential envelope \(e^{-\Delta k^2 x^2}\). The figure below shows an intensity distribution \(\mathcal{I}(k)\) described by a Gaussian (on the left) and a corresponding wave packet \(\psi(x)\) (on the right). Click the button on the right to show the python code underlying the figure.

    @@ -1081,8 +1081,8 @@

    5.4. Wave Motion

    For a de Broglie wave, we know \(E= hf\) and \(p = h/\lambda\). We can rewrite these equations in terms of \(\hbar\) as,

    -
    -(5.20)#\[\begin{align} +
    +(5.20)#\[\begin{align} E &= hf = \hbar(2\pi f) = \hbar \omega, \\ p &= \frac{h}{\lambda} = \hbar\frac{2\pi}{\lambda} = \hbar k. \end{align}\]
    @@ -1098,8 +1098,8 @@

    5.4. Wave Motion

    Equation (5.21) represents a particle with a momentum \(p\) and total energy \(E\). It is plausible to assume that the group velocity of the wave packet can be associated with the particle velocity. The phase velocity is represented by

    -
    -(5.22)#\[\begin{align} +
    +(5.22)#\[\begin{align} v_{\rm ph} = \lambda f = \frac{\omega}{k}, \end{align}\]

    so that \(\omega = kv_{\rm ph}\). Then, the group velocity is related to the phase velocity by

    @@ -1183,8 +1183,8 @@

    5.5.2. Electron Double-Slit Experiment \[ \sin \theta_1 = \frac{\lambda}{d} = \frac{\lambda}{2\times 10^{-6}\ {\rm m}}. \]

    We need to find the wavelength \(\lambda\) of the electrons and then we can determine the distance \(d\) between the two maxima (\(y_1 = D \tan \theta_1\)). From Section 5.2, we can calculate the de Broglie wavelength relativistically by first using Eqn. (2.56) to determine \(pc\) by

    -
    -(5.24)#\[\begin{align} +
    +(5.24)#\[\begin{align} (pc)^2 &= E^2 - E_o^2 = (\text{K.E.} + E_o)^2 - E_o^2, \\ &= (5\times 10^4\ {\rm eV} + 5.11 \times 10^5\ {\rm eV})^ - (5.11 \times 10^5\ {\rm eV})^2, \\ & = (2.32 \times 10^5\ {\rm eV})^2, \\ @@ -1272,8 +1272,8 @@

    5.5.4. Physical Observables5.6. Uncertainty Principle#

    To localize a wave packet over a small region \(\Delta x\), a large range of wave numbers \(\Delta k\) is necessary. For the case of two waves (or a wave packet), Eqn. (5.16) showed that the product of the \(\Delta k\) and \(\Delta x\) is a constant.

    It is impossible to measure the precise values of \(k\) and \(x\) for the same particle, simultaneously. Let’s re-write the wave number \(k\) in terms of the momentum using the relation for the de Broglie wavelength as

    -
    -(5.25)#\[\begin{align} +
    +(5.25)#\[\begin{align} k &= \frac{2\pi}{\lambda} = \frac{2\pi}{h/p} = \frac{p}{\hbar}, \\ \Delta k &= \frac{\Delta p}{\hbar}. \end{align}\]
    @@ -1295,8 +1295,8 @@

    5.6. Uncertainty Principle(5.27) is true for all waves (e.g. water, sound, matter, and light) as a consequence of the de Broglie wavelength. If we want to know the position of a particle accurately, we must accept a large uncertainty with the momentum of the particle (either how fast it is moving or how massive it is). The converse is also true. Classical physics assumed that it is possible to specify (simultaneously and precisely) both the particle’s position and momentum, were it is not the case in the quantum realm. Since \(\hbar \ll 1\), the uncertainty principle becomes important only on the atomic level.

    Consider a particle for which the location is known within a width of \(\ell\) along the \(x\)-axis. The position of the particle is known within a distance \(\Delta x \leq \ell /2\). Applying the uncertainty principle specifies the limits on momentum \(\Delta p_x\) as

    -
    -(5.28)#\[\begin{align} +
    +(5.28)#\[\begin{align} \Delta p_x \geq \frac{\hbar}{2\Delta x} \geq \frac{\hbar}{\ell}. \end{align}\]

    Assume that the mass \(m\) of the particle is precisely known, then \(\Delta p_x = m\Delta v_x\), and

    @@ -1450,14 +1450,14 @@

    5.7. Probability, Wave Functions, and th

    The wave function \(\Psi\) determines the likelihood (or probability) of finding a particle at a particular space at a given time. The value of the wave function \(\Psi\) can have a complex value (i.e., it can contain both real and imaginary numbers). The probability density is represented as \(|\Psi|^2\), which describes the probability of finding the particle in a given unit volume at a given instant of time.

    The wave function \(\Psi(\vec{x},t)\) depends on a spatial vector \(\vec{x} = \left[x,y,z\right]\) as well as a time \(t\). The probability is found by taking the product of the wave function \(\Psi\) with its complex conjugate \(\Psi^*\). The result is \(\Psi^* \Psi dy = |\Psi|^2 dy\), which represents the probability \(P(y)dy\) of observing an electron in the interval \(y\) and \(y+dy\) at a given time on a screen in the double slit experiment. Therefore, the probability is

    -
    -(5.32)#\[\begin{align} +
    +(5.32)#\[\begin{align} P(y)dy = |\Psi(y,t)|^2 dy, \end{align}\]

    since we are only interested in a single dimension \(y\) along the observing screen.

    The electron has to be observed somewhere along the screen (i.e., 100% probability if we add up all the probabilities). We integrate the probability density over all space (i.e., from \(-\infty\) to \(\infty\)). This process is called normalization and is represented mathematically as

    -
    -(5.33)#\[\begin{align} +
    +(5.33)#\[\begin{align} \int_{-\infty}^{\infty} P(y)dy = \int_{-\infty}^{\infty} |\Psi(y,t)|^2 dy = 1. \end{align}\]

    Max Born first proposed the probability interpretation of the wave function in 1926 in his paper Quantum mechanics of collision processes.

    @@ -1502,8 +1502,8 @@

    5.7.1. The Copenhagen Interpretation

    This requirement means that for an integer \(n\) (\(=1,\ 2,\ 3,\ \ldots\)), we have

    -
    -(5.34)#\[\begin{align} +
    +(5.34)#\[\begin{align} \lambda_n &= \frac{2\ell}{n}, \\ k_n &= \frac{2\pi}{\lambda_n} = \frac{n\pi}{\ell}. \end{align}\]
    @@ -1776,7 +1776,7 @@

    5.9. Homework Problems<

    - © Copyright 2022. + © Copyright 2024.

    diff --git a/docs/Chapter_6/quantum-mechanics-part2.html b/docs/Chapter_6/quantum-mechanics-part2.html index 77042fd..16537d6 100644 --- a/docs/Chapter_6/quantum-mechanics-part2.html +++ b/docs/Chapter_6/quantum-mechanics-part2.html @@ -617,8 +617,8 @@

    6.1.1. Normalization and Probability \[ P(x)\ dx = |\Psi(x,t)|^2\ dx = \Psi^*(x,t)\Psi(x,t)\ dx, \]

    where the complex conjugate operator \(^*\) negates the complex parts of a function (i.e., \(i\rightarrow -i\)) and leaves the real parts unchanged. The probability of a particle existing between two points (\(x_1\) and \(x_2\)) is given by

    -
    -(6.3)#\[\begin{align} +
    +(6.3)#\[\begin{align} P = \int_{x_1}^{x_2} \Psi^*\Psi dx. \end{align}\]

    If the wave function represents the probability of a particle existing somewhere, then the sum of all probability intervals must equal unity, or

    @@ -743,8 +743,8 @@

    6.1.3. Time-Independent Schrödinger Wav
    \[ i\hbar \psi(x)\frac{\partial f(t)}{\partial t} = -\frac{\hbar^2 f(t)}{2m}\frac{\partial^2 \psi(x)}{\partial x^2} + V(x)\psi(x)f(t). \]

    Dividing by \(\psi(x) f(t)\), we get

    -
    -(6.5)#\[\begin{align} +
    +(6.5)#\[\begin{align} \frac{i\hbar}{f(t)} \frac{d f(t)}{dt} = -\frac{\hbar^2}{2m}\frac{d^2 \psi(x)}{d x^2} + V(x). \end{align}\]

    Notice that the left-hand side depends only on time, while the right-hand side depends only on spatial coordinates. This allows us to change the partial derivatives to ordinary derivatives because each side depends only on one variable. Each side must be equal to a constant because one variable may change independently of the other. Let’s call this constant \(B\) and set it equal to the left-hand side to get

    @@ -760,8 +760,8 @@

    6.1.3. Time-Independent Schrödinger Wav
    \[ \ln f = \frac{Bt}{i\hbar}, \]

    and we determine \(f\) to be

    -
    -(6.6)#\[\begin{align} +
    +(6.6)#\[\begin{align} f(t) = e^{\frac{Bt}{i\hbar}} = e^{-\frac{iBt}{\hbar}}. \end{align}\]

    If we let \(B = \hbar \omega = E\), then \(f(t) = e^{-i\omega t} = e^{-\frac{iEt}{\hbar}}\). We can now write the time-independent Schödinger equation as

    @@ -772,8 +772,8 @@

    6.1.3. Time-Independent Schrödinger Wav (6.8)#\[\Psi(x,t) = \psi(x)e^{-i\omega t}.\]

    Many important results can be obtained when only the spatial part of the wave function \(\psi(x)\) is needed.

    The probability density used the complex conjugate operator, which now becomes useful. Using Eqn. (6.7), we have

    -
    -(6.9)#\[\begin{align} +
    +(6.9)#\[\begin{align} \Psi^*\Psi &= \psi^2(x)\left(e^{i\omega t}e^{-i\omega t}\right), \\ &= \psi^2(x), \end{align}\]
    @@ -814,23 +814,23 @@

    6.2. Expectation Values \[ \bar{x} = \frac{N_1 x_1 + N_2 x_2 + \cdots}{N_1 + N_2 + \cdots} = \frac{\sum_i N_i x_i}{\sum_i N_i}, \]

    from a series of discrete measurements. For continuous variables, we use the probability \(P(x,t)\) of observing the particle at a particular \(x\). Then the average value of \(x\) is

    -
    -(6.10)#\[\begin{align} +
    +(6.10)#\[\begin{align} \bar{x} = \frac{\int_{-\infty}^\infty xP(x) dx}{\int_{-\infty}^\infty P(x) dx}. \end{align}\]

    In quantum mechanics, we use the probability distribution (i.e., \(P(x)dx = \Psi^*(x,t)\Psi(x,t) dx\)) to determine the average or expectation value. The expectation value \(\langle x \rangle\) is given by

    -
    -(6.11)#\[\begin{align} +
    +(6.11)#\[\begin{align} \langle x \rangle = \frac{\int_{-\infty}^\infty x\Psi^*(x,t)\Psi(x,t) dx}{\int_{-\infty}^\infty \Psi^*(x,t)\Psi(x,t) dx}. \end{align}\]

    The denominator is the normalization equation. If the wave function is normalized, then the denominator is equal to unity. For a normalized wave function \(\Psi\), the expectation value is then given by

    -
    -(6.12)#\[\begin{align} +
    +(6.12)#\[\begin{align} \langle x \rangle = \int_{-\infty}^\infty x\Psi^*(x,t)\Psi(x,t) dx. \end{align}\]

    The same general procedure is applicable to determining the expectation value of any function \(g(x)\) for a normalized wave function \(\Psi(x,t)\) by

    -
    -(6.13)#\[\begin{align} +
    +(6.13)#\[\begin{align} \langle g(x) \rangle = \int_{-\infty}^\infty \Psi^*(x,t)g(x)\Psi(x,t) dx. \end{align}\]

    Any knowledge we might have of the simultaneous value of the position \(x\) and the momentum \(p\) must be consistent with the uncertainty principle. To find the expectation value of \(p\), we need to represent \(p\) in terms of \(x\) and \(t\). Consider the wave function of the free particle, \(\Psi(x,t) = e^{i(kx-\omega t)}\). The spatial derivative is

    @@ -850,19 +850,19 @@

    6.2. Expectation Values\(\hat{Q}\), transforms the function \(f(x)\) by \(\hat{Q}f(x) = g(x)\). The momentum operator \(\hat{p}\) is then represented by

    -
    -(6.14)#\[\begin{align} +
    +(6.14)#\[\begin{align} \hat{p} = -i\hbar \frac{\partial}{\partial x}, \end{align}\]

    where the \(\hat{\ }\) symbol (i.e., “hat”) sign over the letter indicates that it is an operator.

    The momentum operator is not unique, where each of the physical observables has an associated operator that used to find the observable’s expectation value. To compute the expectation value of some physical observable \(Q\), the operator \(\hat{Q}\) must operate on \(\Psi\) before calculating the probability to get

    -
    -(6.15)#\[\begin{align} +
    +(6.15)#\[\begin{align} \langle Q \rangle = \int_{-\infty}^\infty \Psi^*(x,t)\hat{Q}\Psi(x,t) dx. \end{align}\]

    The expectation value of the momentum \(p\) becomes

    -
    -(6.16)#\[\begin{align} +
    +(6.16)#\[\begin{align} \langle p \rangle &= \int_{-\infty}^\infty \Psi^*(x,t)\hat{p}\Psi(x,t) dx, \\ &=-i\hbar \int_{-\infty}^\infty \Psi^*(x,t)\frac{\partial \Psi(x,t)}{\partial x} dx. \end{align}\]
    @@ -875,8 +875,8 @@

    6.2. Expectation Values(6.17)#\[\begin{split}E\left[\Psi(x,t)\right] &= i\hbar \frac{\partial \Psi}{\partial t}, \\ \hat{E}\Psi(x,t) &= i\hbar \frac{\partial }{\partial t}\Psi(x,t).\end{split}\]

    The energy operator \(\hat{E}\) is used to find the expectation value \(\langle E \rangle\) of the energy by

    -
    -(6.18)#\[\begin{align} +
    +(6.18)#\[\begin{align} \langle E \rangle &= \int_{-\infty}^\infty \Psi^*(x,t)\hat{E}\Psi(x,t) dx, \\ &=i\hbar \int_{-\infty}^\infty \Psi^*(x,t)\frac{\partial \Psi(x,t)}{\partial t} dx. \end{align}\]
    @@ -916,8 +916,8 @@

    6.3. Infinite Square-Well Potential

    The potential that gives rise to an infinite square well with a width \(L\) is given by

    -
    -(6.19)#\[\begin{align} +
    +(6.19)#\[\begin{align} V(x) = \begin{cases} \infty &\qquad x\leq 0, x\geq L \\ 0 &\qquad 0<x<L. \end{cases} \end{align}\]

    The particle is constrained to move only between \(x=0\) and \(x=L\), where the particle experiences no forces. The infinite square-well potential can approximate many physical situations, such as the energy levels of simple atomic and nuclear systems.

    @@ -930,24 +930,24 @@

    6.3. Infinite Square-Well Potential \[ \frac{d^2 \psi}{dx^2} = -\frac{2mE}{\hbar^2}\psi = -k^2 \psi, \]

    where the wave number \(k = \sqrt{2mE/\hbar^2}\). A suitable solution to this equation is

    -
    -(6.20)#\[\begin{align} +
    +(6.20)#\[\begin{align} \psi(x) = A\sin kx + B \cos kx, \end{align}\]

    where \(A\) and \(B\) are constants used to normalize the wave function. The wave function must be continuous at the boundaries, which means that \(\psi(0)=\psi(L) = 0\). Let’s apply these boundary conditions as

    -
    -(6.21)#\[\begin{align} +
    +(6.21)#\[\begin{align} \psi(0) &= A \sin (0) + B \cos(0) = B, \\ \psi(L) &= A \sin (kL) + B \cos(kL). \end{align}\]

    For \(\psi(0) = 0\), this implies that \(B=0\). To apply \(\psi(L) = 0\), then \(A\sin(kL) = 0\) and \(A=0\) is a trivial solution. We must have

    -
    -(6.22)#\[\begin{align} +
    +(6.22)#\[\begin{align} kL = n\pi \end{align}\]

    for \(\sin(kL) = 0\) and \(n\) is a positive integer. The wave function is now

    -
    -(6.23)#\[\begin{align} +
    +(6.23)#\[\begin{align} \psi_n(x) = A \sin \left(\frac{n\pi}{L} x \right) \quad (n=1,\ 2,\ 3,\ \ldots). \end{align}\]

    The property that \(d\psi /dx\) must also be continuous so that we can normalize the wave function. The normalization condition is

    @@ -1069,8 +1069,8 @@

    6.3. Infinite Square-Well Potential

    6.4. Finite Square-Well Potential#

    An infinite potential well is not very realistic. The finite-well potential is more realistic and similar to the infinite one. Consider a potential that is zero between \(x=0\) and \(x=L\), but equal to a constant \(V_o\) everywhere else. This is written mathematically as

    -
    -(6.26)#\[\begin{align} +
    +(6.26)#\[\begin{align} V(x) = \begin{cases} V_o \quad & x\leq -L, \quad &\text{region I} \\ @@ -1079,8 +1079,8 @@

    6.4. Finite Square-Well Potential

    First, let’s examine a particle of energy \(E < V_o\) that is classically bound inside the well. Using Eqn. (6.7), we have

    -
    -(6.27)#\[\begin{align} +
    +(6.27)#\[\begin{align} -\frac{\hbar^2}{2m} \frac{d^2\psi}{dx^2} = \left[E-V_o \right]\psi. \quad \text{regions I, III} \end{align}\]

    Let \(\alpha^2 = 2m(V_o-E)/\hbar^2\) (a positive constant), and we find

    @@ -1093,8 +1093,8 @@

    6.4. Finite Square-Well Potential \[ \lim_{x\rightarrow -\infty} \psi = Ae^{-\infty} + Be^{\infty}, \]

    which requires that \(B = 0\). Similarly, we can use a limit to find that \(A=0\) for \(x\geq L\). This allows us to define the wave functions in each region as

    -
    -(6.28)#\[\begin{align} +
    +(6.28)#\[\begin{align} \psi_I(x) &= Ae^{\alpha x}, \quad \text{region I},\ x\leq -l \\ \psi_{III}(x) &= Be^{-\alpha x}. \quad \text{region III},\ x\geq L \end{align}\]
    @@ -1102,21 +1102,21 @@

    6.4. Finite Square-Well Potential \[ \frac{d^2\psi}{dx^2} = -k^2\psi, \]

    where \(k=\sqrt{(2mE)/\hbar^2}\). This differential equation has a solution, which is a combination of sinusoidal functions:

    -
    -(6.29)#\[\begin{align} +
    +(6.29)#\[\begin{align} \psi_{II}(x) = C\cos(kx) + D\sin(kx). \quad \text{region II},\ -L<x<L \end{align}\]

    To determine a valid wave function, it must be continuous at the boundary (e.g., \((\psi_I = \psi_{II})\) and \((d\psi_I/dx = d\psi_{II}/dx)\) at \(x=-L\)). By choosing symmetric boundaries and noting that the potential is an even function (i.e., \(V_ox^0\)), we can simplify \(\psi_{II}\) to include only the even portion (i.e., \(D=0\)). As a result, we find the following conditions

    -
    -(6.30)#\[\begin{align} +
    +(6.30)#\[\begin{align} C\cos(kL) &= A e^{-\alpha L}, \\ kC\sin(kL) &= -\alpha A e^{-\alpha L}, \\ C\cos(kL) &= Be^{-\alpha L}, \\ -kC\sin(kL) &= -\alpha B e^{-\alpha L}. \end{align}\]

    We immediately find that \(B=A\), and we can divide the last two equations to eliminate \(C\) to get,

    -
    -(6.31)#\[\begin{align} +
    +(6.31)#\[\begin{align} k\tan(kL) = \alpha. \end{align}\]

    Recall that \(k\) and \(\alpha\) are both functions of energy \(E\). To solve for \(E\), we perform a variable transformation \(z \equiv kL\) and \(z_o \equiv \frac{L}{\hbar}\sqrt{2mV_o}\). Using the definitions of \(k\) and \(\alpha\), we find that

    @@ -1304,8 +1304,8 @@

    6.4. Finite Square-Well Potential

    The application of the boundary conditions leads to quantized energy values \(E_n\) and associated wave functions \(\psi_n\). Remarkably, the particle has a finite probability of being outside the square well (see Fig. 6.3). Notice that the wave functions joint smoothly at the edges of the well and approach zero exponentially outside the well.

    The particle existing outside the well is prohibited classically, but occurs in quantum mechanics. Because of the exponential decrease of the the wave functions (\(\psi_I\) and \(\psi_{III}\)), the probability of the particle penetrating a distance greater than \(\delta x \approx 1/\alpha\) decreases quickly, where

    -
    -(6.33)#\[\begin{align} +
    +(6.33)#\[\begin{align} \delta x \approx \frac{1}{\alpha} = \frac{\hbar}{\sqrt{2m(V_o-E)}} \end{align}\]

    is called the penetration depth. The fraction of particles that successfully tunnel through the outer walls of the potential well is exceedingly small, but the results have important applications.

    @@ -1313,8 +1313,8 @@

    6.4. Finite Square-Well Potential

    6.5. Three-Dimensional Infinite-Potential Well#

    Like the one-dimensional infinite-potential well, the three-dimensional potential well is expected to have time-independent solutions that can be determined from the time-independent Schrödinger equation. The wave function must depend on all three spatial coordinates, \(\psi =\psi(x,y,z)\). Beginning with the conservation of energy, we multiply by the wave function to get

    -
    -(6.34)#\[\begin{align} +
    +(6.34)#\[\begin{align} \frac{p^2}{2m}\psi + V\psi = E\psi. \end{align}\]

    Then, we use the momentum operator for each dimension because \(\hat{p}^2 = \hat{p}_x^2 + \hat{p}_y^2 + \hat{p}_z^2\). Individually we have

    @@ -1328,13 +1328,13 @@

    6.4. Finite Square-Well Potential (6.35)#\[-\frac{\hbar^2}{2m}\left(\frac{\partial^2 \psi}{\partial x^2} + \frac{\partial^2 \psi}{\partial y^2}+ \frac{\partial^2 \psi}{\partial z^2} \right) + V\psi = E\psi,\]

    which defines the time-independent Schrödinger equation in three dimensions. The expression in parentheses is the Laplacian operator, which is usually written using the shorthand notation as

    -
    -(6.36)#\[\begin{align} +
    +(6.36)#\[\begin{align} \nabla^2 = \frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + \frac{\partial^2 }{\partial z^2}, \end{align}\]

    and we can write Eqn. (6.35) as

    -
    -(6.37)#\[\begin{align} +
    +(6.37)#\[\begin{align} -\frac{\hbar^2}{2m}\nabla^2 \psi + v\psi = E\psi. \end{align}\]
    @@ -1343,14 +1343,14 @@

    6.4. Finite Square-Well Potential

    Consider a free particle inside a box with lengths \(L_1\), \(L_2\), and \(L_3\) along the \(x\), \(y\), and \(z\) axes, respectively. The particle is constrained to be inside the box. (a) Find the wave functions and energies. (b) Find the ground-state wave function for a cube. (c) Find the energies of the ground and first excited state for a cube of sides \(L\).

    (a) Similar to the one-dimensional infinite square well, it is reasonable to try a sinusoidal wave function for each of the dimensions. As a result, we obtain

    -
    -(6.38)#\[\begin{align} +
    +(6.38)#\[\begin{align} \psi(x,y,z) = A\sin(k_1x)\sin(k_2y)\sin(k_3z), \end{align}\]

    which has a normalization constant \(A\) and the quantities \(k_i\) (\(k_1\), \(k_2\), and \(k_3\)) are determined by applying the appropriate boundary conditions.

    The condition that \(\psi = 0\) at \(x=L_1\) requires that \(k_1L_1 = n_1\pi\). Generalizing this constraint and solving for \(k_i\), we get

    -
    -(6.39)#\[\begin{align} +
    +(6.39)#\[\begin{align} k_1 = \frac{n_1\pi}{L_1}, \quad k_2 = \frac{n_2\pi}{L_2}, \quad k_3 = \frac{n_3\pi}{L_3}, \end{align}\]

    using the integers \(n_1\), \(n_2\), and \(n_3\) for quantum numbers in each of the dimensions.

    @@ -1366,23 +1366,23 @@

    6.4. Finite Square-Well Potential \[ \frac{\hbar^2}{2m}\left(k_1^2 + k_2^2 + k_3^2\right)\psi = E\psi. \]

    The \(k_i\) represent a wave number for each dimension, where we can generalize the energies of the one-dimensional case to

    -
    -(6.40)#\[\begin{align} +
    +(6.40)#\[\begin{align} E_n = \frac{\pi^2\hbar^2}{2m}\left(\frac{n_1^2}{L_1^2} + \frac{n_2^2}{L_2^2} + \frac{n_3^2}{L_3^2} \right), \end{align}\]

    where the allowed energies depend on the three quantum numbers. The wave function can also be written in terms of the quantum numbers as

    -
    -(6.41)#\[\begin{align} +
    +(6.41)#\[\begin{align} \psi(x,y,z) = A\sin\left(\frac{n_1\pi x}{L_1}\right) \sin\left(\frac{n_2\pi y}{L_2}\right) \sin\left(\frac{n_3\pi z}{L_3}\right) \end{align}\]

    (b) For a cube \(L_1 = L_2 = L_3 = L\) and the ground state is given by \(n_1 = n_2 = n_3 = 1\). The ground-state wave function is simply

    -
    -(6.42)#\[\begin{align} +
    +(6.42)#\[\begin{align} \psi(x,y,z) = A\sin\left(\frac{\pi x}{L}\right) \sin\left(\frac{\pi y}{L}\right) \sin\left(\frac{\pi z}{L}\right). \end{align}\]

    (c) For a cube the energies are given as

    -
    -(6.43)#\[\begin{align} +
    +(6.43)#\[\begin{align} E_n = \frac{\pi^2\hbar^2}{2mL^2}\left(n_1^2 + n_2^2 + n_3^2\right), \end{align}\]

    and the ground state energy is \(E_{111} = \frac{3\pi^2\hbar^2}{2mL^2}\).

    @@ -1406,8 +1406,8 @@

    6.6. Simple Harmonic Oscillator

    The application of the restoring force introduces simple harmonic motion (SHM). Diatomic molecules or a system of atoms in a solid lattice can be approximated by SHM in a general way. Within a lattice, the force on the atoms depends on the distance \(x\) from some equilibrium position and the potential \(V(x)\) can be represented by a Taylor series as

    -
    -(6.44)#\[\begin{align} +
    +(6.44)#\[\begin{align} V(x) = V_o + V_1(x-x_o) + \frac{V_2}{2}\left(x-x_o\right)^2 + \cdots, \end{align}\]

    where each term of the series has a constant \(V_i\). For \(x-x_o \approx 0\) (i.e., a small perturbation), the higher terms are negligible. The minimum of the potential occurs at the equilibrium position, which means that \(dV/dx = 0\) and the lowest surviving term of the potential is

    @@ -1421,19 +1421,19 @@

    6.6. Simple Harmonic Oscillator

    To study the quantum description of SHM, we insert a potential \(V = kx^2/2\) into Eqn. (6.7) to get

    -
    -(6.45)#\[\begin{align} +
    +(6.45)#\[\begin{align} \frac{d^2 \psi}{dx^2} = -\frac{2m}{\hbar^2}\left(E - \frac{1}{2}\kappa x^2 \right)\psi = \left(\frac{m\kappa}{\hbar^2}x^2 - \frac{2mE}{\hbar^2}\right) \psi. \end{align}\]

    Through a pair of variable transformations,

    -
    -(6.46)#\[\begin{align} +
    +(6.46)#\[\begin{align} \alpha^2 &= \frac{m\kappa}{\hbar^2}, \\ \beta = \frac{2mE}{\hbar^2}, \end{align}\]

    we get

    -
    -(6.47)#\[\begin{align} +
    +(6.47)#\[\begin{align} \frac{d^2 \psi}{dx^2} = (\alpha^2 x^2 - \beta)\psi. \end{align}\]

    The particle is confined to the potential well, and thus, has zero probability of being at \(x= \pm \infty\), which means

    @@ -1660,8 +1660,8 @@

    6.6. Simple Harmonic Oscillator

    In contrast to the particle-in-a-box, the oscillatory behavior is due to the polynomial, which dominates at small \(x\), and the damping occurs due to the Gaussian function, which dominates at large \(x\). Then energy levels are given by

    -
    -(6.49)#\[\begin{align} +
    +(6.49)#\[\begin{align} E_n = \left(n+\frac{1}{2}\right)\hbar \omega = \left(n+\frac{1}{2}\right)\hbar \sqrt{\frac{\kappa}{m}}. \end{align}\]
    @@ -1799,8 +1799,8 @@

    6.6. Simple Harmonic Oscillator

    6.7. Barriers and Tunneling#

    For the finite square-well potential (see Sect. 6.3), there was a finite probability for the wave function to enter the walls of the potential, but it must decay quickly. A potential barrier describes a similar situation with the boundaries inverted. The potential is instead

    -
    -(6.50)#\[\begin{align} +
    +(6.50)#\[\begin{align} V(x) = \begin{cases} 0 \quad & x\leq -L, \quad &\text{region I} \\ @@ -1827,23 +1827,23 @@

    6.7.1. Potential Barrier with

    The particle velocity must decrease for a positive \(V_o\) and increase for a negative \(V_o\). A classical analogy is where a ball is through horizontally over a mountain, where it experiences a pull due to gravity (ignoring air resistance). The pull increases when the ball is directly over the mountain because there is slightly increased mass and this extra force reduces the ball’s speed. Conversely, a ball traveling over a valley experiences slightly less gravity. This is not noticeable for most objects, but NASA’s GRACE has measured the perturbations of Earth’s gravity field using two satellites.

    According to quantum mechanics, the particle will behave differently because of its wavelike character, where the will be differences in the wave number \(k\) in the three regions. The wave numbers are

    -
    -(6.51)#\[\begin{align} +
    +(6.51)#\[\begin{align} k_I &= k_{III} = \frac{\sqrt{2mE}}{\hbar}, \quad &\text{where }V = 0 \\ k_{II} &= \frac{\sqrt{2m(E-V_o)}}{\hbar}. \quad &\text{where }V = V_o \end{align}\]

    Another analogy comes from optics, where light bends (i.e., changes speed) after it penetrates another medium. Actually, some of the light is reflected and the rest is transmitted into the medium. The wave function will exhibit a similar behavior, where there will be an (1) incident wave, (2) reflected wave, and (3) a transmitted wave.

    Classical mechanics allows no reflection if \(E>V_o\) and total reflection for \(E<V_o\). In contrast, quantum mechanics predicts some reflection for \(E\gg V_o\) and some transmission for \(E\ll V_o\).

    The time-independent Schrödinger equation for the three regions are as follows:

    -
    -(6.52)#\[\begin{align} +
    +(6.52)#\[\begin{align} \frac{d^2\psi_I}{dx^2} + \frac{2m}{\hbar^2}E\psi_I &= 0, \quad & \text{Region I } \\ \frac{d^2\psi_{II}}{dx^2} + \frac{2m}{\hbar^2}(E-V_o)\psi_{II} &= 0, \quad & \text{Region II } \\ \frac{d^2\psi_{III}}{dx^2} + \frac{2m}{\hbar^2}E\psi_{III} &= 0. \quad & \text{Region III } \end{align}\]

    The corresponding wave functions for these equations are:

    -
    -(6.53)#\[\begin{align} +
    +(6.53)#\[\begin{align} \psi_I &= Ae^{ik_Ix} + Be^{-ik_Ix}, \quad & \text{Region I } \\ \psi_{II} &= Ce^{ik_{II}x} + De^{-ik_{II}x}, \quad & \text{Region II } \\ \psi_{III} &= Fe^{ik_Ix} + Ge^{-ik_Ix}. \quad & \text{Region III } @@ -1861,15 +1861,15 @@

    6.7.1. Potential Barrier with

    The probability of particles being reflected \(R\) or transmitted \(T\) is determined by the ratio of the appropriate \(\psi^*\psi\), or

    -
    -(6.55)#\[\begin{align} +
    +(6.55)#\[\begin{align} R &= \frac{|\psi_I(\text{reflected})|^2}{|\psi_I(\text{incident})|^2} = \frac{B^*B}{A^*A}, \\[5pt] T &= \frac{|\psi_{III}(\text{transmitted})|^2}{|\psi_1(\text{incident})|^2} = \frac{F^*F}{A^*A}. \end{align}\]

    Because the particles must be either reflected or transmitted, we must have \(R+T = 1\) (i.e., the total probability must equal unity).

    The values of \(R\) and \(T\) are found by solving for \(C\) and \(D\) in terms of \(F\). Then those results are substituted to find \(A\) and \(B\) in terms of \(F\). This a long process of tedious algebra, where the transmission probability is

    -
    -(6.56)#\[\begin{align} +
    +(6.56)#\[\begin{align} T = \left[ 1 + \frac{V_o^2\sin^2(k_{II}L)}{4E(E-V_o)}\right]^{-1}. \end{align}\]

    Notice when \(k_{II}L = n\pi\), then the second term goes to zero, resulting in a transmission probability equal to 1. Also, it is possible for particles to be reflected at both \(x=-L\) and \(x=L\). Their path difference back toward the \(-x\) direction is \(2L\). For an integral number of wavelengths inside the potential barrier, the incident and reflected wave functions are precisely out of phase and cancel completely.

    @@ -2005,24 +2005,24 @@

    6.7.2. Potential Barrier with

    There are only a few changes to the equations already presented in Sect. 6.7.1. In region II, the wave function becomes

    -
    -(6.57)#\[\begin{align} +
    +(6.57)#\[\begin{align} \psi_{II}(x) = Ce^{\kappa x} + De^{-\kappa x},\quad \text{where } \kappa = \sqrt{\frac{2m(V_o-E)}{\hbar^2}}. \end{align}\]

    The parameter \(\kappa\) is a positive, real number because \(V_o > E\). The application of boundary conditions and a lot of tedious algebra relate the coefficients of the wave functions.

    The equations for the reflection and transmission probabilities are unchanged, but the results will be modified by replacing \(ik_{II}\rightarrow \kappa\). Tunneling is the effect where the particle penetrates the barrier and emerges on the other side. The transmission probability is

    -
    -(6.58)#\[\begin{align} +
    +(6.58)#\[\begin{align} T = \left[ 1 + \frac{V_o^2 \sinh^2(kL)}{4E(V_o-E)}\right]^{-1}. \end{align}\]

    The only difference from the previous transmission coefficient is the replacement of sine with hyperbolic sine (\(\sinh\)). When \(kL \gg 1\) the transmission probability reduces to

    -
    -(6.59)#\[\begin{align} +
    +(6.59)#\[\begin{align} T = \frac{16E}{V_o}\left(1 -\frac{E}{V_o}\right)e^{-2\kappa L}. \end{align}\]

    Tunneling can be understood using the uncertainty principle, where inside the barrier the wave function \(\psi_{II}\) is dominated by the \(e^{-\kappa x}\) term, and the probability density \(|\psi_{II}|^2 \approx e^{-2\kappa x}\). Over a small interval \(\Delta x = 1/\kappa\), the probability density of observing the particle has decreased substantially (\(e^{-2} \approx 0.14\)). The uncertainty in the momentum is \(\Delta p \geq \hbar/\Delta x = \hbar \kappa\). The minimum kinetic energy must be

    -
    -(6.60)#\[\begin{align} +
    +(6.60)#\[\begin{align} K_{\rm min} = \frac{(\Delta p)^2}{2m} = \frac{\pi^2 \kappa^2}{2m} = V_o - E. \end{align}\]

    The violation allowed by the uncertainty principle is exactly equal to the negative kinetic energy required. The particle is allowed by quantum mechanics and the uncertainty principle to penetrate into a classically forbidden region.

    @@ -2240,7 +2240,7 @@

    6.8. Homework Problems<

    - © Copyright 2022. + © Copyright 2024.

    diff --git a/docs/Chapter_7/hydrogen-atom.html b/docs/Chapter_7/hydrogen-atom.html index b6ef55b..086e290 100644 --- a/docs/Chapter_7/hydrogen-atom.html +++ b/docs/Chapter_7/hydrogen-atom.html @@ -533,8 +533,8 @@

    7.2.1. Separation of Variables (7.4)#\[\frac{\partial \psi}{\partial r} = fg\frac{\partial R}{\partial r}, \qquad \frac{\partial \psi}{\partial \theta} = Rg\frac{\partial f}{\partial \theta}, \qquad \text{and}\quad \frac{\partial^2 \psi}{\partial \phi^2} = Rf\frac{\partial^2 g}{\partial \phi^2}.\]

    We substitute substitute these results into Eq. (7.2) to get

    -
    -(7.5)#\[\begin{align} +
    +(7.5)#\[\begin{align} -\frac{\hbar^2}{2\mu}\left[\frac{fg}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial R}{\partial r}\right) + \frac{Rg}{r^2\sin \theta}\frac{\partial}{\partial \theta}\left( \sin \theta \frac{\partial f}{\partial \theta} \right) + \frac{Rf}{r^2\sin^2 \theta}\frac{\partial^2 g}{\partial \phi^2} \right] + V(Rfg) &= E(Rfg). \end{align}\]

    Next we multiply both sides by \(r^2 \sin^2 \theta/ (Rfg)\) and rearrange to have

    @@ -566,13 +566,13 @@

    7.2.2. Solution of the Radial Equation (7.12)#\[\frac{d^2 R}{d r^2} + \frac{2}{r}\frac{d R}{d r} + \frac{2\mu}{\hbar^2}\left[E+\frac{e^2}{4\pi \epsilon_o r}\right]R = 0.\]

    Similar to the solution for the azimuthal equation, we will try a solution with an exponential. With some prior experience in solving differential equations, we try a solution having the form

    -
    -(7.13)#\[\begin{align} +
    +(7.13)#\[\begin{align} R = Ae^{-r/a_o}, \end{align}\]

    where \(A\) is a normalization constant and \(a_o\) is a constant that must cancel out the dimension of \(r\) with a length. The first and second derivatives are

    -
    -(7.14)#\[\begin{align} +
    +(7.14)#\[\begin{align} \frac{dR}{dr} = -\frac{A}{a_o}e^{-r/a_o} = -\frac{R}{a_o}, \qquad \text{and} \qquad \frac{d^2R}{dr^2} = \frac{R}{a_o^2}. \end{align}\]

    Substituting these derivatives into Eq. (7.12) yields

    @@ -580,13 +580,13 @@

    7.2.2. Solution of the Radial Equation(7.15)#\[\begin{split}\frac{R}{a_o^2} - \frac{2R}{ra_o} + \frac{2\mu}{\hbar^2}\left[E+\frac{e^2}{4\pi \epsilon_o r}\right]R &= 0, \\ \left(\frac{2\mu e^2}{4\pi \epsilon_o \hbar^2} - \frac{2}{a_o}\right)\frac{1}{r} + \left(\frac{1}{a_o^2} + \frac{2\mu E}{\hbar^2}\right) & = 0.\end{split}\]

    Trying to solve Eq. (7.15) for \(r\) algebraically is rather complicated, where we’ll use the more trivial solution that each of the terms in parentheses are themselves equal to zero. By setting the first term in parentheses to zero and solving for \(a_o\), we find

    -
    -(7.16)#\[\begin{align} +
    +(7.16)#\[\begin{align} a_o = \frac{4\pi \epsilon_o \hbar^2}{\mu e^2}, \end{align}\]

    and that \(a_o\) is equal to the Bohr radius (see Sect. 4.4). Then we can set the second term in parentheses to zero and solve for \(E\) to find

    -
    -(7.17)#\[\begin{align} +
    +(7.17)#\[\begin{align} E = -\frac{\hbar^2}{2\mu a_o^2} = -E_o, \end{align}\]

    which is the ground state energy from the Bohr model.

    @@ -639,8 +639,8 @@

    7.2.2. Solution of the Radial Equation

    7.2.3. Solution of the Angular and Azimuthal Equations#

    In the azimuthal equation, its solution can be expressed in exponential form, but the angular equation also contains the quantum number \(m_\ell\). Thus the solutions to the angular and azimuthal equations are linked. It is customary to group these solution together into so-called spherical harmonics defined as

    -
    -(7.19)#\[\begin{align} +
    +(7.19)#\[\begin{align} Y(\theta,\ \phi) = f(\theta)g(\phi), \end{align}\]

    where the \(f(\theta)\) is a Legendre polynomial of order \(\ell\). See Table 7.2 for a listing of the normalized spherical harmonics up to \(\ell = 3\).

    @@ -728,8 +728,8 @@

    7.2.3. Solution of the Angular and Azimu

    Show that the hydrogen wave function \(\psi_{211}\) is normalized.

    A normalized function (e.g., \(\psi_{n\ell m_\ell}\)) has an area under the curve equal to unity, which is given mathematically as

    -
    -(7.20)#\[\begin{align} +
    +(7.20)#\[\begin{align} 1 &= \int \psi^*_{n\ell m_\ell} \psi_{n\ell m_\ell} d\tau, \\ & = \int \psi^*_{211} \psi_{211} r^2 \sin \theta\ dr\ d\theta\ d\phi, \end{align}\]
    @@ -822,8 +822,8 @@

    7.3.2. Orbital Angular Momentum Quantum
    \[ \vec{L} = \vec{r} \times \vec{p},\]

    with a magnitude \(L = mv_{\rm orb}r\) and the \(v_{\rm orb}\) represents the electron’s orbital velocity perpendicular to its orbital radius \(r\). The quantum number \(\ell\) is related to the magnitude of the orbital angular momentum number \(L\) by,

    -
    -(7.21)#\[\begin{align} +
    +(7.21)#\[\begin{align} L = \sqrt{\ell(\ell + 1)}\hbar. \end{align}\]

    The dependence on \(\ell(\ell + 1)\) is a wave phenomenon that results from applying the boundary conditions on \(\psi\). The quantum result disagrees with the Bohr’s theory, where \(L = n\hbar\). This will force us to discard Bohr’s semiclassical “planetary” model when considering the finer structures of the atom.

    @@ -846,8 +846,8 @@

    7.3.3. Magnetic Quantum Number

    The solution to the Schrödinger equation for the angular equation \(f(\theta)\) specifies that \(\ell\) is an integer, and therefor the magnitude of \(\vec{L}\) is quantized.

    The azimuthal angle \(\phi\) measures a rotation about the \(z\)-axis. The solution of the azimuthal equation \(g(\phi)\) specifies that \(m_\ell\) is an integer and is related to \(L_z\) (i.e., the \(z\)-component of the angular momentum) by

    -
    -(7.22)#\[\begin{align} +
    +(7.22)#\[\begin{align} L_z = m_\ell \hbar. \end{align}\]

    The relationship of \(L,\ L_z,\ \ell,\) and \(m_\ell\) is displayed in Fig. 7.1 for the value \(\ell=2\). Because \(L_z\) is quantized, only certain orientations of \(\vec{L}\) are possible and each orientation corresponds to a different \(m_\ell\). This phenomenon is called space quantization because only certain orientations of \(\vec{L}\) are allowed in space.

    @@ -914,8 +914,8 @@

    7.3.3. Magnetic Quantum Number The simultaneous knowledge of the same component of position and momentum is forbidden by the uncertainty principle.

    Only the magnitude of \(\vec{L}\) and \(L_z\) may be specified simultaneously. The values of \(L_x\) and \(L_y\) must be consistent with \(L^2 = L_x^2 + L_y^2 + L_z^2\) but cannot be specified individually. The angular momentum vector \(\vec{L}\) never points in the \(z\) direction because \(L = \sqrt{\ell(\ell _ 1)}\hbar\) and \(L > L_z^{\rm max} = \ell \hbar\).

    The values of \(L_z\) range from \(-\ell\) to \(\ell\) in steps of \(1\), for a total of \(2\ell + 1\) allowed values. There is nothing special about the Cartesian directions, and thus we expect the average angular momentum components squared in the three directions to be the same (i.e., \(\langle L_x^2\rangle = \langle L_y^2\rangle = \langle L_z^2\rangle\)). the average value of \(\langle L^2 \rangle\) is equal to three times the average value of any one of the components. To find \(\langle L^2 \rangle\), we use \(L_z\) to get

    -
    -(7.23)#\[\begin{align} +
    +(7.23)#\[\begin{align} \langle L^2 \rangle &= 3 \langle L_z^2 \rangle = \frac{3}{2\ell + 1} \sum_{m_\ell = - \ell}^\ell (m_\ell \hbar)^2, \\ & = \frac{3}{2\ell + 1} \left[\frac{\ell}{3}(\ell+1)(2\ell+1)\hbar^2\right], \\ &= \ell(\ell+1)\hbar^2. @@ -974,21 +974,21 @@

    7.4. Magnetic Effects on Atomic Spectra

    The normal Zeeman effect can be explained by considering the atom to behave like a small magnet. By the 1920s, the fine structure of atomic spectral lines from hydrogen and other elements had already been identified. Fine structure refers to the splitting of a spectral line into more closely spaced lines.

    As a rough model, think of an electron circulating around the nucleus as a circular current loop. The current loop has a magnetic moment \(\mu = IA\), which depends on the area of the current loop \(A\) and the movement of the electron through a current \(I = dq/dt\). The current can be calculated as the electron charge \(-e\) divided by the period \(T\) to make one revolution \(T = 2\pi r/v\). Thus the magnetic moment is:

    -
    -(7.24)#\[\begin{align} +
    +(7.24)#\[\begin{align} \mu &= IA = \frac{q}{T}A, \\ &= -\frac{e\pi r^2}{2\pi r/v},\\ & = -\frac{1}{2}erv, \\ & = -\frac{eL}{2m}, \end{align}\]

    and can be written in terms of the orbital angular momentum \(L = mvr\). Both the magnetic moment \(\vec{\mu}\) and angular momentum \(\vec{L}\) are vectors so that

    -
    -(7.25)#\[\begin{align} +
    +(7.25)#\[\begin{align} \vec{\mu} = -\frac{e}{2m}\vec{L}. \end{align}\]

    In the absence of an external magnetic field to align them, the magnetic moments \(\vec{\mu}\) of atoms point in random directions. In classical electromagnetism, if a magnetic dipole is placed in an external field, then the dipole will experience a torque \(\vec{\tau} = \vec{\mu} \times \vec{B}\) that tends to align the dipole with the magnetic field. The dipole has a potential energy \(V_B\) in the field given by

    -
    -(7.26)#\[\begin{align} +
    +(7.26)#\[\begin{align} V_B = -\vec{\mu} \cdot \vec{B}. \end{align}\]

    According to classical physics, the magnetic moment will align itself with the external magnetic field to minimize energy as long as the system can change its potential energy.

    @@ -997,8 +997,8 @@

    7.4. Magnetic Effects on Atomic Spectra

    There is a similarity to the case of a spinning top in a gravitational field. A child’s spinning top precesses about the gravitational field, where the spin axis rotates about the direction of the gravitational force. (i.e., vertical). The gravitational field is not parallel to the angular momentum, and the force of gravity results in a precession of the top about the field direction.

    The angular momentum is aligned with the magnetic moment, and the torque (due to the magnetic moment) causes a precession of \(\vec{\mu}\) about the magnetic field, not an alignment. The magnetic field establishes a preferred direction in space along which we customarily define the \(z\)-axis. Then we have

    -
    -(7.27)#\[\begin{align} +
    +(7.27)#\[\begin{align} \mu_z = \frac{e\hbar}{2m}m_\ell = -\mu_B m_\ell, \end{align}\]

    where a unit of the magnetic moment \(\mu_B\) is called the Bohr magneton. From the quantization of \(L_z\) and that \(L>m_\ell \hbar\), we cannot have \(|\vec{\mu}| = \mu_z\), or the magnetic moment can’t align itself exactly in the \(z\)-direction. Therefore, the magnetic moment has only certain quantized orientations.

    @@ -1042,14 +1042,14 @@

    7.5. Intrinsic SpinTo explain the experimental data, Goudsmit and Uhlenbeck proposed that the electron must have an intrinsic spin quantum number \(s= 1/2\). The spinning electron reacts similarly to the orbiting electron in a magnetic field. Therefore, we should try to find quantities analogous to the angular momentum variable \(L\), \(L_z\), \(\ell\), and \(m_\ell\). By analogy, there will be \(2s+1 = 2(1/2) + 1 = 2\) components of the spin angular momentum \(\vec{S}\). Thus the magnetic spin quantum number \(m_s\) has only two values \(m_s = \pm 1/2\). The electron’s spin will be oriented either “up” or “down” in a magnetic field, and the electron can never be spinning with its magnetic moment \(\mu\) exactly along the \(z\) axis (or the direction of the external field \(\vec{B}\)).

    Each atomic state is described by three quantum numbers (\(n\), \(\ell\), and \(m_\ell\)) plus an additional quantum number \(m_s \pm 1/2\) that accounts for the spin. The states are degenerate in energy unless an external magnetic field is applied and the energies are split. Then the energy levels introduced by the magnetic field have removed the energy degeneracy.

    The intrinsic spin angular momentum vector \(\vec{S}\) has a magnitude \(S = \sqrt{s(s+1)}\hbar\), where the magnetic moment \(\vec{\mu}_s = -(e/m)\vec{S} = -(2\mu_B/\hbar)\vec{S}\). Incorporating the special theory of relativity into the Schrödinger equation (i.e., the Dirac equation) introduces a second factor of \(\mu_B\). The relationship between the magnetic moment and the angular momentum vector is called the gyromagnetic ratio and is designated by the letter \(g\), along with an appropriate subscript \(\ell\) or \(s\). In terms of the gyromagnetic ratio:

    -
    -(7.29)#\[\begin{align} +
    +(7.29)#\[\begin{align} \vec{\mu}_\ell &= -\frac{g_\ell \mu_B}{\hbar}\vec{L} = -\frac{\mu_B}{\hbar}\vec{L} \\ \vec{\mu}_s &= -\frac{g_s \mu_B}{\hbar}\vec{S} = -\frac{-2\mu_B}{\hbar}\vec{S}. \end{align}\]

    The Stern-Gerlach produce two distinct lines due to the quantum number \(m_s\). If the atoms were in a \(\ell = 0\) state, there would be now splitting due to \(m_\ell\), but there is still space quantization due to the intrinsic spin. A similar expression for potential energy from Eq. (7.28) becomes

    -
    -(7.30)#\[\begin{align} +
    +(7.30)#\[\begin{align} V_s = -\vec{\mu} \cdot \vec{B} = \frac{e}{m}\vec{S} \cdot \vec{B}. \end{align}\]

    @@ -1084,23 +1084,23 @@

    7.6.1. Selection Rules\((4,\ 2,\ -1,\ -1/2) \rightarrow (2,\ 1,\ 0,\ 1/2)\)

    To check whether a transition is allowed, we compare the proposed transiton with the selection rules:

    -
    -(7.31)#\[\begin{align} +
    +(7.31)#\[\begin{align} \Delta n &= \text{anything}, \\ \Delta \ell &= \pm 1, \\ \Delta m_\ell &= 0,\ \pm 1. \end{align}\]

    In (a), \(\Delta \ell = \Delta m_\ell = 1\), which are allowed. Since this transition is allowed, we calculate transition energy using Eq. 4.23 from Chapter 4. Using \(E_o = 13.6\ {\rm eV}\), we have

    -
    -(7.32)#\[\begin{align} +
    +(7.32)#\[\begin{align} \Delta E &= E_3 - E_2 = -13.6\ {\rm eV}\left(\frac{1}{3^2}-\frac{1}{2^2}\right),\\ &= -13.6\ {\rm eV} \left(\frac{2^2-3^2}{2^2 3^2} \right) = 1.89\ {\rm eV}. \end{align}\]

    The above result corresponds to the absorption of an \(1.89\) eV photon.

    In (b), the change in \(\ell\) is \(0\), which is not allowed.

    In (c), \(\Delta \ell = 1-2 = -1\) and \(\Delta m_\ell = 0-(-1) = 1\), which are both allowed. From our selection rules, the \(\Delta n = -2\) and \(\Delta m_s\) do not affect whether a transition is allowed. We calculate the transition energy as

    -
    -(7.33)#\[\begin{align} +
    +(7.33)#\[\begin{align} \Delta E &= E_2-E_4 = -13.6\ {\rm eV}\left(\frac{1}{2^2}-\frac{1}{4^2}\right),\\ &= -13.6\ {\rm eV} \left(\frac{4^2-2^2}{2^2 4^2} \right) = -2.55\ {\rm eV}. \end{align}\]
    @@ -1134,31 +1134,31 @@

    7.8. Total Angular Momentum

    7.8.1. Single-Electron Atoms#

    For an atom with orbital angular momentum \(\vec{L}\) and spin angular momentum \(\vec{S}\), the total angular momentum \(J\) is given through vector addition, or

    -
    -(7.34)#\[\begin{align} +
    +(7.34)#\[\begin{align} \vec{J} = \vec{L} + \vec{S}. \end{align}\]

    The magnitude \(J\) and the \(z\)-component \(J_z\) are quantized just as \(L,\ L_z,\ S,\ S_z\) are also quantized. The quantum numbers \(j\) and \(m_j\) are also analogous to the component angular momenta. For a single electron hydrogen atom,

    -
    -(7.35)#\[\begin{align} +
    +(7.35)#\[\begin{align} J &= \sqrt{j(j+1)}\hbar, \\ J_z &= m_j \hbar. \end{align}\]

    The magnetic number \(m_j\) will be half-integral like \(m_s\) and range from \(-j\) to \(j\). The quantizations of the magnitudes of \(\vec{L},\ \vec{S},\ \text{and}\ \vec{J}\) are all similar:

    -
    -(7.36)#\[\begin{align} +
    +(7.36)#\[\begin{align} L &= \sqrt{\ell(\ell +1)}\hbar, \\ S &= \sqrt{s(s+1)}\hbar, \\ J &= \sqrt{j(j+1)}\hbar. \end{align}\]

    The total angular momentum quantum number for the single electron can only have the values

    -
    -(7.37)#\[\begin{align} +
    +(7.37)#\[\begin{align} j = \ell \pm s, \end{align}\]

    which can only be \(\ell + 1/2\) or \(\ell -1/2\). The notation commonly used to describe these states is

    -
    -(7.38)#\[\begin{align} +
    +(7.38)#\[\begin{align} nL_j \end{align}\]

    where \(n\) is the principal quantum number, \(j\) the total angular momentum quantum number, and \(L\) is an uppercase letter \((S,\ P,\ D,\ \ldots)\) representing the orbital angular momentum quantum number.

    @@ -1179,14 +1179,14 @@

    7.8.1. Single-Electron Atoms \[ E = \frac{hc}{\lambda}. \]

    For a small* splitting, we approximate \(\Delta E\) using a differential, or

    -
    -(7.39)#\[\begin{align} +
    +(7.39)#\[\begin{align} \frac{dE}{d\lambda} &= \frac{d}{d\lambda}\left(\frac{hc}{\lambda} \right), \\ dE &= -hc \frac{d\lambda}{\lambda^2}. \end{align}\]

    Letting \(dE \rightarrow \Delta E\) and \(d\lambda \rightarrow \Delta \lambda\), we find

    -
    -(7.40)#\[\begin{align} +
    +(7.40)#\[\begin{align} |\Delta E| = hc \frac{|\Delta \lambda|}{\lambda^2}, \end{align}\]

    after taking the absolute values.

    @@ -1234,8 +1234,8 @@

    7.8.2. Many-Electron Atoms\(d\) subshell should have all the same value of \(m_s\). This requires that each electron have a different \(m_\ell\), where the allowed \(m_\ell\) values are \(-2,\ -1,\ 0,\ 1,\ 2\). By rule 2, the first two electrons to occupy a \(d\) subshell should have \(m_\ell = \pm 2\) and \(m_\ell = \pm 1\).

    In additon to spin-orbit interactions, there are spin-spin and orbital-orbital interactions.

    Assume the angular momemnta \(\vec{L}_1,\ \vec{S}_1\) and \(\vec{L}_2,\ \vec{S}_2\) for a two electron atom, where \(1\) and \(2\) label the electrons. The total angular momentum \(\vec{J}\) is the vector sum of the four angular momenta:

    -
    -(7.41)#\[\begin{align} +
    +(7.41)#\[\begin{align} \vec{J} = \vec{L}_1 + \vec{L}_2 + \vec{S}_1 + \vec{S}_2. \end{align}\]

    There are two schemes for combining the four angular momenta to form \(J\), which are called LS coupling and jj coupling. Which scheme to use depends on the relative strengths of the various interactions.

    @@ -1243,8 +1243,8 @@

    7.8.2. Many-Electron Atoms

    7.8.3. LS Coupling#

    The LS coupling scheme (also called Russell-Saunders coupling) is used for most atoms when the magnetic field is weak. The angular momenta combine to form a total angular momenta:

    -
    -(7.42)#\[\begin{align} +
    +(7.42)#\[\begin{align} \vec{L} = \vec{L}_1 + \vec{L}_2, \\ \vec{S} = \vec{S}_1 + \vec{S}_2. \end{align}\]
    @@ -1257,8 +1257,8 @@

    7.8.3. LS Coupling

    For a given value of \(L\), there are \(2S+1\) values of \(J\) because \(J\) goes from \(L-S\) to \(L+S\) for \(L>S\). For \(L<S\), there are fewer than \(2S+1\) possible values of \(J\).

    The value \(2S+1\) is called the multiplicity of the state. The spectroscopic (or term) symbols for the atom becomes

    -
    -(7.43)#\[\begin{align} +
    +(7.43)#\[\begin{align} n^{2S+1}L_j. \end{align}\]

    For two electrons we have singlet states \((S=0)\) and triplet states \((S=1)\).

    @@ -1266,8 +1266,8 @@

    7.8.3. LS Coupling

    7.8.4. jj Coupling#

    This coupling scheme predominates for the heavier elements, where the nuclear charge causes the spin-orbit interactions to be as strong as the forces between the individual \(\vec{S}_i\) and \(\vec{L}_i\). The coupling order becomes

    -
    -(7.44)#\[\begin{align} +
    +(7.44)#\[\begin{align} \vec{J}_1 &= \vec{L}_1 + \vec{S}_1, \\ \vec{J}_2 &= \vec{L}_2 + \vec{S}_2, \\ \vec{J} &= \sum_i \vec{J}_i. @@ -1289,15 +1289,15 @@

    7.8.4. jj CouplingWhat are the \(L\), \(S\), and \(J\) values for the first few excited states of helium?

    The lowest excited states of helium must be \(1s^12s^1\) or \(1s^12p^1\), where one electron is promoted to either the \(2s^1\) or \(2p^1\) subshell. We expect the excited states of \(1s^12s^1\) to be lower than those of \(1s^12p^1\) because the \(2s^1\) subshell is lower in energy than the \(2p^1\) subshell.

    The possibilities are

    -
    -(7.45)#\[\begin{align} +
    +(7.45)#\[\begin{align} 1s^12s^1 & \quad L = 0\ \text{for the }s\ \text{subshell},\\ & \quad \text{If }S=0,\ \text{then } J=0,\\ & \quad \text{If }S=1,\ \text{then } J=1, \end{align}\]

    with \(S=1\) being lowest in energy. The lowest excited state is \(^3S_1\) and then comes \(^1S_0\).

    -
    -(7.46)#\[\begin{align} +
    +(7.46)#\[\begin{align} 1s^12p^1 & \quad L = 1\ \text{for the }p\ \text{subshell},\\ & \quad \text{If }S=0,\ \text{then } J=1,\\ & \quad \text{If }S=1,\ \text{then } J=0,\ 1,\ 2. @@ -1460,7 +1460,7 @@

    7.9. Homework Problems<

    - © Copyright 2022. + © Copyright 2024.

    diff --git a/docs/Chapter_8/statistical-physics.html b/docs/Chapter_8/statistical-physics.html index c6d7b0b..ad83843 100644 --- a/docs/Chapter_8/statistical-physics.html +++ b/docs/Chapter_8/statistical-physics.html @@ -537,18 +537,18 @@

    8.2. Maxwell Velocity Distribution

    where \(d^3\vec{v} = dv_x\ dv_y\ dv_z\), or a volume element in the phase space. The product of a velocity distribution funciton with a phase space volume plays a role analogous to the probability density \(\Psi^*\Psi\) in quantum theory.

    Maxwell proved that the probability distribution funciton is proportional to \(e^{-mv^2/(2kT)}\), where \(m\) is the molecular mass, \(v\) is the molecular speed, \(k\) is Boltzmann’s constant, and \(T\) is the absolute temperature. Then, we may write

    -
    -(8.1)#\[\begin{align} +
    +(8.1)#\[\begin{align} f(\vec{v})\ d^3\vec{v} = Ce^{-\frac{1}{2}\beta mv^2}\ d^3\vec{v}, \end{align}\]

    where \(C\) is aproportionality factor and \(\beta \equiv (kT)^{-1}\); not to be confused with \(\beta = v/c\) from relativity. We can expand the above expression as

    -
    -(8.2)#\[\begin{align} +
    +(8.2)#\[\begin{align} f(\vec{v})\ d^3\vec{v} = Ce^{-\frac{1}{2}\beta m(v_x^2+v_y^2+v_z^2)}\ d^3\vec{v}, \end{align}\]

    and use the properties of exponents to rewrite as three factors that each contain one of the three velocity components. They are defined as

    -
    -(8.3)#\[\begin{align} +
    +(8.3)#\[\begin{align} g(v_x)\ dv_x &\equiv C^\prime e^{-\frac{1}{2}\beta mv_x^2}\ dv_x, \\ g(v_y)\ dv_y &\equiv C^\prime e^{-\frac{1}{2}\beta mv_y^2}\ dv_y, \\ g(v_z)\ dv_z &\equiv C^\prime e^{-\frac{1}{2}\beta mv_z^2}\ dv_z, @@ -595,24 +595,24 @@

    8.2. Maxwell Velocity Distribution

    Recalling that \(a=\beta m/2\), we have

    -
    -(8.6)#\[\begin{align} +
    +(8.6)#\[\begin{align} C^\prime &= \frac{1}{2}\sqrt{\frac{4a}{\pi}}, \\ &= \frac{1}{2}\sqrt{\frac{2\beta m}{\pi}} = \sqrt{\frac{\beta m}{2\pi}}, \end{align}\]

    and

    -
    -(8.7)#\[\begin{align} +
    +(8.7)#\[\begin{align} g(v_x)\ dv_x = \sqrt{\frac{\beta m}{2\pi}} e^{-\frac{1}{2}\beta mv_x^2}\ dv_x. \end{align}\]

    With this distribution we can calculate \(\bar{v}_x\) (mean value of \(v_x\)) as

    -
    -(8.8)#\[\begin{align} +
    +(8.8)#\[\begin{align} \bar{v}_x = \int_{-\infty}^\infty v_x\ g(v_x)\ dv_x = C^\prime \int_{-\infty}^\infty v_x e^{-\frac{1}{2}\beta mv_x^2}\ dv_x = 0 \end{align}\]

    because \(v_x\) is an odd function. The result makes sense because in a random distribution of velocities one woul expect the velocity components to be evenly distributed about \(v_x = 0\). The mean value of \(v_x^2\) is

    -
    -(8.9)#\[\begin{align} +
    +(8.9)#\[\begin{align} \bar{v}_x^2 &= \int_{-\infty}^\infty v_x^2\ g(v_x)\ dv_x, \\ &= 2C^\prime \int_{0}^\infty v_x^2 e^{-\frac{1}{2}\beta mv_x^2}\ dv_x. \end{align}\]
    @@ -627,14 +627,14 @@

    8.2. Maxwell Velocity Distribution

    Alternatively, one could use an integral table with exponential functions. The final result via substitution is

    -
    -(8.10)#\[\begin{align} +
    +(8.10)#\[\begin{align} \bar{v}_x^2 &= 2C^\prime I_2 = 2 \left(\frac{\beta m}{2\pi}\right)^{1/2} \frac{\sqrt{\pi}}{4}\left(\frac{2}{\beta m}\right)^{3/2}, \\ &= \frac{1}{\beta m} = \frac{kT}{m}. \end{align}\]

    There’s nothing special about the \(x\)-direction, so the results for the \(x\)-, \(y\)-, and \(z\)-velocity components are identical. The three components can be summed to find the mean translational kinetic energy \(\overline{K}\) of a molecule:

    -
    -(8.11)#\[\begin{align} +
    +(8.11)#\[\begin{align} \overline{K} = \frac{1}{2}m \left(\bar{v}_x^2 + \bar{v}_y^2 + \bar{v}_z^2\right) = \frac{1}{2}m \left(\frac{3kT}{m}\right) = \frac{3}{2}kT. \end{align}\]

    This is one of the principal results of kinetic theory.

    @@ -853,8 +853,8 @@

    8.4. Maxwell Speed Distribution

    The space between \(r\) and \(r+dr\) is a spherical shell, and thus, we must integrate over that volume to transform to the radial distribution. These two distributions are related by the volume of spherical shell, or \(4\pi r^2\). We may write

    -
    -(8.12)#\[\begin{align} +
    +(8.12)#\[\begin{align} F(r)dr = f(x,\ y,\ z) 4\pi r^2\ dr. \end{align}\]

    Returning to the problem of the speed distribution \(F(v)\) from the velocity distribution \(f(\vec{v})\). We need to integrate over a spherical shell in the velocity phase space, where we replace \(r\rightarrow v\) from the previous example. The desired speed distribution is

    @@ -870,8 +870,8 @@

    8.4. Maxwell Speed Distribution

    the most probable speed \(v_{\rm mp}\), the mean speed \(\bar{v}\), and the root-mean-square speed \(v_{\rm rms}\) are all slightly different from each other.

    To find the most probable speed, we simply find maximum speed in the probabilty curve, or

    -
    -(8.14)#\[\begin{align} +
    +(8.14)#\[\begin{align} \frac{dF}{dv} &= 4\pi C \frac{d}{dv}\left[e^{-\frac{1}{2}\beta m v^2}v^2 \right], \\ &= 2ve^{-\frac{1}{2}\beta m v^2} - \left(\beta m v\right)v^2 e^{-\frac{1}{2}\beta m v^2}. \end{align}\]
    @@ -880,8 +880,8 @@

    8.4. Maxwell Speed Distribution

    Most probable speed \(v_{\rm mp}\)

    -
    -(8.15)#\[\begin{align} +
    +(8.15)#\[\begin{align} 2v_{\rm mp}e^{-\frac{1}{2}\beta m v_{\rm mp}^2} &= \left(\beta m v_{\rm mp}\right)v_{\rm mp}^2 e^{-\frac{1}{2}\beta m v_{\rm mp}^2}, \\ v_{\rm mp} &= \sqrt{\frac{2}{\beta m}} = \sqrt{\frac{2kT}{m}}. \end{align}\]
    @@ -906,8 +906,8 @@

    8.4. Maxwell Speed Distribution

    See a list of integrals containing exponential functions and using \((k=2,\ n=4,\ a=\beta m/2)\) to find

    -
    -(8.17)#\[\begin{align} +
    +(8.17)#\[\begin{align} \overline{v^2} &= 4\pi C \int_0^\infty v^4e^{-\frac{1}{2}\beta m v^2}dv = 4\pi C \left(\frac{3!!}{8a^2}\sqrt{\frac{\pi}{a}}\right), \\ &= 4\pi \left(\frac{\beta m}{2\pi}\right)^{3/2} \frac{3}{2(\beta m)^2} \sqrt{\frac{2\pi}{\beta m}}, \\ &= \frac{3(\beta m)^{3/2}}{(\beta m)^{5/2} } = \frac{3}{\beta m}, \\ @@ -918,8 +918,8 @@

    8.4. Maxwell Speed Distribution

    Root-mean-square speed \(v_{\rm rms}\)

    -
    -(8.18)#\[\begin{align} +
    +(8.18)#\[\begin{align} v_{\rm rms} = \left(\overline{v^2}\right)^{1/2} = \sqrt{\frac{3kT}{m}} = \sqrt{\frac{3}{2}}v_{\rm mp}. \end{align}\]

    A particle moving with the mean squared speed has a kinetic energy

    @@ -927,8 +927,8 @@

    8.4. Maxwell Speed Distribution

    in keeping with our basic law of kinetic theory.

    The standard deviation of the molecular speeds \(\sigma_v\) is

    -
    -(8.19)#\[\begin{align} +
    +(8.19)#\[\begin{align} \sigma_v &= \left(\overline{v^2} - \bar{v}^2\right)^{1/2} = \left(\frac{3kT}{m} - \frac{8kT}{m\pi}\right)^{1/2}, \\ &= \left[\frac{(3\pi-8)kT}{m\pi}\right]^{1/2},\\ &= \left[3-\left(\frac{8}{\pi}\right)\right]^{1/2} \sqrt{\frac{kT}{m}}, \\ @@ -943,8 +943,8 @@

    8.4. Maxwell Speed Distribution \[ \bar{v} = \frac{2}{\sqrt{\pi}} \sqrt{\frac{2kT}{m}}, \]

    which can be applied to each type of gas. The only difference between results will be due to the molar mass \(m\). The molar mass of \(\rm H_2\) is \(2.01568\ {\rm u}\), which is found by multiplying the mass of hydrogen by two. The conversion from atomic mass units \({\rm u}\) to \(\rm kg\) is \(1.660539 \times 10^{-27}\ {\rm kg/u}\). Let’s re-write the mean molecular speed so that all the physical constants are combined and so that we can use atomic mass units directly. This gives

    -
    -(8.20)#\[\begin{align} +
    +(8.20)#\[\begin{align} \bar{v} = \left(2\sqrt{\frac{2k}{\pi u}}\right)\sqrt{\frac{T\ \text{(in K)}}{m\ \text{(in u)}}} = 145.5081\sqrt{\frac{T}{m}}. \end{align}\]

    Then, we compute the mean molecular speed for molecular hydrogen \(\rm H_2\) as

    @@ -1006,8 +1006,8 @@

    8.4. Maxwell Speed Distribution

    What fraction of the molecules in an ideal gas in equilibrium has speeds with \(\pm1\%\) of the most probable speed \(v_{\rm mp}\)?

    The Maxwell speed distribution function provides the probability of finding a particle within an interval of speeds. The fraction of molecules within the a given speed interval is equal to the integrated probability over the interval. Mathematically, this is expressed by the number of molecules at a particular speed \(N(v)\):

    -
    -(8.21)#\[\begin{align} +
    +(8.21)#\[\begin{align} P(\pm1\%) = \frac{N(1.01v_{\rm mp})-N(0.99v_{\rm mp})}{N} = \int_{0.99v_{\rm mp}}^{1.01v_{\rm mp}} F(v)dv. \end{align}\]

    The indefinite integral introduces the error function, which is beyond the scope of this course. However, we can obtain an approximate solution by calculating \(F(v_{\rm mp})\) and multiplying by \(dv \approx \Delta v = 0.02v_{\rm mp}\). This solution works for a small window, where wider intervals are easily evaluated using numerical methods (e.g., Simpson’s rule).

    @@ -1086,8 +1086,8 @@

    8.5. Classical and Quantum Statistics

    8.5.1. Classical Distributions#

    Because energy levels are fundamental in quantum theory, we rewrite the Maxwell speed distribution in terms of energy rather than velcocity. For a monatomic gas, the energy is all translational kinetic energy. We can use the following relations

    -
    -(8.22)#\[\begin{align} +
    +(8.22)#\[\begin{align} E &= \frac{1}{2}mv^2, \\ dE &= mv dv, \\ dv &= \frac{dE}{\sqrt{2mE}}. @@ -1097,8 +1097,8 @@

    8.5.1. Classical Distributions

    Maxwell-Boltzmann energy distribution

    -
    -(8.23)#\[\begin{align} +
    +(8.23)#\[\begin{align} F(v)\ dv &= 4\pi C e^{-\frac{1}{2}\beta mv^2}v^2\ dv, \\ &= 4\pi C e^{-\beta E}\left( \frac{2E}{m}\right) \frac{dE}{\sqrt{2mE}}, \\ &= \frac{8\pi C}{\sqrt{2m^3}} e^{-\beta E}\sqrt{E}\ dE. @@ -1108,13 +1108,13 @@

    8.5.1. Classical Distributions

    Maxwell-Boltzmann factor

    -
    -(8.24)#\[\begin{align} +
    +(8.24)#\[\begin{align} F_{\rm MB} = Ae^{-\beta E}, \end{align}\]

    where \(A\) is a normalization constant. The energy distribution for a classical system will then have the form

    -
    -(8.25)#\[\begin{align} +
    +(8.25)#\[\begin{align} n(E) = g(E)F_{\rm MB}, \end{align}\]

    where \(n(E)\) is a distribution that represents the number of particles within a bin of energy from \(E\) to \(E+dE\).

    @@ -1187,8 +1187,8 @@

    8.5.2. Quantum Distributions

    Fermi-Dirac distribution

    -
    -(8.26)#\[\begin{align} +
    +(8.26)#\[\begin{align} n(E) &= g(E)F_{\rm FD},\ \qquad \text{where } \\ F_{\rm FD} &= \frac{1}{B_{\rm FD}\ e^{\beta E}+1}. \end{align}\]
    @@ -1197,8 +1197,8 @@

    8.5.2. Quantum Distributions

    Bose-Einstein distribution

    -
    -(8.27)#\[\begin{align} +
    +(8.27)#\[\begin{align} n(E) &= g(E)F_{\rm BE},\ \qquad \text{where } \\ F_{\rm BE} &= \frac{1}{B_{\rm BE}\ e^{\beta E}-1}. \end{align}\]
    @@ -1241,8 +1241,8 @@

    8.5.2. Quantum Distributions

    Assume that the Maxwell-Boltmann distribution is valid in a gas of atomic hydrogen. What is the relative number of atoms in the ground state and first excited state at: \(293\ {\rm K}\), \(5000\ {\rm K}\), and \(10^6\ {\rm K}\)?

    The relative number of atoms between the ground and first excited states can be determined by using the ratio of \(n(E_2)\) to \(n(E_1)\). Mathematically this is given as

    -
    -(8.28)#\[\begin{align} +
    +(8.28)#\[\begin{align} \frac{n(E_2)}{n(E_1)} &= \frac{g(E_2)}{g(E_1)} e^{-\beta(E_2-E1)}, \end{align}\]

    where the density of states \(g(E_1) = 2\) for the ground state (spin up or down) and \(g(E_2) = 8\) for the first excited state (see Electron shells). For atomic hydrogen, \(E_2-E_1 = 10.2\ {\rm eV}\). Putting this together we can write

    @@ -1304,19 +1304,19 @@

    8.6. Fermi-Dirac Statistics8.6.1. Introduction to Fermi-Dirac Theory#

    The Fermi-Dirac distribution helps us understand how collections of fermions behave, one of its most useful applications is for the problem of electrical conducion.

    First, we consider the role of the factor \(B_{\rm FD}\). Like any other normalization factor, one could simply integrate over all allowed values of \(F_{\rm FD}\). The factor \(B_{\rm FD}\) is temperature dependent because of the parameter \(\beta\ (=1/kT)\). We express this temperature dependence as

    -
    -(8.29)#\[\begin{align} +
    +(8.29)#\[\begin{align} B_{\rm FD} = e^{-\beta E_{\rm F}}, \end{align}\]

    where \(E_{\rm F}\) is the Fermi energy. Then we can rewrite the Fermi-Dirac distribution as

    -
    -(8.30)#\[\begin{align} +
    +(8.30)#\[\begin{align} F_{\rm FD} = \frac{1}{e^{\beta(E-E_{\rm F})}+1}. \end{align}\]

    Notice what happens when \(E=E_{\rm F}\), \(F_{\rm FD} = 1/2\) (exactly). As a result, it is common to define the Fermi energy as the energy at which \(F_{\rm FD} = 1/2\).

    Consider the temperature dependence of \(F_{\rm FD}\), where in the limit as \(T \rightarrow 0\):

    -
    -(8.31)#\[\begin{align} +
    +(8.31)#\[\begin{align} F_{\rm FD} &= \begin{cases} 1, & \text{for } E < E_{\rm F}, \\ 0, & \text{for } E > E_{\rm F}. \end{cases} \end{align}\]