From 800b82eb82628170e6dfea10159be346cd094a6c Mon Sep 17 00:00:00 2001 From: Billy Quarles <4674360+saturnaxis@users.noreply.github.com> Date: Fri, 14 Jun 2024 15:54:32 -0400 Subject: [PATCH] updated Ch 11 --- Chapter_11/nuclear-physics.ipynb | 147 ++++++- docs/Chapter_11/nuclear-physics.html | 362 ++++++++++++------ .../_sources/Chapter_11/nuclear-physics.ipynb | 147 ++++++- docs/searchindex.js | 2 +- 4 files changed, 537 insertions(+), 121 deletions(-) diff --git a/Chapter_11/nuclear-physics.ipynb b/Chapter_11/nuclear-physics.ipynb index 3609254..e02cd44 100644 --- a/Chapter_11/nuclear-physics.ipynb +++ b/Chapter_11/nuclear-physics.ipynb @@ -1525,7 +1525,152 @@ "cell_type": "markdown", "metadata": {}, "source": [ - "## Fusion" + "## Fusion\n", + "The only primary souce in widespread use that is *not* derived from the Sun is nuclear power. Energy emitted by stars arises from *nuclear fusion* reations in which light nuclei fuse together. This process contrasts with nuclear fission, in which large nuclei divide.\n", + "\n", + "When $\\rm ^{236}U^*$ fissions, it divides into nuclei with a larger binding energy per nucleon, thereby releasing energy. If two light nuclei fuse together, they alos form a nucleus with a larger binding energy per nucleon and energy is released.\n", + "\n", + "The most energy is released if two isotopes of hydrogen (deuteron and triton) fuse toghether in the reaction\n", + "\n", + "\\begin{align}\n", + "{\\rm ^2H} + {\\rm ^3H} &\\rightarrow n + {\\rm ^4He}. \\qquad (Q=17.6\\ {\\rm MeV})\n", + "\\end{align}\n", + "\n", + "About $3.5\\ {\\rm MeV}$ per nucleon is released because of the stron binding of $\\rm ^4He$. Less than $1\\ {\\rm MeV}$ per nucleon is released in fission.\n", + "\n", + "\n", + "### Formation of Elements\n", + "\n", + "In the first few minutes of the universe, the light elements of hydrogen and helium were formed. Millions of years later, the heavier elements were formed in stars through nuclear fusion. Thre are two main cycles for producing energy in stars.\n", + "\n", + "The first is the **proton-proton chain**, which converts 4 protons into an $\\alpha$ particle. As stars form due to gravitational attraction, the heat (average speed of the protons) increases substantially so they can overcome their Coulomb repulsion (with the help of quantum tunneling) and fuse by the following reactions:\n", + "\n", + "\\begin{align}\n", + "{\\rm ^1H} + {\\rm ^1H} &\\rightarrow {\\rm ^2H} + \\beta^+ + \\nu. \\qquad (Q=0.42\\ {\\rm MeV})\n", + "\\end{align}\n", + "\n", + "This reaction produces a deuteron $(\\rm ^2H)$ and is a special kind of weak-interaction beta decay process. It is extremely slow, because only 1 collision in about $10^{26}$ produces a reaction. This is good, otherwise the Sun would explode!\n", + "\n", + "The deuterons that accumulate can combine with a proton $\\rm ^1H$ to produce $\\rm ^{3}He$:\n", + "\n", + "```{math}\n", + ":label: pp_chain_I\n", + "{\\rm ^2H} + {\\rm ^1H} &\\rightarrow {\\rm ^3He} + \\gamma. \\qquad (Q=5.49\\ {\\rm MeV})\n", + "```\n", + "\n", + "The ${\\rm ^3He}$ can then combine to produce ${\\rm ^4He}$ through:\n", + "\n", + "\\begin{align}\n", + "{\\rm ^3He} + {\\rm ^3He} &\\rightarrow {\\rm ^4He} + {\\rm ^1H} + {\\rm ^1H}. \\qquad (Q=12.85\\ {\\rm MeV}).\n", + "\\end{align}\n", + "\n", + "```{note}\n", + "The first two reactions which create the deuteron and helium-3 must occur more than once to produce the enough matter for the final reaction that produces helium-4. \n", + "```\n", + "\n", + "A total of six ${\\rm ^1H}$ are requred to produce ${\\rm ^4He}$ and two ${\\rm ^1H}$. This process consumes four protons. The total $Q$ for 6 ${\\rm ^1H}$ produce ${\\rm ^4H}$ is $24.7\\ {\\rm MeV}$, where an additional $2\\ {\\rm MeV}$ come from teh annihilation of two electron-positiron pairs for a total of $26.7\\ {\\rm MeV}$.\n", + "\n", + "The proton-proton chain is slow because Eq. {eq}`pp_chain_I` limits the entire process. As the reaction proceeds, the star's temperature increases and eventually carbon-12 nuclei are formed through by the **triple-alpha process** (i.e., converting 3 ${\\rm ^4He}$ into a ${\\rm ^{12}C}$).\n", + "\n", + "Another cycle can produce the ${\\rm ^4He}$, if enough **carbon** is produced (or already present) near the stellar core. The series of reactions responsible are called the **CNO cycle**:\n", + "\n", + "\\begin{align}\n", + "{\\rm ^1H} + {\\rm ^{12}C} &\\rightarrow {\\rm ^{13}N} + \\gamma, \\\\\n", + "{\\rm ^{13}N} &\\rightarrow {\\rm ^{13}C} + \\beta^+ + \\nu, \\qquad (t_{1/2} = 9.96\\ {\\rm min}) \\\\\n", + "{\\rm ^1H} + {\\rm ^{13}C} &\\rightarrow {\\rm ^{14}N} + \\gamma, \\\\\n", + "{\\rm ^1H} + {\\rm ^{14}N} &\\rightarrow {\\rm ^{15}O} + \\gamma, \\\\\n", + "{\\rm ^{15}O} &\\rightarrow {\\rm ^{15}N} + \\beta^+ + \\nu, \\qquad (t_{1/2} = 2.04\\ {\\rm min}) \\\\\n", + "{\\rm ^1H} + {\\rm ^{15}N} &\\rightarrow {\\rm ^{12}C} + {\\rm ^4He}.\n", + "\\end{align}\n", + "\n", + "```{note}\n", + "Four ${\\rm ^1H}$ and one ${\\rm ^{12}C}$ are required to produce ${\\rm ^4He}$ and ${\\rm ^{12}C}$. The {\\rm ^{12}C} nucleus serves as a catlyst.\n", + "```\n", + "\n", + "The proton-proton chain is probably responsible for most of our Sun's energy production, but the CNO cycle is a much more rapid fusion reaction. It requires higher temperatures (${\\sim}2\\times 10^7\\ {\\rm K}$) than are present in the Sun, because of the higher Coulomb barrier of ${\\rm ^{12}C}$ compared to the ${\\rm ^1H}$ for the protons.\n", + "\n", + "A hydrostatic equilibrium exists in the Sun between the gravitational attraction that contracts a star and a gas presssure that pushes out due to the fusion process. As the lighter nuclides are consumed to produce heavier nuclides, the gravitatioanl attraction succeeds in contracting the star's mass into a smaller volume and the temperature increaes.\n", + "\n", + "A higher temperature allows the nulcides with higher $Z$ to fuse. This process continues until a large part of the Sun's mass is converted to iron. Then the Sun collapses under its own graviational attraction to become a white dwarf. For more massive stars, they can contract more which allows them to become a neutron or black hole, but this is sensitive to the star's initial mass.\n", + "\n", + "### Nuclear Fusion on Earth\n", + "\n", + "Some scientists beleive that controlled nuclear fusion is ultimatlely our best source of terrestrial energy. Among the several possible fusion reactions, three of the simplest involve the three isotopes of hydrogen:\n", + "\n", + "\\begin{align}\n", + "{\\rm ^2H} + {\\rm ^2H} &\\rightarrow n + {\\rm ^3He}, \\qquad (Q=3.3\\ {\\rm MeV}) \\\\\n", + "{\\rm ^2H} + {\\rm ^2H} &\\rightarrow p + {\\rm ^3He}, \\qquad (Q=4.0\\ {\\rm MeV}) \\\\\n", + "{\\rm ^2H} + {\\rm ^3H} &\\rightarrow n + {\\rm ^4He}. \\qquad (Q=17.6\\ {\\rm MeV})\n", + "\\end{align}\n", + "\n", + "Deuterium $({\\rm ^2H})$ exists in vast quantities in seawater. Assuming that there are $10^{21}\\ {L}$ of water on Earth, the natural abundance of deuterium $(0.015\\%)$ gives $10^{43}$ deuterons. These deuterons (when fused together to make $\\rm ^{3}He$) would produce over $10^{30}\\ {\\rm J}$ of energy. This is enough to support present world energy consumption for a few billion years.\n", + "\n", + "Three main conditions are necessary for controlled nuclear fusion:\n", + "\n", + "1. The temperature must be hot enough to allow the ions (e.g., $d$ and $t$) to overcome the Coulomb barrier and fuse thier nuclei together. This requires a temperature of $100-200$ million $\\rm K$.\n", + "2. The ions must be confined to a small volume to allow the ions to fuse. A suitable ion density is $2-3 \\times 10^{20}\\ {\\rm ions/m^3}$.\n", + "3. The ions must be held together in close proximity at high temperature long enough to avoid plasma cooling, where a suitable time is $1-2\\ {\\rm s}$.\n", + "\n", + "The suitable values given above assume magnitic confinement. The product of the plasma density $n$ and the containment time $\\tau$ must have a minimum value at a sufficiently hgih temperature to initiate fusion and produce as much energy as it consumes. The minimum value is\n", + "\n", + "\\begin{align}\n", + "n\\tau \\geq 3 \\times 10^{20}\\ {\\rm s/m^3}.\n", + "\\end{align}\n", + "\n", + "The above relation is calle dthe [Lawson criterion](https://en.wikipedia.org/wiki/Lawson_criterion). A triple product of $n\\tau T$ called the *fusion product* is also sometimes use:\n", + "\n", + "\\begin{align}\n", + "n\\tau T \\geq 5\\times 10^{21}\\ {\\rm s\\cdot keV/m^3}.\n", + "\\end{align}\n", + "\n", + "The factor $Q$ is used to represent the ratio fo the power produced in the fusion reation to the power required to produce the fusion (heat). **This $Q$ factor is not to be confused with the $Q$ value for the release of binding energy.**\n", + "\n", + "The breakeven point is $Q=1$, and ignition occurs for $Q\\gg 1$. For controlled fusion produced in the laboratory, temperatures equivalent to $kT = 20\\ {\\rm keV}$ are satisfactory. For uncontrolled fusion (i.e., a H-bomb), high temperatures and densities are acheived over a very brief time by using a fission bomb. Fusion bombs do not produce radiation effects nearly as severe as those of fission bombs, because the primary products are not dangerously radioactive.\n", + "\n", + "### Controlled Thermonuclear Reactions\n", + "\n", + "A controlled thermonuclear reaction of nuclear fusion in the laboratory is one of the primary goals of science and engineering. Scientists do not expect to reach this goal for several more decades. The first fusion reaction will likely be the $D + T$ reaction. The tritium will be derived from two possible reactions:\n", + "\n", + "\\begin{align}\n", + "n + {\\rm ^6Li} &\\rightarrow {\\rm ^3H} + {\\rm ^4He}, \\\\\n", + "n + {\\rm ^7Li} &\\rightarrow {\\rm ^3H} + {\\rm ^4He} + n. \n", + "\\end{align}\n", + "\n", + "The lithium is required to generate the tritium. It is also used as the heat transfer medium and a neutron radiation shield. \n", + "\n", + "For $Q=1$, a product of a few times $10^{21}\\ {\\rm s\\cdot keV/m^3}$ will be required for a commercial reactor using $D+T$. There are two major schemes to control thermonuclear reactions:\n", + "\n", + "1. Magnetic confinement fusion (MCF), and\n", + "2. Inertial confinement fusion (ICF).\n", + "\n", + "```{margin}\n", + "**Magnetic Confinement of Plasma**\n", + "```\n", + "\n", + "The primary effort of research laboratories around the world for several years has been a device called the [tokamak](https://en.wikipedia.org/wiki/Tokamak). As many as six separate magnetic fields can be used to contain and heat the plasma.\n", + "\n", + "In a schematic cross section of a typical magnetic containment vessel, \n", + "\n", + "- The center is the location of the hot plasma where the $D+T$ reaction takes place. \n", + "- The plasma is surrounded by vaccum to keep out impurities that would poison the reaction.\n", + "- A wall surrounds the plasma and vacuum region, which would be subjected to intense radiation. The plasma must be kept from touching the enclosure.\n", + "- A lithium blanket is the next layer, which absorbs neutrons to breed more tritium.\n", + "- Next is a radiation shield to prevent radiation from reaching the magnets, which may be superconducting in a commerical reactor.\n", + "\n", + "The heating of the plasma to sufficiently high temperatures begins with the resistive heating from the electric current flowing in the plasma. This is insufficient to attain the high ignition temperature. As a result, there are two other schemes to add additional heat: (1) injection of high-energy $(40-120\\ {\\rm keV})$ neutral fuel atoms that interact with the plasma, and (2) radio-frequency (RF) induction heating of the plasma (similar to a microwave oven).\n", + "\n", + "The most significant fusion project in the near future is a large tokamak fusion reactor called the [**ITER**](https://en.wikipedia.org/wiki/ITER) (International Thermonuclear Experimental Reactor). It is currently under construction in France and is expected to generate self-sustained fusion power of $500\\ {\\rm MW}$ for $1000\\ {\\rm s}$.\n", + "\n", + "```{margin}\n", + "**Inertial Confinement**\n", + "```\n", + "The concept of intertial confinement fusion is to use an intense high-powered beam of heavy ions or a laser called a *driver* to implode a pea-sized target composed of $D+T$ to a density and temperature high enough to cause fusion ignition. \n", + "\n", + "Several US-based institutions are doing research and development in laser fusion. \n", + "\n", + "The [National Ignitition Facility](https://en.wikipedia.org/wiki/National_Ignition_Facility) (NIF) at Lawrence Livermore National Lab fires 192 lasers in to a high-Z cylinder, which produces x-rays. These x-rays heat the small fuel pellet containing the $D+T$. It produced $1.3\\ \\rm MJ$ of UV laser energy in 2010, just short of the $1.5\\ {\\rm MJ}$ needed for ignition. \n", + "\n", + "Sandia National Lab used a device called a [*Z-pinch*](https://en.wikipedia.org/wiki/Z-pinch) that uses a huge jolt of current to produe a powerful magnetic field that squeezes ions inot implosion and heats the plasma. Fracne is building a device called [Laser Megajoule](https://en.wikipedia.org/wiki/Laser_M%C3%A9gajoule) with similar objectives as the NIF.\n" ] }, { diff --git a/docs/Chapter_11/nuclear-physics.html b/docs/Chapter_11/nuclear-physics.html index a096a9d..2251fc3 100644 --- a/docs/Chapter_11/nuclear-physics.html +++ b/docs/Chapter_11/nuclear-physics.html @@ -508,7 +508,12 @@
Atomic masses are measured in atomic mass units, which are designated by the symbol \(\rm u\). Atomic mass units are defined in terms of the mass of the isotope \(\rm ^{12}C\) whose atomic mass is defined to be exactly \(12\ \rm u\). The atomic mass unit works out to be
-Both neutrons and protons are known collectively as nucleons and are constructed of other particles called quarks*. Neutrons are slightly more massive than protons.
@@ -637,14 +642,14 @@The nuclear radius is approximated by a spherical charge distribution with
-where \(r_o \approx 1.2\ {\rm fm}\) and measurements range from \(1-1.5\ {\rm fm}\); Note \(1\ {\rm fm} = 10^{-15}\ {\rm m}\). However, the term fermi is also used interchangeably with femtometer.
Robert Hofstadter and his colleagues (in the 1950s) performed the first precision electron-scattering measurements of the nuclear charge distribution using electron energies from \(100-500\ {\rm MeV}\). To probe the shape of most nuclei, we need a particle with a short wavelength. The de Broglie wavelength of a \(500\)-\(\rm MeV\) electron is \({\sim}2.5\ {\rm fm}\). These measurements are approximately described for all the lightest nuclei by the Fermi distribution for the nuclear charge density \(\rho(r)\) with the following form
-The shape of the distribution shown below illustrates the distance \(R\) at which the nuclear density has dropped to \(50\%\) of its central value, and \(t = 4.4a\) is the surface thickness, measureed from \(90\%-10\%\) of the central density.
@@ -722,8 +727,8 @@If we approximate the nuclear shape as a sphere, we have:
-The proton’s intrinsic magnetic moment points in the same direction as its intrinsic spin angular momentum because of the positive charge proton. This is contrasted with the negatively charged electron, where the spin and magnetic moment point in opposite directions.
Nuclear magnetic moments are measured in units of the nuclear magneton \(\mu_{\rm N}\), which is analogous to the Bohr magneton for electrons, by the relation
-Let us determine how strongly the neutron and proton are bound together in a deuteron. The deuteron mass is \(2.013553\ {\rm u}\), and the mass of a deuterium atom (adding the electron) is \(2.014102\ {\rm u}\). The difference in masses is just the mass of an electron. The electron binding energy is so mall that it can be neglected.
The deuteron nucleus is bound by an energy \(B_d\), which represents mass-energy. The mass of a deuteron is then
-The deuteron mass is less than the sum of the masses of a neutron and proton by just the nuclear binding energy \(B_d\). If we add an electron mass to each side of the equation, we have
-The left hand side \((m_d + m_e)\) is the atomic deuterium mass \(M({\rm ^2H})\) and \(m_p + m_e\) is the atomic hydrogen mass (if we neglect the small amount of binding energy). Then we have
-and we can use atomic masses.
@@ -787,22 +792,22 @@The binding energy \(B_d\) of the deuteron is determined using:
-Using \({\rm u} = 931.5\ {\rm Mev/c^2}\), we convert the above mass-difference to energy by
-Neglecting the atomic electron binding energy \((13.6\ {\rm eV})\) was justified, given that the nuclear binding energy is \(2.2\ {\rm MeV}\) (or \({\sim}10^6\) times greater). Even for heavier nuclei, we normally neglect the electron binding energies.
The binding energy of any nucleus \(\rm ^A_ZX\) is the energy required to separate the nucleus into free neutron and protons. It can be determined using the atomic masses by
-We can check our result for the \(2.22-{\rm MeV}\) binding energy of the deuteron by using a nuclear reaction. We scatter gamma rays (i.e., photons) from deuterium gas and look for the breakup of a deuteron into a neutron and proton:
-This type of reaction is called photodisintegration or a photonuclear reaction, because a photon causes the target nucleus to change form. The mass-energy relation is
-where \(hf\) is the incident photon energy, \(K_n\) is the neutron kinetic energy, and \(K_p\) is the proton kinetic energy. If we want to find the minimum energy required for the photodisintegration, we let \(K_n = K_p = 0\). We then find
-The above equation is not quite correct, because momentum must also be conserved in the reaction (\(K_n\) and \(K_p\) can’t both be zero). The precise relation is
-This value of \(hf_{\rm min}\) is almost exactly \(B_d\), the deuteron binding energy, because the second term is so small. Experiment shows that a photon energy less than \(2.22\ {\rm MeV}\) cannot dissociate a deuteron.
@@ -841,8 +846,8 @@The most straightforward technique to measure nuclear forces are based on scattering experiments. In scattering neutrons from protons, a deuteron is sometimes formed in the nuclear reaction:
-The angular distribution of neutrons elastically scattered by protons (e.g., neutron + proton and proton + proton scattering) reveals that the nuclear potential is shaped with a deep well followed second potential. In the case of proton + proton scattering the second potential is the electrostatic interaction through the Coulomb potential.
@@ -857,8 +862,8 @@If the binding energy of a nuclide \(B\) is positive, then the nuclide is stable against dissociating into free neutrons and protons. More generally, a nucleus containing \(A\) nucleons is stable if its mass is smaller than that of nay other possible combination of \(A\) nucleons. The binding energy of a nucleus \(^A_ZX\) against dissociation into any other possible combination of nucleons (e.g., \(R\) and \(S\)) is
-The energy required to remove one proton (or neutron) from a nuclide is called the proton (or neutron) separation energy, where the above equation is useful for finding this energy. Even if \(B\) is negative, there may be other reasons why the nucleus is stable.
@@ -873,8 +878,8 @@We can understand this through the strength of the nuclear force that is independent of whether the particles are \(nn\), \(np\), or \(pp\). Equal numbers of neutrons and protons may give them most attractive internucleon nuclear force, but the Coulomb force must be considered as well. As the number of protons increases, the Coulomb force between all the protons becomes stronger until eventually affects the binding energy.
The electrostatic energy required to contain a charge \(Ze\) eveny spread throughout a sphere of radius \(R\) can be calculated by determining the work required to bring the charge inside the sphere from infinity (i.e., potential energy). This energy is
-For a single proton, the self-energy is
@@ -888,8 +893,8 @@Show that Eq. (11.19) can be written as
-and use this equation to calculate the total Coulomb energy of \(\rm ^{238}_{\ \ 92}U\).
@@ -957,8 +962,8 @@The general form to law of radioactivity is the same for all decays because it is a statistical process. Given a sample of radioactive material we measure the disintegrations (or decays per unit time), which we define as activity. If we have \(N\) unstable atoms of a material, the activity \(R\) is given by
-where we insert the minus sign to make \(R\) positive; \(dN/dt\) is negative because the total number \(N\) decreases with time. The SI unit of activity is the becquerel \((1\ {\rm Bq} = 1\ {\rm decays/s})\). In the past the unit was the curie \((\rm Ci)\) which is \(3.7\times 10^{10}\ {\rm decays/s}\).
@@ -971,13 +976,13 @@The number \(dN\) of nuclei decaying during the time interval \(dt\) is
-This is the radioactive decay law, and it applies to all decays. The exponetial decay rate is consistent with experimental observation. The activity \(R\) is
-where \(R_o\) is the initial activity at \(t=0\). The activity of a radioactive sample falls of exponentially.
It is common to refer to the mean lifetime \(\tau\) (or half-life \(t_{1/2}\)) rather than its decay constant. The half-life is the time it takes one-half of the radioactive nuclei to decay. Mathematically, this is
-Half-life
-Mean lifetime
-The python code below creates a plot showing the fraction of remaining nuclei \(N(t)/N_o\) as a function of the half-life \(t_{1/2}\) for a sample of radioactive nuclides. The mean lifetime \(\tau\) is indicated as well.
@@ -1078,8 +1083,8 @@The nucleus \(\rm ^4He\) is particularly stable, with a binding energy of \(28.3\ {\rm MeV}\). The combination of two neutrons and tow protons is strong because of the pairing effects. If the last two protons and neutrons in a nucleus are bound by less than \(28.3\ {\rm MeV}\), then the emission of an \(\alpha\) particle (alpha decay) is energetically possible. For alpha decay, we have
-In alpha decay, the parent nucleus reverts to a daughter nucleus that is down 2 units in neutron number \(N\) and to the left \(2\) units in atomic number \(Z\). \(\alpha decay\) leaves the daugter nucleas farther from the line of stability than the parent.
Unstable nuclei can move closer to the line of stability by undergoing beta \((\beta)\) decay. The simplest example is the decay of a free neutron
-Electrons cannot exist within the nucleus, when \(\beta\) decay occurs the \(\beta^-\) (electron) is created at the time of the decay. Note that \(\rm ^{14}C\) is unstable, where it has an excess of neutrons. We expect the beta decay of \(\rm ^{14}C\) to form \(\rm ^{14}N\) by
-The emitted electron \((\beta^-)\) from the beta decay of \(\rm ^{14}C\) produces a continuous energy spectrum up to a maximum energy in contrast to the \(\alpha\) decay process that produces a monoenergetic spectrum. The measurement of a energy spectrum for beta decay was a problem for many years.
@@ -1158,15 +1163,15 @@Neutrinos interact weakly with ordinary matter, and they pass right throught the Earth with little chance of being absorbed. They have no charge and do not interact electromagnetically. They are not affected by the strong force either. Beta decay is the product of the weak interaction, where neutrinos are created.
There are also antineutrinos (denoted by \(\bar{\nu}\)) as well as neutrinos. The beta decay of a free neutron and of \(\rm ^{14}C\) are correctly written as
-where the antineutrino has \(-1/2\) spin.
The general forms of \(\beta^-\) decay and its disintegration energy are written as
-For unstable nuclides with too many protons (i.e., to the right of the line of stability) another type of beta decay can occur with a positively charged electron (positron). In beta decay the electron and positron are denoted as \(\beta^-\) and \(\beta^+\), respectively.
Current experimental evidence suggests that a free proton does not decay, where the lower limit for the half life is much longer than the age of the universe \(t_{1/2}> 10^{27}\ {\rm yr}\). But a proton bound within the nucleus can transumtate if energy is taken from the nucleus and the result is a more stable nucleus. The isotope \(\rm ^{14}O\) is unstable and decays through \(\beta^+\) decay by:
-The general forms of \(\beta^+\) decay and its disintegration energy are given as
-There is another form of beta decay. Inner \(K\)- and \(L\)-shell electrons are tightly bound and their (classical) orbits are highly elliptical. As a result, these electrons spend a reasonable amount of time passing through the nucleus, which increases the possibility of atomic electron capture. A proton in the nucleus absorbs the electron, which produces a neutron and a neutrino in the process. The reaction for a proton is
-The general forms of electron capture and its disintegration energy are given as
-Uranium-230 can alpha decay to the ground state or any of the low-lying excited states of \(\rm ^{226}Th\). If the decay proceeds to an excited energy state of energy \(E_x\) rather than the ground state, then the disintegration energy \(Q\) (from the excited state to the ground state) can be determined. Let the disintegration energy to the ground state be \(Q_o\), such that the transition to the excited state \(E_x\) is related to the disintegration energy \(Q\) by
-The distintegration value \(Q\) for the \(\alpha\) decay of \(\rm ^{230}U\) to \(\rm ^{226}Th\) can be determined using the energy of an excited state \(E_x = 0.072\ {\rm MeV}\) and \(Q_o = 5.992\ {\rm MeV}\). Therefore the disintegration energy \(Q\) is
@@ -1221,19 +1226,19 @@A nucleus has excited states, but the exitation energies tend to be much larger (many \(\rm keV\) or even \(\rm MeV\)) due to the stronger nuclear interaction. One way for the nucleus to get rid of the extra energy is to emit a photon (i.e., \(\gamma\) ray) and undergo a transition to some lower energy state.
The gamma ray difference of the upper energy state \(E_u\) and the lower one \(E_l\) is
-To conserve momentum, the nucleus must absorb some of this energy difference. However, a nuclues at rest is usually a very good approximation.
The decay of an excited state of \({^A}X^*\) (using \(*\) for the excited state) to the ground state is denoted by
-where a transition between two excited states is denoted by
-The excited energies (or levels) are characteristic of each nuclide. The gamma rays are normally emitted soon after the nucleus is created in an excited state.
@@ -1405,8 +1410,8 @@Radioactive \(\rm ^{14}C\) is produced in out atmosphere by the bombardment of \(\rm ^{14}N\) by neutrons by cosmic rays through
-A natural equilibrium of \(\rm ^{14}C\) to \(\rm ^{12}C\) exists for molecules of \(\rm CO_2\) from the atmosphere, where all living organisms use or breathe that \(\rm CO_2\). When living organisms die, their intake of \(\rm ^{14}C\) ceases and the ratio \({\rm ^{14}C}/{\rm ^{12}C}\ (=R)\) decreases as \(\rm ^{14}C\) decays.
@@ -1419,19 +1424,19 @@Rutherford used \(7.7\)-\({\rm MeV}\) alpha particles from the decay of \({\rm ^{214}_{\ \ 84}Po}\) to bombard a nitrogen target, where he observed the protons being emitted. He was not certain of the exact nuclear reaction taking place, but he convinced himself that a nuclear transformation took place. The reaction was
-The first nucleus written is the projectile, where the second is the target and is normally at rest. These two nuclei interact and undergo a transmutation to one or more final particles. The detected particle is normally listed first after the arrow, and the residual nucleus listed last.
A short hand way of writing the above reaction is
-The gneraly reaction \(x + X \rightarrow y + Y\) is written in shorthand as
-Isotopes of hydrogen are written using \(p,\ d,\ t\) in place of the more cumbersome \(\rm ^1H\), \(\rm ^2H\), and \(\rm ^3H\). The stable helium-4 nucleus is denoted by \(\alpha\). These shorthand symbols are used when they are the projectile or detected particle.
@@ -1450,8 +1455,8 @@The probability of the particle begin scattered is proportional to the cross section times the total number of target nuclei \(N_s\). We compute a normalized product as
@@ -1480,8 +1485,8 @@Scattering Probability
-Differential cross sections are determined by the number of particles scattered into a small solid angle \(d\Omega\) (measured in units of steradian \(\rm sr\)) surrounding the scattering angle \(\theta\). In spherical coordinates, the solid angle is \(d\Omega = \sin{\theta}\ d\theta d\phi\).
The differential cross section can also be written as \(d\sigma/d\Omega\), which measures the number of particles per differential solid angle. Integrating the differential cross section over the entire range of scattering angles yields the total cross section \(\sigma_T\),
-The cross sections depend on the incident kinetic energy, as well as other properties (e.g., spins). They are traditionally measured in units of barns (\(\rm b\)) with \(1\ {\rm barn} \equiv 10^{-28}\ {\rm m^{2}} = 100\ {\rm fm^2}\).
@@ -1505,8 +1510,8 @@The conservation of energy for the reaction is
-If we rearrange the above equation to separate the masses and kinetic energies, we find an expression similar to the disintegration energy:
@@ -1526,8 +1531,8 @@The minimum kinetic energy needed to initiate a nuclear reaction is the threshold energy \(K_{th}\), where the particles \(y\) and \(Y\) will be at rest in the \(cm\) system . Particle \(y\) will be moving in the lab system.
The speed of the center of mass is given by
-The speeds of \(x\) and \(X\) in the center-of-mass system (denoted by prime \(^\prime\)) are
@@ -1537,8 +1542,8 @@Since \(v_y^\prime = v_Y^\prime = 0\) at threshold, we have (using Eq. (11.53))
@@ -1552,8 +1557,8 @@Niels Bohr proposed that nuclear reactions take place through formation and decay of a compound nucleus. This compound nucleus is a composite of the projectile and target nuclei, usually in a high state of excitation.
A compound nucleus \(CN\) is formed by \(x\) and \(X\) in the entrance channel with an excitation energy
-In the \({\rm ^{12}C}(\alpha,\ n){\rm ^{15}O}\) reaction, enough energy is available from just the masses to leave \(\rm ^{16}O^*\) in an excited state with \(7.2\ {\rm MeV}\) when an \(\alpha\) particle and carbon-12 join together. The kinetic energy available in the center-of-mass \(K_{cm}^\prime\) is
-which can excite the compound nucleus to even higher excitation energies than that from just the masses.
@@ -1589,8 +1594,8 @@This explains why \({\rm ^{12}C}(\alpha,\ n){\rm ^{15}O}\) has such a large cross section. The reaction populated a resonance near \(E({\rm ^{16}O^*}) = 18.2\ {\rm MeV}\). The quantum numbers of \({\rm ^{12}C}(\alpha,\ n){\rm ^{15}O}\) exit channel select this energy level in \(\rm ^{16}O^*\) to be populated, where as the quantum numbers for the \({\rm ^{12}C}(\alpha,\ p){\rm ^{15}N}\) exit channel do not.
The uncertainty principal can be used to relate the energy width of a particular nuclear state \(\Gamma\) to its lifetime \(\tau\), or
-If the width of a measured nuclear state is \(\Gamma\) (using an excitation function), then the above equation can be used to determine its lifetime. Ground states for stable nuclei have zero energy width, and hence an infinite lifetime.
@@ -1630,8 +1635,8 @@Fission may also be induced by a nuclear reaction. A neutron abosrbed by a heavy nucleus forms a highly excited compound nucleus that may quickly fission. An example of induced fission is
-The only primary souce in widespread use that is not derived from the Sun is nuclear power. Energy emitted by stars arises from nuclear fusion reations in which light nuclei fuse together. This process contrasts with nuclear fission, in which large nuclei divide.
+When \(\rm ^{236}U^*\) fissions, it divides into nuclei with a larger binding energy per nucleon, thereby releasing energy. If two light nuclei fuse together, they alos form a nucleus with a larger binding energy per nucleon and energy is released.
+The most energy is released if two isotopes of hydrogen (deuteron and triton) fuse toghether in the reaction
+About \(3.5\ {\rm MeV}\) per nucleon is released because of the stron binding of \(\rm ^4He\). Less than \(1\ {\rm MeV}\) per nucleon is released in fission.
+In the first few minutes of the universe, the light elements of hydrogen and helium were formed. Millions of years later, the heavier elements were formed in stars through nuclear fusion. Thre are two main cycles for producing energy in stars.
+The first is the proton-proton chain, which converts 4 protons into an \(\alpha\) particle. As stars form due to gravitational attraction, the heat (average speed of the protons) increases substantially so they can overcome their Coulomb repulsion (with the help of quantum tunneling) and fuse by the following reactions:
+This reaction produces a deuteron \((\rm ^2H)\) and is a special kind of weak-interaction beta decay process. It is extremely slow, because only 1 collision in about \(10^{26}\) produces a reaction. This is good, otherwise the Sun would explode!
+The deuterons that accumulate can combine with a proton \(\rm ^1H\) to produce \(\rm ^{3}He\):
+The \({\rm ^3He}\) can then combine to produce \({\rm ^4He}\) through:
+Note
+The first two reactions which create the deuteron and helium-3 must occur more than once to produce the enough matter for the final reaction that produces helium-4.
+A total of six \({\rm ^1H}\) are requred to produce \({\rm ^4He}\) and two \({\rm ^1H}\). This process consumes four protons. The total \(Q\) for 6 \({\rm ^1H}\) produce \({\rm ^4H}\) is \(24.7\ {\rm MeV}\), where an additional \(2\ {\rm MeV}\) come from teh annihilation of two electron-positiron pairs for a total of \(26.7\ {\rm MeV}\).
+The proton-proton chain is slow because Eq. (11.64) limits the entire process. As the reaction proceeds, the star’s temperature increases and eventually carbon-12 nuclei are formed through by the triple-alpha process (i.e., converting 3 \({\rm ^4He}\) into a \({\rm ^{12}C}\)).
+Another cycle can produce the \({\rm ^4He}\), if enough carbon is produced (or already present) near the stellar core. The series of reactions responsible are called the CNO cycle:
+Note
+Four \({\rm ^1H}\) and one \({\rm ^{12}C}\) are required to produce \({\rm ^4He}\) and \({\rm ^{12}C}\). The {\rm ^{12}C} nucleus serves as a catlyst.
+The proton-proton chain is probably responsible for most of our Sun’s energy production, but the CNO cycle is a much more rapid fusion reaction. It requires higher temperatures (\({\sim}2\times 10^7\ {\rm K}\)) than are present in the Sun, because of the higher Coulomb barrier of \({\rm ^{12}C}\) compared to the \({\rm ^1H}\) for the protons.
+A hydrostatic equilibrium exists in the Sun between the gravitational attraction that contracts a star and a gas presssure that pushes out due to the fusion process. As the lighter nuclides are consumed to produce heavier nuclides, the gravitatioanl attraction succeeds in contracting the star’s mass into a smaller volume and the temperature increaes.
+A higher temperature allows the nulcides with higher \(Z\) to fuse. This process continues until a large part of the Sun’s mass is converted to iron. Then the Sun collapses under its own graviational attraction to become a white dwarf. For more massive stars, they can contract more which allows them to become a neutron or black hole, but this is sensitive to the star’s initial mass.
+Some scientists beleive that controlled nuclear fusion is ultimatlely our best source of terrestrial energy. Among the several possible fusion reactions, three of the simplest involve the three isotopes of hydrogen:
+Deuterium \(({\rm ^2H})\) exists in vast quantities in seawater. Assuming that there are \(10^{21}\ {L}\) of water on Earth, the natural abundance of deuterium \((0.015\%)\) gives \(10^{43}\) deuterons. These deuterons (when fused together to make \(\rm ^{3}He\)) would produce over \(10^{30}\ {\rm J}\) of energy. This is enough to support present world energy consumption for a few billion years.
+Three main conditions are necessary for controlled nuclear fusion:
+The temperature must be hot enough to allow the ions (e.g., \(d\) and \(t\)) to overcome the Coulomb barrier and fuse thier nuclei together. This requires a temperature of \(100-200\) million \(\rm K\).
The ions must be confined to a small volume to allow the ions to fuse. A suitable ion density is \(2-3 \times 10^{20}\ {\rm ions/m^3}\).
The ions must be held together in close proximity at high temperature long enough to avoid plasma cooling, where a suitable time is \(1-2\ {\rm s}\).
The suitable values given above assume magnitic confinement. The product of the plasma density \(n\) and the containment time \(\tau\) must have a minimum value at a sufficiently hgih temperature to initiate fusion and produce as much energy as it consumes. The minimum value is
+The above relation is calle dthe Lawson criterion. A triple product of \(n\tau T\) called the fusion product is also sometimes use:
+The factor \(Q\) is used to represent the ratio fo the power produced in the fusion reation to the power required to produce the fusion (heat). This \(Q\) factor is not to be confused with the \(Q\) value for the release of binding energy.
+The breakeven point is \(Q=1\), and ignition occurs for \(Q\gg 1\). For controlled fusion produced in the laboratory, temperatures equivalent to \(kT = 20\ {\rm keV}\) are satisfactory. For uncontrolled fusion (i.e., a H-bomb), high temperatures and densities are acheived over a very brief time by using a fission bomb. Fusion bombs do not produce radiation effects nearly as severe as those of fission bombs, because the primary products are not dangerously radioactive.
+A controlled thermonuclear reaction of nuclear fusion in the laboratory is one of the primary goals of science and engineering. Scientists do not expect to reach this goal for several more decades. The first fusion reaction will likely be the \(D + T\) reaction. The tritium will be derived from two possible reactions:
+The lithium is required to generate the tritium. It is also used as the heat transfer medium and a neutron radiation shield.
+For \(Q=1\), a product of a few times \(10^{21}\ {\rm s\cdot keV/m^3}\) will be required for a commercial reactor using \(D+T\). There are two major schemes to control thermonuclear reactions:
+Magnetic confinement fusion (MCF), and
Inertial confinement fusion (ICF).
The primary effort of research laboratories around the world for several years has been a device called the tokamak. As many as six separate magnetic fields can be used to contain and heat the plasma.
+In a schematic cross section of a typical magnetic containment vessel,
+The center is the location of the hot plasma where the \(D+T\) reaction takes place.
The plasma is surrounded by vaccum to keep out impurities that would poison the reaction.
A wall surrounds the plasma and vacuum region, which would be subjected to intense radiation. The plasma must be kept from touching the enclosure.
A lithium blanket is the next layer, which absorbs neutrons to breed more tritium.
Next is a radiation shield to prevent radiation from reaching the magnets, which may be superconducting in a commerical reactor.
The heating of the plasma to sufficiently high temperatures begins with the resistive heating from the electric current flowing in the plasma. This is insufficient to attain the high ignition temperature. As a result, there are two other schemes to add additional heat: (1) injection of high-energy \((40-120\ {\rm keV})\) neutral fuel atoms that interact with the plasma, and (2) radio-frequency (RF) induction heating of the plasma (similar to a microwave oven).
+The most significant fusion project in the near future is a large tokamak fusion reactor called the ITER (International Thermonuclear Experimental Reactor). It is currently under construction in France and is expected to generate self-sustained fusion power of \(500\ {\rm MW}\) for \(1000\ {\rm s}\).
+ +The concept of intertial confinement fusion is to use an intense high-powered beam of heavy ions or a laser called a driver to implode a pea-sized target composed of \(D+T\) to a density and temperature high enough to cause fusion ignition.
+Several US-based institutions are doing research and development in laser fusion.
+The National Ignitition Facility (NIF) at Lawrence Livermore National Lab fires 192 lasers in to a high-Z cylinder, which produces x-rays. These x-rays heat the small fuel pellet containing the \(D+T\). It produced \(1.3\ \rm MJ\) of UV laser energy in 2010, just short of the \(1.5\ {\rm MJ}\) needed for ignition.
+Sandia National Lab used a device called a Z-pinch that uses a huge jolt of current to produe a powerful magnetic field that squeezes ions inot implosion and heats the plasma. Fracne is building a device called Laser Megajoule with similar objectives as the NIF.
+