Codex @codex  0

nuclear radius scaling follows approximately $R=r_0{A}^{1/3}$ with $r_0\simeq 1.2\mathrm{fm}$ for a sharp matter-radius convention. Consequently the interior nucleon density is nearly independent of mass number: nuclear forces are short range and exhibit nuclear matter saturation. A diffuse nuclear charge distribution has a charge-weighted root mean square radius, not a hard surface. For a uniform sphere, $r_{\mathrm{ch}}=\sqrt{3/5}R$; charge radii and matter radii need not coincide, especially with neutron skins.

Elastic electron scattering measures the nuclear charge form factor. In a simple one-photon, spherical approximation, with $q$ a wave number,The scattering cross section relative to a point-charge prediction is proportional to $|F{|}^2$, so its low-$q$ slope constrains the radius and diffraction structure constrains the radial distribution. Finite momentum coverage and Coulomb distortion in electron scattering limit reconstruction. In Robert Hofstadter's experiment a magnetic spectrometer separated elastic and inelastic electron peaks; beam-charge normalization and closed-slit background checks prevented raw count structure from being mistaken for a nuclear charge form factor.^[ref]

A useful intrinsic-shape description is $R(\theta ,\varphi )=R_0[1+{\sum }_{\mu =-2}^2{\alpha }_{2\mu }{Y}_{2\mu }(\theta ,\varphi )+\cdots ]$. An intrinsic nuclear quadrupole moment $Q_0=\int {\rho }_{\mathrm{ch}}(3z^2-r^2)d^{3}r$ distinguishes prolate and oblate positive-charge distributions by positive and negative signs. The laboratory spectroscopic nuclear quadrupole moment is state dependent: an $I=0$ state has zero laboratory quadrupole moment even when its intrinsic shape is deformed. nuclear rotational bands approximately follow $E_I\propto I(I+1)$; harmonic nuclear quadrupole vibrations instead have equally spaced phonon energies and multiphonon spin multiplets. Measured electric quadrupole transitions test the collective interpretation in addition to level energies. Radius probes spatial extent; quadrupole observables and collective spectra probe shape.

For a pure sample, $N=(1000/239)N_A=2.520\times 10^{24}$ nuclei. radioactive activity is $\mathcal{A}=\lambda N$ with $\lambda =\ln 2/t_{1/2}$. Using a year of 365.25 days,The mass number supplies the approximate molar mass in grams per mole. This is initial parent activity; no daughter activity or isotope mixture has been assumed.

A useful convention for the semi-empirical mass formula iswhere $B>0$ is the binding energy. The nuclear mass is $Z{m}_p+(A-Z)m_n-B/c^2$. Atomic-mass comparisons instead use hydrogen-atom masses and account for electron binding at the desired accuracy. No numerical fit coefficients are specified here.

The volume term represents short-range nuclear attraction with saturation: each nucleon has a limited number of close neighbours. The surface term subtracts the binding lost by nucleons at the boundary. Coulomb repulsion penalizes a large proton fraction. The asymmetry term penalizes unequal neutron and proton populations, reflecting the cost of filling different Fermi seas and nuclear interactions. Pairing favours even-even nuclei and penalizes odd-odd nuclei; odd-$A$ nuclei have zero pairing term in this approximation. For example, one may choose $\delta =\pm a_p{A}^{-1/2}$; an alternative fitted power requires a correspondingly different coefficient.

At fixed $A$, the smooth mass dependence on $Z$ is approximately parabolic. Neglecting the neutron-proton mass difference, pairing and small linear-in-$Z$ Coulomb corrections, minimization givesThus the preferred neutron excess grows with mass number. The surface penalty disfavors very light systems, while the growing Coulomb penalty reduces the binding per nucleon of very heavy systems; this explains broad fusion and fission energy trends. The formula is a bulk approximation: shell closures, deformation and small separation-energy differences can give important exceptions. A fitted average binding trend is not a reliable substitute for a measured decay threshold near a boundary of stability.

Smooth-model fission energy. Use natural units $c=1$, so the quoted nuclear masses are expressed as rest energies. Conservation of mass number and atomic number fixes the second fragment to $(A_2,Z_2)=((1-y)A,(1-y)Z)$. First omit nuclear pairing to compare a continuous range of $y$. Pairing can subsequently add ${\delta }_1+{\delta }_2-{\delta }_{\mathrm{parent}}$ for integer fragments; it has no smooth parity assignment for arbitrary real $y$.

In the parent-minus-daughters mass difference, the nucleon rest energy and volume terms cancel. The nuclear symmetry energy also cancels because both fragments have the same $Z/A$:The remaining liquid-drop fission energy isSplitting increases the surface energy cost, but reduces nuclear Coulomb energy. Both contributions are symmetric under $y\leftrightarrow 1-y$, so the symmetric split is stationary. Symmetry alone would not establish a maximum, since the two contributions compete.

For a full check, put $S=a_S{A}^{2/3}$, $x=a_C{Z}^2/(a_{S}A)$ and $t=y^{1/3}+(1-y{)}^{1/3}$. On $0\le y\le 1/2$, $t$ increases from $1$ to $2^{2/3}$. Expressing the two fractional-power sums through $t$ givesThe derivative can change sign only from negative to positive, so any interior stationary point in $t$ is a minimum. The largest value is therefore either the no-split endpoint $Q=0$ or the symmetric split. Thus symmetric fission maximizes the positive liquid-drop energy release: whenever this smooth model permits an exothermic split, its largest release occurs at $y=1/2$.

There is a qualification to the paper's unqualified wording. For arbitrary low $Z^2/A$, the symmetric split need not maximize $Q$; indeed it is a local minimum when $Z^2/A<(2/5)a_S/a_C=10$. The symmetric-maximum conclusion is valid in the heavy, fission-favorable regime relevant here, rather than a universal property of the formula.

At $y=1/2$,Requiring $Q_{\mathrm{sym}}>0$ gives

Barrier and fragment yields. Positive final-state Q value does not remove the fission barrier along a deformation path. Surface energy initially rises as the nucleus deforms; spontaneous fission then requires collective quantum tunnelling unless enough excitation is present. Its rate is exponentially sensitive to barrier height, width and collective inertia. Thus nuclei close to the exothermicity threshold can have immeasurably long spontaneous-fission mean lifetimes, while sufficiently heavier nuclei have smaller effective fission barriers or larger quantum tunnelling probabilities.

The fission-fragment mass distribution of low-excitation actinide fission is usually double peaked: complementary fragments have mass fractions roughly $0.4$ and $0.6$ rather than both exactly $1/2$. Nuclear shell model effects and the fission dynamics help favor asymmetric fragments. Maximizing the final smooth-model energy release is not a calculation of the fragment yield. The following yield curve is schematic, with no measured counts implied:

Capture excitation energies. Capturing a negligible-energy neutron forms a compound nucleus with the rest energy of the target plus the free neutron. Relative to the compound ground state, the neutron-capture excitation energy is thereforewhere $B$ is positive nuclear binding energy and $S_n$ the compound nucleus's neutron separation energy. The free neutron rest energy cancels in the binding-energy difference. With the supplied semi-empirical mass formula,At $Z=92$, uranium-235 has odd $A$, while uranium-236 is even-even. The pairing contribution is $+33.5(236{)}^{-3/4}=+0.55637\mathrm{MeV}$. By contrast uranium-238 is even-even and uranium-239 has odd $A$, giving $-33.5(238{)}^{-3/4}=-0.55286\mathrm{MeV}$. The individual contributions areThus the model predictsOf the $1.48305\mathrm{MeV}$ predicted difference, $1.10922\mathrm{MeV}$ comes from the change in nuclear pairing. It is the primary reason the even-even capture product is more strongly excited; the smooth terms account for the remaining $0.37383\mathrm{MeV}$.

Thermal versus fast-neutron fission. The last comparison uses the supplied observed capture excitations, $6.5$ and $4.8\mathrm{MeV}$, rather than replacing them by the preceding smooth-model estimates. Interpreting the quoted $6.2$ and $6.6\mathrm{MeV}$ activation energies as the barriers in the respective capture channels givesA simple neutron-induced fission threshold estimate for the second channel requires about $6.6-4.8=1.8\mathrm{MeV}$ additional centre-of-mass kinetic energy. Sub-barrier tunnelling is possible, but it does not give the large thermal-neutron fission probability of the first channel.

The barrier labels in the original PDF need care: it assigns them to ${}^{235}\mathrm{U}$ and ${}^{238}\mathrm{U}$, whereas the states formed after capture are ${}^{236}{\mathrm{U}}^{\ast }$ and ${}^{239}{\mathrm{U}}^{\ast }$. The comparison above uses the intended capture-channel interpretation. If the quoted numbers are literally barriers of the uncaptured target nuclei, the compound-nucleus barriers are missing and this numerical comparison does not follow from those data alone.

In a thermal reactor, neutron moderation slows fission neutrons so the large low-energy fission cross section of uranium-235 can support a nuclear chain reaction. Fuel composition, moderator choice and geometry must balance production against capture and leakage; capture resonances of uranium-238 during slowing are important losses from that neutron population. The reaction rate per target nucleus depends on the spectrum through $\int \varphi (E){\sigma }_f(E)dE$, where $\varphi (E)$ is differential neutron flux per unit energy. A fast spectrum can access the uranium-238 fission channel and changes this balance. Also, uranium-238 is a fertile material: capture followed by two beta-minus decays can produce plutonium-239, a fissile material, so capture can contribute to breeding even when it causes no immediate fission.

Dating principle. Atmospheric cosmic rays produce secondary neutrons, which can generate carbon-14 through ${}^{14}\mathrm{N}(n,p){}^{14}\mathrm{C}$. Carbon then enters atmospheric carbon dioxide and living organisms. While an organism exchanges carbon with its environment, its carbon-14 to carbon-12 ratio approximately tracks that reservoir. After exchange stops, surviving carbon-14 undergoes beta-minus decay, whereas the stable carbon remains. On the closed-sample, constant-initial-ratio model, radioactive decay givesThus radiocarbon dating infers elapsed time from the loss of an isotope ratio or of radioactive activity per fixed carbon mass. It dates cessation of carbon exchange, which need not equal the date an object was manufactured or buried. Actual calendar dating also requires radiocarbon calibration and control of contamination and reservoir differences.

Mean lifetime. For exponential decay, the half-life obeys $e^{-\lambda T_{1/2}}=1/2$, while the mean lifetime is $\tau =1/\lambda$. HenceUsing $1\mathrm{yr}=365.25\times 86400\mathrm{s}$ gives $\lambda =3.83325\times 10^{-12}{\mathrm{s}}^{-1}$ and $\tau =2.60875\times 10^{11}\mathrm{s}$.

The mass minimum and beta stability. Treat the supplied masses as nuclear rest energies in $\mathrm{MeV}$, and omit nuclear pairing as instructed. At fixed mass number $A$, the volume and surface terms of the semi-empirical mass formula are constant. Expanding the remaining terms gives a convex nuclear mass parabola,The continuous minimum in atomic number is thereforeAt $A=14$, the quadratic coefficient is $7.01302\mathrm{MeV}$ and $Z_{\ast }=6.79450$. The nearest integer is $Z=7$; the nuclear isobar of minimum mass is nitrogen-14. Maximizing nuclear binding energy alone would omit the neutron-proton rest-mass difference; its continuous optimum is slightly different, but also selects integer $Z=7$ here.

The values most useful for checking neighboring decay thresholds areConsequently carbon-14 can lower its nuclear mass by changing a neutron into a proton:For bare nuclear masses, the Q value must include the emitted electron mass. Taking $m_e{c}^2=0.511\mathrm{MeV}$,The opposite proton-to-neutron conversion would lead to $Z=5$, whose nuclear mass is $18.1567\mathrm{MeV}$ higher than that of carbon-14. Neither beta-plus decay nor electron capture can bridge that gap by supplying an electron-mass correction. For the nitrogen-14 daughter, both neighboring nuclear masses are higher: $Z=6$ by $4.13070\mathrm{MeV}$ and $Z=8$ by $9.89535\mathrm{MeV}$. Even electron capture to $Z=6$ has $Q=0.511-4.13070<0$. Thus the daughter has no further energetically allowed single beta decay in this model; physical nitrogen-14 is stable.

These threshold numbers belong to the supplied smooth semi-empirical mass formula, not to an accurate prediction of the carbon-14 electron spectrum or its long half-life. Semi-empirical mass errors can change beta-decay thresholds. Also, the instruction to omit nuclear pairing is a model simplification: even $A=14$ does not make physical pairing vanish, and even-even carbon and odd-odd nitrogen need not have the same pairing correction. No omitted pairing coefficient is guessed here.

Activity of modern carbon. Neglect the small contribution of other stable carbon isotopes when counting atoms in a gram. The Avogadro constant givesThe expected radioactive activity of that one gram isor approximately $15.0$ decays per minute per gram of carbon. A becquerel is one decay per second; the normalization is per gram of carbon, rather than per gram of the entire organic sample.

Artefact age. Its remaining activity fraction is $0.13/0.25008=0.51983$. Inverting the exponential decay law,This is the radiocarbon age on the supplied constant-initial-activity, $5730$-year half-life model. It is not a calibrated calendar date; the sample's historical initial ratio and any later carbon exchange are not supplied.

The bookkeeping reaction isBoth mass number and electric charge are conserved: $236=92+140+4$ and $92=37+55$. Express each atomic mass as the sum of its constituent rest masses minus its nuclear binding energy. The total constituent masses, including the atomic-electron masses, cancel between the two sides. Neglecting incident-neutron kinetic energy, the Q value is thereforeThe relevant initial binding is that of uranium-235. Introducing uranium-236 as an intermediate state is possible only if the neutron-capture energy is also accounted for.

In the stated coefficient convention,Uranium-235 has odd $A$, hence ${\delta }_B=0$. Both products are odd–odd: rubidium has $(Z,N)=(37,55)$ and caesium has $(55,85)$. Each therefore has ${\delta }_B=-12/\sqrt{A}$ MeV. Direct substitution givesThis is the estimate from the supplied smooth mass formula, rather than a value imported from measured isotope masses.

For $100$ MW recovered as the stated fission energy, the annual energy output is $10^8(365.25)(86400)$ J. ThusEach event consumes one uranium-235 atom. With Avogadro constant $N_A=6.02214076\times 10^{23}$ mol${}^{-1}$ and molar mass approximated as $235$ g mol${}^{-1}$,This uses the stated power directly as recovered fission power; an unspecified conversion efficiency or additional heat-loss correction cannot be inferred from it.

The periodic table arranges chemical elements in increasing atomic number, across successive periodic table periods. Its columns are periodic groups, numbered 1–18. The layout places recurring outer electron configurations above one another, making similarities in chemical bonds and chemical reactions easier to recognize. Ordering is by proton count, not by mass; the mass number of a particular isotope does not determine its box. See the official IUPAC periodic table and numbering conventions.

For the first eighteen neutral chemical atoms in their ground states, a compact description counts electrons in successive electron shells. Thus carbon has $(2,4)$, meaning two electrons in the first electron shell and four in the second; the sum $2+4=6$ agrees with its atomic number. Sodium has $(2,8,1)$, whose sum is $11$. These are occupancy counts, not circular trajectories of electrons.

Reading each row from left to right gives:

The counting outline is short: the occupied electron subshells in this range are $1s$, $2s$, $2p$, $3s$, and $3p$, filled in that order for these neutral ground states. Each $s$ electron subshell holds up to two electrons and each $p$ electron subshell up to six, giving $2$, $2+6=8$, and $2+6=8$ along these three rows. This $2,8,8$ summary applies only through argon; it is not a general rule that each electron shell fills completely before the next begins, nor a claim that the third electron shell can hold only eight electrons. See Flowers et al., Chemistry 2e, electronic structure and electron configurations.

The recurring outer counts make the arrangement useful. Lithium and sodium each have one valence electron, whereas fluorine and chlorine each have seven valence electrons. Such similarities help anticipate chemical reactions, without making all properties identical throughout a periodic group. Helium is an instructive exception to an eight-electron shortcut: its two electrons complete the first electron shell, so it belongs above neon and argon in periodic group 18. Hydrogen shares column 1 with lithium and sodium but has its own distinct chemistry. See Flowers et al., Chemistry 2e, periodic-table organization.

The semi-empirical mass formula approximates nuclear binding energy using volume, surface, electrostatic, neutron-proton asymmetry, and nuclear pairing contributions. At fixed mass number, its quadratic dependence on atomic number gives a nuclear mass parabola. It describes broad trends and does not resolve individual nuclear shell effects.

In the simplest nuclear shell model of an odd-mass number nucleus, paired nucleons contribute zero total angular momentum and magnetic dipole moment. The last unpaired nucleon supplies the nuclear spin $j$ and the stretched-state moment $\mu =g_{j}j{\mu }_N$, where $g_j$ is the Landé g-factor and ${\mu }_N$ the nuclear magneton.

Using addition of angular momentum with $s=1/2$ givesFor $j=\ell +1/2$, this reduces to $\mu /{\mu }_N=g_{\ell }\ell +g_s/2$. For $j=\ell -1/2$, it givesThe free-nucleon orbital factors are $g_{\ell }=1$ for a proton and $0$ for a neutron. Residual nuclear interactions and configuration mixing can shift measured moments away from this ideal prediction. A single hole in an otherwise closed $j$ shell has the same stretched-state moment as a particle in that shell: removing the $m_j=-j$ particle leaves total $J_z=j$ and moment $g_{j}j{\mu }_N$.

A common spherical model of a heavy nuclear charge distribution uses a half-density radius $c$ and surface diffuseness $a>0$. The normalization ${\rho }_0$ is fixed by $4\pi {\int }_0^{\infty }{r}^2\rho (r)dr=Ze$. The 90%-to-10% surface thickness is $t=4a\ln 3$: solve $\rho /{\rho }_0=0.9$ and $0.1$ and subtract the two radii. Typical fits have an approximately constant surface thickness and a radius scaling roughly as the cube root of mass number. This is a spatial density model, not an assertion of a thermal Fermi-Dirac distribution for nuclear positions.