Past exam of the nuclear and particle physics course of the University of Oxford / 2022 / 3

Let $M_P$ and $M_D$ denote neutral atomic rest masses for the appropriate parent and daughter, and initially neglect electronic binding energy corrections and daughter excitation. The Q values areFor beta-minus decay, the daughter nucleus has one extra positive charge. Its neutral atom contains one more electron than the neutral parent, cancelling the emitted electron's rest mass in the atomic accounting. For beta-plus decay, the neutral parent contains one more electron than the neutral daughter; together with the emitted Positron, this gives the $2m_e$ subtraction. Thus beta-plus decay requires an atomic mass difference exceeding $2m_e{c}^2$, whereas electron capture need not.

For electron capture, the captured bound electron is already included in $M_P$. Capture from a shell leaving an atomic hole of excitation energy $E_{\mathrm{hole}}$, and a nuclear excited state of energy $E_x$, gives approximatelyThe hole's energy later appears as atomic radiation or Auger electrons. Subtract $E_x$ for either beta decay as well. Exact endpoints also depend on daughter ionization and small atomic binding energy changes; the boxed expressions are the conventional leading atomic-mass formulas.

For a parent at rest, momentum conservation gives equal and opposite daughter and alpha particle momenta. In the nonrelativistic recoil approximation,Approximating nuclear masses by their mass numbers, $M_D\simeq (A-4)m_u$ and $m_{\alpha }\simeq 4m_u$, givesThe printed phrase “atomic number $A$” must here mean mass number. Atomic number alone does not determine the recoil fraction. For example, ${}^{210}\mathrm{Po}$ and ${}^{212}\mathrm{Po}$ have the same atomic number $84$, but the approximate fractions are $206/210$ and $208/212$, respectively.

For completeness, exact relativistic two-body decay kinematics giveswhere $M_P=M_D+m_{\alpha }+Q/c^2$. Expanding for $Q\ll M_P{c}^2$ recovers the recoil-sharing expression.

Model alpha decay as a preformed alpha particle making repeated escape attempts from a short-range nuclear potential well, then undergoing quantum tunnelling through the external Coulomb potential. Write the daughter atomic number as $Z_D$, the nuclear radius as $R$, and the relative reduced mass as $\mu$. For zero orbital angular momentum, the WKB approximation givesSet $v=\sqrt{2Q/\mu }$ and $r=r_2{\sin }^2\theta$. ThenWhen $R/r_2\ll 1$, the leading term isHere $P_{\alpha }$ is the alpha preformation probability and ${\nu }_{\mathrm{attempt}}$ the attempt frequency. The stated proportionality retains only the dominant barrier exponent. It assumes a thick classically forbidden barrier, nonrelativistic relative motion, a spherical unscreened external Coulomb potential, and slowly varying prefactors. Finite $R$, changes in alpha preformation probability, and a centrifugal barrier for nonzero orbital angular momentum produce significant corrections. The precise charge in the barrier is $Z_D=Z_P-2$; using the parent charge is an approximation for large $Z_P$. The measured alpha particle speed approximates $v$ only when daughter recoil is negligible.

For a closed unit-branching radioactive decay chain, the rate equations areThe first gives $N_a=N_{a0}{e}^{-{\lambda }_{a}t}$. Multiply the second by its integrating factor $e^{{\lambda }_{b}t}$:Integrate from zero to $t$. For ${\lambda }_a\ne {\lambda }_b$,soThe coefficients can individually be negative; the physical population is nonnegative because the production integral is nonnegative. For equal nuclear decay constants ${\lambda }_a={\lambda }_b=\lambda$, the finite limit isA sum of two constant-coefficient copies of the same exponential function cannot represent this limit; the $t$ factor is essential.

For $N_{b0}=0$, differentiate the distinct-nuclear decay constant result:Equating the production and loss terms gives $e^{({\lambda }_b-{\lambda }_a)t}={\lambda }_b/{\lambda }_a$, henceFor positive nuclear decay constants and $N_{a0}>0$, this is positive regardless of their ordering. Initially ${\dot{N}}_b>0$, while $N_b\to 0$ at late times; the stationary point is the unique maximum. At that time ${\lambda }_a{N}_a={\lambda }_b{N}_b$. If the nuclear decay constants are equal, differentiating $\lambda N_{a0}t{e}^{-\lambda t}$ gives $\boxed{t_{\max }=1/\lambda }$, also the continuous limit of the expression above.

A stable terminal product obeys ${\dot{N}}_c={\lambda }_b{N}_b$. Conservation of the number of chain members gives $N_a+N_b+N_c=N_{a0}+N_{c0}$ when $N_{b0}=0$. Therefore, for distinct nuclear decay constants,Set $N_{c0}=0$ for an initially absent terminal product. For equal nuclear decay constants,Both expressions have ${\dot{N}}_c(0)=0$ because the intermediate population initially vanishes; their leading growth is $\frac{1}{2}{N}_{a0}{\lambda }_a{\lambda }_b{t}^2$. For positive nuclear decay constants, $N_c$ increases monotonically to $N_{c0}+N_{a0}$.