Curium-244 decays entirely by emission to the ground state of plutonium-240

PHYSICS · Multi-part · Fermi University Physics · Quantum Mechanics · Approximation methods · Approximation methods

The Nucleus Resolved, Ch Decay rates from the golden rule: kinetics closed and alpha decay finished

Problem

Curium-244 decays entirely by α\alpha emission to the ground state of plutonium-240, with Qα=5.902 MeVQ_\alpha = 5.902\,\mathrm{MeV}. Model the decay as ℓ=0\ell = 0 tunnelling through a point-charge Coulomb barrier (Zd=94Z_d = 94, Ad=240A_d = 240), with the nuclear radius given by the touching-sphere prescription R=1.2(Ad1/3+41/3) fmR = 1.2(A_d^{1/3} + 4^{1/3})\,\mathrm{fm}. Take the reduced mass of the α\alpha-daughter pair as μredc2=4AdAd+4×931.494 MeV\mu_{\rm red}c^2 = \dfrac{4A_d}{A_d+4}\times 931.494\,\mathrm{MeV} (mass numbers standing in for the true atomic masses, as in the barrier estimate carried over from the first course) and the α\alpha particle's rest energy as mαc2=3727.4 MeVm_\alpha c^2 = 3727.4\,\mathrm{MeV}. Use ℏc=197.327 MeV fm\hbar c = 197.327\,\mathrm{MeV\,fm} and e2/4πϵ0=1.43996 MeV fme^2/4\pi\epsilon_0 = 1.43996\,\mathrm{MeV\,fm}. The measured half-life is T1/2meas=18.1 yT_{1/2}^{\rm meas} = 18.1\,\mathrm{y} (1 y=3.156×107 s1\,\mathrm{y} = 3.156\times10^7\,\mathrm{s}). ((a)) With κ(r)=2μred(V(r)−E)/ℏ\kappa(r) = \sqrt{2\mu_{\rm red}(V(r)-E)}/\hbar and V(r)=ZdZαe2/4πϵ0r=bQα/rV(r) = Z_dZ_\alpha e^2/4\pi\epsilon_0 r = bQ_\alpha/r, where b=ZdZαe2/(4πϵ0Qα)b = Z_dZ_\alpha e^2/(4\pi\epsilon_0 Q_\alpha) is the outer turning point, show that 2G=2∫Rbκ dr=22μredc2Qαℏc b[arccos⁡R/b−(R/b)(1−R/b)].\begin{aligned}2G &= 2\int_R^b \kappa\,\mathrm{d}r\\ &= \frac{2\sqrt{2\mu_{\rm red}c^2Q_\alpha}}{\hbar c}\,b\Big[\arccos\sqrt{R/b} - \sqrt{(R/b)(1-R/b)}\Big].\end{aligned} Evaluate bb, 2G2G and the penetrability P=e−2GP = e^{-2G}. ((b)) The α\alpha particle carries kinetic energy Tα=QαAd/(Ad+4)T_\alpha = Q_\alpha A_d/(A_d+4) and (non-relativistic) speed vv from Tα=12mαv2T_\alpha = \tfrac12 m_\alpha v^2. Using the assault frequency νknock=v/2R\nu_{\rm knock} = v/2R, write the golden-rule decay constant λ=νknockP\lambda = \nu_{\rm knock}P, find the calculated half-life, and compare it with T1/2measT_{1/2}^{\rm meas}. ((c)) In λ=(2π/ℏ)∣Mfi∣2ρ(Ef)\lambda = (2\pi/\hbar)|M_{fi}|^2\rho(E_f), state which of νknock\nu_{\rm knock} and PP is standing in for the squared matrix element and which for the density of final states, and say why the identification is only approximate. ((d)) At r=br=b the barrier is locally linear over the Airy length ℓ=(ℏ2/2μredF)1/3\ell = (\hbar^2/2\mu_{\rm red}F)^{1/3} with F=∣dV/dr∣r=b=Qα/bF = |\mathrm{d}V/\mathrm{d}r|_{r=b} = Q_\alpha/b. Evaluate ℓ\ell, compare it with the width b−Rb-R of the forbidden region, and use the size of this ratio to say, order of magnitude, how tightly the WKB approximation itself lets the calculated T1/2T_{1/2} be trusted.

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