The Nucleus Resolved, Ch Decay rates from the golden rule: kinetics closed and alpha decay finished
Problem
Curium-244 decays entirely by α emission to the ground state of plutonium-240, with Qα=5.902MeV. Model the decay as ℓ=0 tunnelling through a point-charge Coulomb barrier (Zd=94, Ad=240), with the nuclear radius given by the touching-sphere prescription R=1.2(Ad1/3+41/3)fm. Take the reduced mass of the α-daughter pair as μredc2=Ad+44Ad×931.494MeV (mass numbers standing in for the true atomic masses, as in the barrier estimate carried over from the first course) and the α particle's rest energy as mαc2=3727.4MeV. Use ℏc=197.327MeVfm and e2/4πϵ0=1.43996MeVfm. The measured half-life is T1/2meas=18.1y (1y=3.156×107s).
((a)) With κ(r)=2μred(V(r)−E)/ℏ and V(r)=ZdZαe2/4πϵ0r=bQα/r, where b=ZdZαe2/(4πϵ0Qα) is the outer turning point, show that 2G=2∫Rbκdr=ℏc22μredc2Qαb[arccosR/b−(R/b)(1−R/b)]. Evaluate b, 2G and the penetrability P=e−2G.
((b)) The α particle carries kinetic energy Tα=QαAd/(Ad+4) and (non-relativistic) speed v from Tα=21mαv2. Using the assault frequency νknock=v/2R, write the golden-rule decay constant λ=νknockP, find the calculated half-life, and compare it with T1/2meas.
((c)) In λ=(2π/ℏ)∣Mfi∣2ρ(Ef), state which of νknock and P 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=b the barrier is locally linear over the Airy length ℓ=(ℏ2/2μredF)1/3 with F=∣dV/dr∣r=b=Qα/b. Evaluate ℓ, compare it with the width b−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 T1/2 be trusted.