Answer to Question #108643 in Atomic and Nuclear Physics for Arjun

Question

238
92U decays spontaneously by α emission to 234
90Th. Calculate the total energy released by
this process and the recoil velocity of the 234
90Th nucleus. The atomic masses are 238.050783
u for 238
92U and 234.043595 u for 234
90Th.

Answer 1

The decay energy $Q$ is given by conservation of mass-energy as

$\frac{Q}{c^2}=M_U-M_{Th}-M_\alpha$.

We get

$\frac{Q}{c^2}=238.050783-234.043595-4.002603=0.004585u$

If multiplied by the conversion factor $931.5$ $MeV/c^2$ , then $Q\approx4.27$ $MeV$.

So, the total energy released by this process $Q\approx4.27$ $MeV$.

If the parent nucleus is at rest when it decays, the daughter nucleus and the  particle must have equal and opposite momenta. If $p$ is the magnitude of the momentum of either particle, the decay energy is

$Q=\frac{p^2}{2M_D}+\frac{p^2}{2M_\alpha}=\frac{p^2}{2M_D}(1+\frac{M_D}{M_\alpha})=E_D(1+\frac{M_D}{M_\alpha})$.

$E_D=\frac{Q}{1+\frac{M_D}{M_\alpha}}=\frac{Q}{1+\frac{A-4}{4}}=Q\frac{4}{A}=4.27\frac{4}{238}\approx0.072$ $MeV$.

$E_D=\frac{M_Dv^2}{2}\to v=\sqrt{\frac{2E_D}{M_D}}=\sqrt{\frac{2\cdot 0.072\cdot 1.6\cdot10^{-19}}{234\cdot 1.66\cdot10^{-27}}}\approx243200$ $m/s$.