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

Question #108643
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.
1
Expert's answer
2020-04-09T09:35:04-0400

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".


















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