NEET-XII-Physics
46: The Nucleus
- #17228Th emits an alpha particle to reduce to 224Ra. Calculate the kinetic energy of the alpha particle emitted in the following decay:
Th228→Ra224*+αRa224*→224Ra+Υ(217 keV).Atomic mass of 228Th is 228.028726 u, that of 224Ra is 224.020196 u and that of
H24is 4.00260 u.Ans : Given:
Atomic mass of 228Th, m(228Th) = 228.028726 u
Atomic mass of 224Ra, m(224Ra) = 224.020196 u
Atomic mass of `` {}_{2}{}^{4}\,\mathrm{\,H\,}``, m(`` {}_{2}{}^{4}\,\mathrm{\,H\,}``) = 4.00260 u
Mass of 224Ra = 224.020196 × 931 + 0.217 MeV = 208563.0195 MeV
Kinetic energy of alpha particle, K = `` \left[m\left({}^{228}\,\mathrm{\,Th\,}\right)-\left[m\left({}^{224}\,\mathrm{\,Ra\,}\right)+m\left({}_{2}{}^{4}\,\mathrm{\,H\,}\right)\right]\right]``c2
= (228.028726 × 931) - [(208563.0195 + 4.00260 × 931]
= 5.30383 MeV = 5.304 MeV
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