Nuclear Physics And Radioactivity Question 475
Question: A nucleus at rest undergoes a decay emitting an $ \alpha - $ particle of de-Broglie wavelength $ \lambda =5.76\times {{10}^{-15}}m. $ If the mass of the daughter nucleus is $ 223.610 $ amu and that of the $ \alpha - $ particle is $ 4.002,amu, $ determine the mass of the parent nucleus inamu. $ (1,amu=931.470MeV/c^{2}]. $
Options:
A) $ 227.62amu $
B) $ 112.11amu $
C) $ 90.3amu $
D) $ 23.8amu $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] By conservation of momentum, we have $ 0={{\vec{P}}{\alpha }}+{{\vec{P}}{d}}\Rightarrow {{\vec{P}}{\alpha }}=-{{\vec{P}}{d}} $ or $ {P_{\alpha }}=P_{d}=P $ The kinetic energy released in the process $ K={K_{\alpha }}+K_{p}=\frac{P^{2}}{2{m_{\alpha }}}+\frac{P^{2}}{2m_{d}} $ $ =\frac{P^{2}}{2{m_{\alpha }}}( 1+\frac{{m_{\alpha }}}{m_{d}} )=\frac{{{(h/\lambda )}^{2}}}{2{m_{\alpha }}}( 1+\frac{{m_{\alpha }}}{m_{d}} ) $ After substituting the given values, we get $ K=6.25MeV $ If $ m_{P} $ is the mass of the parent nucleus, then $ K+({m_{\alpha }}+m_{d})c^{2}=m_{p}c^{2} $ or $ 6.25+(223.61+4.002)c^{2}=m_{p}c^{2} $ After simplifying, we get $ m_{p}=227.62 $ amu.
BETA
