Atoms And Nuclei Question 405
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:
-
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.
