A) \[\frac{{{h}^{2}}{{v}^{2}}}{2M{{c}^{2}}}\]
B) Zero
C) hv
D) \[{{\operatorname{Mc}}^{2}}-hv\]
Correct Answer: A
Solution :
\[\operatorname{Momentum} of photon=\frac{hv}{e}\] From the law of conversation of linear momentum \[{{\operatorname{P}}_{nuclear}} ={{P}_{photon}}\] \[\operatorname{Mv}=\frac{hv}{e}\] \[v=\frac{hv}{Me}\] \[\operatorname{K}.E of nucleus=\frac{1}{2}M{{\left[ \frac{hv}{Mc} \right]}^{2}}=\frac{{{h}^{2}}{{v}^{2}}}{2M{{c}^{2}}}\]
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