A) \[\frac{1}{8}\]
B) \[\frac{3}{8}\]
C) \[\frac{5}{8}\]
D) \[\frac{7}{8}\]
Correct Answer: B
Solution :
\[{{K}_{particle}}=\frac{1}{2}m{{v}^{2}},\] also \[\lambda =\frac{h}{mv}\] \[\Rightarrow \] \[{{K}_{particle}}=\frac{1}{2}\,\left( \frac{h}{v\,\lambda } \right)\cdot \,{{v}^{2}}=\frac{vh}{2\lambda }\] \[{{K}_{photon}}=\frac{hc}{\lambda }\] \[\therefore \] \[\frac{{{K}_{particle}}}{{{K}_{photon}}}=\frac{v}{2c}=\frac{2.25\times {{10}^{8}}}{2\times 3\times {{10}^{8}}}=\frac{3}{8}\]
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