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