A) \[\frac{1}{4}\]
B) \[\frac{1}{2}\]
C) 2
D) 4
Correct Answer: A
Solution :
\[{{K}_{e}}=\frac{1}{2}m{{v}^{2}}\] and \[\lambda =\frac{h}{mv}\] \[{{K}_{e}}=\frac{1}{2}\left( \frac{h}{\lambda v} \right)\,.\,{{v}^{2}}=\frac{vh}{2\lambda }\] \[{{K}_{p}}=\frac{hc}{\lambda }\] \[\therefore \] \[\frac{{{K}_{e}}}{{{K}_{p}}}=\frac{v}{2\,c}\] \[=\frac{1.5\times {{10}^{8}}}{2\times 3\times {{10}^{8}}}=\frac{1}{4}\]
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