A) \[v\sqrt{\frac{3}{4}}\]
B) \[v\sqrt{\frac{4}{3}}\]
C) less than \[v\sqrt{\frac{4}{3}}\]
D) greater than \[v\sqrt{\frac{4}{3}}\]
Correct Answer: D
Solution :
\[\frac{1}{2}m{{v}^{2}}=\frac{hc}{\lambda }-\phi \] \[\frac{1}{2}mv{{'}^{2}}=\frac{hc}{(3\lambda /4}-\phi =\frac{4hc}{3\pi }-\phi \] Clearly, \[v'>\sqrt{\frac{4}{3}}v\]
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