A) 4
B) \[\frac{1}{4}\]
C) 2
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
According to Einsteins equation \[\frac{hc}{\lambda }={{W}_{0}}+\frac{1}{2}mv_{\max }^{2}\] Assume \[{{W}_{0}}\] to be negligible in comparison to\[\frac{hc}{\lambda }\]. \[ie,\] \[v_{\max }^{2}\propto \frac{1}{\lambda }\Rightarrow {{v}_{\max }}\propto \frac{1}{\sqrt{\lambda }}\] On increasing wavelength \[\lambda \] to\[4\lambda ,\,\,{{v}_{\max }}\] becomes half.
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