A) 4
B) \[\frac{1}{4}\]
C) 2
D) \[\frac{1}{2}\]
Correct Answer: D
Solution :
\[\frac{hc}{\lambda }={{W}_{0}}+\frac{1}{2}mv_{\max }^{2}\] Assuming \[{{W}_{0}}\] to be negligible in comparison to \[\frac{hv}{\lambda }\] i.e. \[v_{\max }^{2}\propto \frac{1}{\lambda }\Rightarrow {{v}_{\max }}\propto \frac{1}{\sqrt{\lambda }}\]. (On increasing wavelength l to 4l, vmax becomes half).
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