A) \[\lambda '=\frac{\lambda }{2}\]
B) \[\lambda '=2\lambda \]
C) \[\frac{\lambda }{2}<\lambda '<\lambda \]
D) \[\lambda '>\lambda \]
Correct Answer: C
Solution :
\[E=\frac{hc}{\lambda }-{{W}_{0}}\] and \[2E=\frac{hc}{\lambda '}-{{W}_{0}}\] \[\Rightarrow \frac{\lambda '}{\lambda }=\frac{E+{{W}_{0}}}{2E+{{W}_{0}}}\Rightarrow \lambda '=\lambda \left( \frac{1+{{W}_{0}}/E}{2+{{W}_{0}}/E} \right)\] Since \[\frac{(1+{{W}_{0}}/E)}{(2+{{W}_{0}}/E)}>\frac{1}{2}\] so \[\lambda '>\frac{\lambda }{2}\]
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