Dual Nature Of Light Question 271
Question: Light of wavelength $ \lambda $ strikes a photo-sensitive surface and electrons are ejected with kinetic energy E. If the kinetic energy is to be increased to 2E, the wavelength must be changed to $ \lambda ’ $ where [MP PET 1997]
Options:
A) $ \lambda ‘=\frac{\lambda }{2} $
B) $ \lambda ‘=2\lambda $
C) $ \frac{\lambda }{2}<\lambda ‘<\lambda $
D) $ \lambda ‘>\lambda $
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Answer:
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 ( \frac{1+W_{0}/E}{2+W_{0}/E} ) $ Since $ \frac{(1+W_{0}/E)}{(2+W_{0}/E)}>\frac{1}{2} $ so $ \lambda ‘>\frac{\lambda }{2} $
BETA
