NEET Solved Paper 2016 Question 29
Question: When a metallic surface is illuminated with radiation of wavelength $ \lambda , $ the stopping potential is V. If the same surface is illuminated with radiation of wavelength $ 2\lambda , $ the stopping potential is $ \frac{V}{4}. $ The threshold wavelength for the metallic surface is :-
Options:
A) $ 4\lambda $
B) $ 5\lambda $
C) $ \frac{5}{2}\lambda $
D) $ 3\lambda $
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Answer:
Correct Answer: D
Solution:
$ eV=\frac{hc}{\lambda }-\frac{hc}{{\lambda_0}} $ ..(i)
$ eV/4=\frac{hc}{2\lambda }-\frac{hc}{{\lambda_0}} $ ..(ii)
From equation (i) and (ii)
$ \Rightarrow $ $ 4=\frac{\frac{1}{\lambda }-\frac{1}{{\lambda_0}}}{\frac{1}{2\lambda }-\frac{1}{{\lambda_0}}} $ On solving $ {\lambda_0}=3\lambda $
BETA
