Dual Nature Of Light Question 396
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:
[d] According to Einstein’s photoelectric effect, $ eV=\frac{hc}{\lambda }-\frac{hc}{{\lambda_{0}}} $ ?.(i) $ eV/4=\frac{hc}{2\lambda }-\frac{hc}{{\lambda_{0}}} $ ?(ii) Dividing equation (i) by (ii) by
$ \Rightarrow 4=\frac{\frac{1}{\lambda }-\frac{1}{{\lambda_{0}}}}{\frac{1}{2\lambda }-\frac{1}{{\lambda_{0}}}} $ on solving we get, $ {\lambda_{0}}=3\lambda $
BETA
