### 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 $

## Show Answer

#### 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 $