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 metallic surface is

Options

A. 5 `lambda`
B. `(5)/(2)lambda`
C. `3lambda`
D. `4lambda`

Answer

C

Explanation

For first case,
`eV=(hc)/(lambda)-(hc)/(lambda_(0))`
For second case
`e(V)/(4)=(hc)/(2lambda)-(hc)/(lambda_(0))`
`therefore eV=(2hc)/(lambda)-(4hc)/(lambda_(0))` ……..(2)
By comparing (1) and (2)
`therefore (hc)/(lambda)-(hc)/(lambda_(0))=(2hc)/(lambda)-(4hc)/(lambda_(0))`
`therefore (hc)/(lambda)-(2hc)/(lambda)=(hc)/(lambda_(0))-(4hc)/(lambda_(0))`
`therefore -(hc)/(lambda)=-(3hc)/(lambda_(0))`
`therefore -(hc)/(lambda)=-(3hc)/(lambda_(0))`
`therefore lambda_(0)=3lambda`