Find the frequency of light which ejects electrons from a metal surface, fully stopped by a retarding potential of 3V. The photoelectric effect begins in this metal at frequency `6xx10^(14)s^(-1)`

Options

A. `10^(5)`
B. `10^(16)`
C. `1.324xx10^(15)`
D. none of these.

Answer

C

Explanation

`hv=E_(k)+w " " :. E_(k)=hv-w`
If `V_(s)` is retarding potential then
`E_(k)=eV_(s) and w=hv_(0)`
`:.eV_(s)=hv-hv_(0)`
or `v=(eV_(s))/(h)+v_(0)`
Given `V_(s)=3V, v_(0)=6xx10^(-14)s^(-1)`
`:.` Required frequency v is
`v=(1*6xx10^(-19)xx3)/(6*62xx10^(-34))+6xx10^(14)`
`v=1*324xx10^(15)` per sec.