Light of wavelength 330 nm falling on a piece of metal ejects electrons with sufficient energy which requires voltage `V_(0)` to prevent them from reaching a collector. In the same setup, light of wavelength 220 nm, ejects electrons which requires twice the voltage `V_(0)` to stop them in reaching a collector. The numerical value of voltage `V_(0)` is `(15)/(x)V` . Find value of x.

Answer

`15//8 V`

Explanation

`E_(330)=12400/3300=3.7575 eV`
`E_(220)=12400/2200=5.636 eV`
Let work function be `phi`
`eV_(0)=(3.7575-phi)e, 2eV_(0) (5.6363-phi)e`
`V_(0)=1.87 V`