A light beam of wavelength `400 nm` is incident on a metal of work- function `2.2 eV`. A particular electron absorbs a photon and makes 2 collisions before coming out of the metal
(a) Assuming that 10% of existing energy is lost to the metal in each collision find the final kinetic energy of this electron as it comes out of the metal.
(b) Under the same assumptions find the maximum number of collisions, the electron should suffer before it becomes unable to come out of the metal.

Answer

`4`

Explanation

`E_(lambda)="Energy of the photon" =(hc)/(lambda)=(1.24xx10^(4))/(4000)eV`
`KE " " "of" " "e=3.1 eV`
For `e^(-)` not to come out, its energy should be less then `2.2 eV`.
i.e., `(3.1)xx(0.9)^(n) lt 2.2 n gt 4`