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.

Explanation

(a) When `t = 400nm`
Energy of photon `= hc / lambda = 1240/ 400 = 3.1eV `
This energy is given to electon :
` But for the first collision , Energy lost
` = 3.1eV xx 10%`
` = 0.31eV`
for second collision, Energy lsot ` = 3.1 ev xx 10% `
`= 0.31eV `
Total energy lost in two collison
`= 0.31 +0.31 = 0.62 eV `
K.E of photoelectron whe it comes out of metal ,
` rArr hc / lambda - work ftmction - Energy lsot due to collision
` = (3.1 - 2.2 - 0.62 )eV `
` = 0.31eV `
(b) Similarly for the 3rd collision . the energy lost
` = 0.31eV`
It just comes out of the metal Hence in the fourth collision ebecomes unable to come out of the metal.