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.

Options

A. 2
B. 6
C. 4
D. 8

Answer

C

Explanation

Energy of photon `E=(hc)/(lamda)=(1.24xx10^(3))/(400)=3.1 eV`
Remaining energy `=3.1-0.31=2.79eV`
Energy lost is first collision is
`(3.1)xx((10)/(100))=0.31eV`
Remaining energy is
`3.1-0.31=2.79 eV`
Energy lost in second collision is
`(2.79)xx((10)/(100))=0.279 eV`
Total energy lost two collisions is
`(0.31)+(0.279)eV=0.589 eV` So, from conservation of energy, we have
`(hc)/(lamda)=phi+KE_(max)+` energy lost in two collision
`3.1=2.2+KE_(max)+0.589`
`KE_(max)=0.31eV`
Total energy after second collision is `(2.79-0.279)=2.511 eV`
Energy lost in third collision is
`2.511xx(10)/(100)=0.2511 eV`
Remaining energy `=(2.511-0.2155)=2.2500 eV`
Energy lost in fourth collision
`=(2.2599xx(10)/(100))=0.2259 eV`
Remaining energy `=(2.2599-0.2250)=2.034 eV`
After the fourth collision, the electron doen not have enough energy to overcome the work function, so it cannot come out.