Question

In a photoelectric effect experiment, a metal surface of work function $2.2eV$ is illuminated by light of wavelength $400nm$. If electron makes atleast two collisions before getting emitted out and in each collision 10% of energy is lost, then
(Use $hc=1240eV.nm$)

• A. Maximum kinetic energy of photo-electrons is $0.311eV$
• B. If electron makes 3 collisions instead of 2, it cannot come out
• C. Kinetic energy of photo-electron may be $0.06eV$
• D. Electron can make maximum of 3 collisions before coming out

Answer 1

Answer: (A), (C), (D)

The correct options are
A Maximum kinetic energy of photo-electrons is $0.311eV$
C Kinetic energy of photo-electron may be $0.06eV$
D Electron can make maximum of 3 collisions before coming out

Energy of photon $=\frac{hc}{\lambda }=\frac{1240}{400}=3.1eV$
So, the energy given by the photon to an electron is $3.1eV$
Energy after 1st collision = 3.1 $\times$ 0.9 eV = 2.79 eV
Energy after 2nd collision = 0.9 $\times$ 2.79 eV = 2.511 eV,
which is greater than work function hence, electrons can be ejected. $K{E}_{max}=2.511-2.2=0.311eV$

Energy after 3rd collision = 0.9 $\times$ 2.511 eV = 2.26
Still electron can come out. $K{E}_{max}=2.26-2.2=0.06eV$

Energy after 4th collision = 0.9 $\times$ 2.26 = 2.03 eV
Electron cannot come out.