Question

Photons of wave length 248 nm fall on a metal surface whose work function is 2.2 eV. Assume that each
photoelectron inside the metal lattice may come out of the surface or collide with the lattice before coming out. In each collision with the lattice, it loses 20% of its existing energy. The kinetic energy of an ejected electron can be

• A. 4.8 eV
• B. 2.24 eV
• C. 1.85 eV
• D. 3.9 eV

Answer 1

The correct option is B 2.24 eV

Energy of incident photon $E_i=\frac{1240}{248}=5eV$

$K{E}_{max}$ of electron = $E_i–\varphi =5-2.2=2.8eV$

Lost of energy during $1^{st}$ collision = 20% of 2.8 eV = 0.56 eV

K.E of electron after $1^{st}$ collision = 2.8 - 0.56 = 2.24 eV
KE of electron after $2^{nd}$ collision = 2.24 – 0.448 = 1.792 eV