Question

The Kinetic energy of a photoelectron emitted on shining a light of wavelength 6.2 X 10^-6 m on a metal surface of work function 0.1 eV is _________________

(a) 0.01 eV

(b) 0.02 eV

(c) 0.1 eV

(d) 1 eV

Answer 1

Correct choice is (c) 0.1 eV

The explanation is: Kinetic Energy of photoelectrons = hv – Φ0

= (frac{hc}{lambda})– Φ0

Here, λ = 6.2 X 10^-6m

Φ0 = 0.1 eV = 1.6 X 10^-20J

Kmax = 3.2 X 10^-20J – 1.6 X 10^-20J

= 1.6 X 10^-20J

= 0.1 eV.