Question

• Light of wavelength 4000 Å is incident on a metal
plate whose work function is 2 eV. What is the
maximum KE of emitted photoelectron?
(a) 0.5 eV
(b) 1.1 eV
(c) 1.5 eV
(d) 2.0 eV
​

Answer 1

Answer:

B

Explanation:

Wavelength of light, λ = 4000 A° = 400 nm

Energy of incident photon E = hc/λ

Or E = 1240/λ (in nm) eV = 1240/400 = 3.1 eV

Maximum kinetic energy of photoelectrons

K.E max = E−ϕ where ϕ=2 eV

⟹ K.E max = 3.1−2 = 1.1 eV