051220 | Dual Nature of Radiation and Matter

Photons of frequency ν are incident on the surfaces of metals A & B of threshold frequencies  3/4 ν and  2/3 ν respectively. The ratio of the maximum kinetic energy of electrons emitted from A to that from B is a) 2:3 b) 4:3 c) 3:4 d) 3:2 [2020 | Delhi | Section - A | SET - 2, Q.8]

Solution :
We know,
K_max = hν-hν₀
K_max =h(ν-ν₀)

Now
K_A=h(ν-3ν/4)
K_A=hν/4

K_B=h(ν-2ν/3)
K_B=h(ν/3)
K_B=hν/3

K_A/K_B =[hν/2]/[2hν/3]
K_A/K_B=3/4
K_A : K_B = 3:4