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Question

Physics Question on Waves

Let K1 and K2 be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength λ1 and λ2, respectively are incident on a metallic surface. If λ1 = 3λ2 then:

A

K1>K23K_1> \frac{K_2}{3}

B

K1<K23K_1< \frac{K_2}{3}

C

K1=K23K_1=\frac{K_2}{3}

D

K2<=K13K_2<=\frac{K_1}{3}

Answer

K1<K23K_1< \frac{K_2}{3}

Explanation

Solution

K1=hcλ1−Φ\frac{ℎc}{λ_1}−Φ=hc3λ2−Φ\frac{ℎc}{3λ_2}−Φ….(i)
and
K2=hcλ2−Φ\frac{ℎc}{λ_2}−Φ….(ii)
from (i) and (ii) we can say
3K1 = K2 – 2φ
K1<K23K1<\frac{K2}{3}