Question

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. $K_1>\frac{K_2}{3}$
• B. $K_1< \frac{K_2}{3}$
• C. $K_1=\frac{K_2}{3}$
• D. $K_2=\frac{K_1}{3}$

Answer 1

Answer: (B)

$K_1< \frac{K_2}{3}$

Answer 2

The correct answer is (B) : $K_1< \frac{K_2}{3}$
$K_1=\frac{hc}{λ_1}−\phi=\frac{hc}{3λ_2}−\phi….(i)$
and
$K_2=\frac{hc}{λ_2}−\phi….(ii)$
from (i) and (ii) we can say
3K1 = K2 – 2φ
$K_1<\frac{K_2}{3}$