Question-34 Photoelectric emission is observed from a metallic surface for frequencies \nu_1 and \nu_2 of the incident light rays (\nu_1>\nu_2). If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1:k, then the threshold frequency of metallic surface is
(A) \frac{k\nu_2-\nu_1}{k-1}
(B) \frac{\nu_2-\nu_1}{k}
(C) \frac{\nu_1-\nu_2}{k-1}
(D) \frac{k\nu_1-\nu_2}{k-1}
Answer : (D)
Explanation:
hν₁ = hν₀ + \frac{1}{2}mu_1^2 ….(i)
hν₂ = hν₀ + \frac{1}{2}mu_2^2 ….(ii)
Given: \frac{1}{2}mu_1^2 = \left(\frac{1}{K}\right)\frac{1}{2}mu_2^2
From (i),
hν₁ = hν₀ + \frac{1}{2K}mu_2^2 ….(iii)
\frac{1}{2}mu_2^2 = Khν_1 - Khν_0 ….(iv)
From (ii) and (iv)
hν₂ − hν₀ = Khν₁ − Khν₀
⇒ ν₀ (1 − K) = ν₂ − Kν₁
∴ \nu_0 = \frac{K\nu_1 - \nu_2}{K - 1}