Question

The threshold frequency of a metal is 1 × 15 s^–1. The ratio of maximum kinetic energies of the photoelectrons when the metal is irradiated with radiations of frequencies 1.5 × 10¹⁵ s^–1 and 2.0 × 10¹⁵ s^–1 respectively, would be

• A. 3 : 4
• B. 1 : 2
• C. 2 : 1
• D. 4 : 3

Answer 1

Answer: (B)

1 : 2

Answer 2

from KE = h(n–n₀)

$\frac{KE_{1}}{KE_{2}}$= $\frac{\nu_{1}–\nu_{0}}{\nu_{2}–\nu_{0}}$

Given n₁ = 1.5 × 10¹⁵ s^–1

n₂ = 2.0 × 10¹⁵ s^–1

n₀ = 10¹⁵ s^–1