Question

The threshold frequency of a metal is 1 × 10¹⁵ 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

$\frac{KE_{1}}{KE_{2}}$ = $\frac{h(\nu_{1} - \nu_{0})}{h(\nu_{2}–\nu_{0})}$