The threshold frequency of a metal is 10¹⁵ s^-1. The ratio of maximum kinetic energies of the photoelectrons when the metal is irradiated with radiations of frequencies 1.5
x 10¹⁵ s^-1 and 2 x 10¹⁵ s^-1 respectively would be

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

2 : 1

3 : 4

4 : 3

1 : 2

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$\begin{array}{l} From KE = h (v - v_{0}) \\ \begin{matrix} \frac{{KE}_{1}}{{KE}_{2}} = \frac{v_{1} - v_{0}}{v_{2} - v_{0}} \\ Given v_{1} = 1.5 \times 10^{15} s^{- 1} \\ v_{2} = 2 \times 10^{15} s^{- 1} \\ v_{0} = 10^{15} s^{- 1} \end{matrix} \\ \frac{{KE}_{1}}{{KE}_{2}} = \frac{1.5 \times 10^{15} - 10^{15}}{2 \times 10^{15} - 10^{15}} \\ \frac{{KE}_{1}}{{KE}_{2}} = \frac{(1.5 - 1) 10^{15}}{(2 - 1) 10^{15}} = \frac{0.5}{1} = \frac{1}{2} = 1 : 2 \end{array}$