The threshold frequency of a metal is $10^{15} s^{- 1}$ . The ratio of maximum kinetic energies of the photoelectrons, when the metal is made to strike with radiations of frequencies $1.5 \times 10^{15} s^{- 1}$ and $2.0 \times 10^{15} s^{- 1}$ respectively is

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4:3

1:2

3:4

2:1

answer is B.

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Detailed Solution

$\frac{{(K . E)}_{1}}{{(K . E)}_{2}} = \frac{v_{1} - v_{o}}{v_{1} - v_{o}} \frac{{(K . E)}_{1}}{{(K . E)}_{2}} = \frac{(1.5 \times 10^{15} - 1 \times 10^{15})}{(2.0 \times 10^{15} - 1 \times 10^{15})} = \frac{1}{2}$

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