The threshold frequency of metal is 10¹⁵ S^-1. The ratio of maximum kinetic energies of the photo electrons, when the metal is made to strike with radiations of frequencies is $1.5 \times 10^{15} S^{- 1}$ and $2.0 \times 10^{15} S^{- 1}$ respectively is.

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2 : 1

1 : 2

4 : 3

3 : 4

answer is B.

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Detailed Solution

A Given,
Threshold frequency of metal $(V_{O}) = 10^{15} S^{- 1}$
First frequency $(V_{1}) = 1.5 \times 10^{15} S^{- 1}$
Second frequency $(V_{2}) = 2 \times 10^{15} S^{- 1}$
$\frac{{(K E_{\max})}_{1}}{{(K E_{\max})}_{2}} = ?$
Using the equation of photo electric effect.
$h v = h v_{0} + K E \Rightarrow K E = h (v - v_{0})$
$\begin{array}{l} ∴ \frac{{(K E_{max})}_{1}}{{(K E_{max})}_{2}} = \frac{h (1.5 \times 10^{15} - 1 \times 10^{15})}{h (2 \times 10^{15} - 1 \times 10^{15})} \\ \frac{K E_{1}}{K E_{2}} = \frac{0.5 \times 10^{15}}{1 \times 10^{15}} \\ \frac{K E_{1}}{K E_{2}} = \frac{1}{2} or K E_{1} : K E_{2} = 1 : 2 . \end{array}$