The threshold frequency of metal is $f_{0}$ . When the light of frequency $2 f_{0}$ is incident on the metal plate, the maximum velocity of photoelectron is $v_{1}$ . When the frequency of incident radiation is increased to $5 f_{0}$ , the maximum velocity of photoelectrons emitted is $v_{2}$ . The ratio of $v_{1}$ to $v_{2}$ is

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$\frac{v_{1}}{v_{2}} = \frac{1}{4}$

$\frac{v_{1}}{v_{2}} = \frac{1}{2}$

$\frac{v_{1}}{v_{2}} = \frac{1}{8}$

$\frac{v_{1}}{v_{2}} = \frac{1}{16}$

answer is A.

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Detailed Solution

$K_{\max} = h υ - ϕ$

$\frac{1}{2} m v_{1}^{2} = h f_{0}$

$\frac{1}{2} m v_{2}^{2} = 4 h f_{0}$

$\frac{v_{1}}{v_{2}} = \frac{1}{2}$

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