Photoelectric emission is observed from a metallic surface for frequencies $v_{1}$ and $v_{2}$ of the incident light rays $(v_{1} > v_{2})$ . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is

\frac{v_{1} - v_{2}}{k - 1}

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\frac{k v_{1} - v_{2}}{k - 1}

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\frac{k v_{2} - v_{1}}{k - 1}

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\frac{v_{2} - v_{1}}{k}

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Solution

The correct option is B $\frac{k v_{1} - v_{2}}{k - 1}$
By using $h v - h v_{0} = k_{m a x}$
$\Rightarrow h (v_{1} - v_{0}) = k_{1}$ ….(i)
And $h (v_{2} - v_{0}) = k_{2}$ ….(ii)
$\Rightarrow \frac{v_{1} - v_{0}}{v_{2} - v_{0}} = \frac{k_{1}}{k_{2}} = \frac{1}{k}$ , Hence $v_{0} = \frac{k v_{1} - v_{2}}{k - 1}$ .

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Q. Photoelectric emission is observed from a metallic surface for frequencies

v_{1}

and

v_{2}

of the incident light rays

(v_{1} > v_{2})

. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is

Photoelectric emission is observed from a metallic surface for frequencies $v_{1}$ and $v_{2}$ and of the incident light rays $(v_{1} > v_{2})$ . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is