Photoelectric emission is observed from a metallic surface for frequencies n 1 and n 2 of the incident light rays n 1 n 2 . If the maximum values of kinetic energy of the photoelectrons emitted in t

Question

Photoelectric emission is observed from a metallic surface for frequencies $n_{1}$ and $n_{2}$ of the incident light rays $(n_{1} > n_{2})$ . 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

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Answer:

B

Solution:

$\frac{K E_{1}}{K E_{2}} = \frac{h n_{1} - h n_{0}}{h n_{2} - h n_{0}} = \frac{1}{k}$

$n_{0} = \frac{k n_{1} - n_{2}}{k - 1}$

TestHub · Accepted Answer

K E 1 K E 2 = h n 1 - h n 0 h n 2 - h n 0 = 1 k n 0 = k n 1 - n 2 k - 1

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