Photoelectric emission is observed from a metallic surface for frequencies n1 and n2 of the incident light rays (n1 n2) . 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

Question

Photoelectric emission is observed from a metallic surface for frequencies n1 and n2 of the incident light rays (n1 > n2) . 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 (A) (n1 - n2/k - 1) (B) (k n1 - n2/k - 1) (C) (k n2 - n1/k - 1) (D) (n2 - n1/k)

Tardigrade Admin · Accepted Answer

(K E1/K E2)=(h n1 - h n0/h n2 - h n0)=(1/k) n0=(k n1 - n2/k - 1)

$\frac{n _{1} - n _{2}}{k - 1}$

$\frac{k n _{1} - n _{2}}{k - 1}$

$\frac{k n _{2} - n _{1}}{k - 1}$

$\frac{n _{2} - n _{1}}{k}$

$\frac{K E _{1}}{K E _{2}} = \frac{h n _{1} - h n _{0}}{h n _{2} - h n _{0}} = \frac{1}{k}$
$n_{0} = \frac{k n _{1} - n _{2}}{k - 1}$

k−1n1​−n2​​

k−1kn1​−n2​​

k−1kn2​−n1​​

kn2​−n1​​

KE2​KE1​​=hn2​−hn0​hn1​−hn0​​=k1​ n0​=k−1kn1​−n2​​

$\frac{n _{1} - n _{2}}{k - 1}$

$\frac{k n _{1} - n _{2}}{k - 1}$

$\frac{k n _{2} - n _{1}}{k - 1}$

$\frac{n _{2} - n _{1}}{k}$

$\frac{K E _{1}}{K E _{2}} = \frac{h n _{1} - h n _{0}}{h n _{2} - h n _{0}} = \frac{1}{k}$
$n_{0} = \frac{k n _{1} - n _{2}}{k - 1}$