Question

Photoelectric emission is observed from a surface for frequencies $v_{1}$, and $v_{2}$ of the incident radiation ($v_{1}, > v_{2}$). If the maximum kinetic energies of the photoelectrons in two cases are in ratio $1 : K$ then the threshold frequency $v_o$ is given by

• A. $\frac{v_{2}-v_{1}}{K-1}$
• B. $\frac{Kv_{1}-v_{2}}{K-1}$
• C. $\frac{Kv_{2}-v_{1}}{K-1}$
• D. $\frac{v_{2}-v_{1}}{K}$

Answer 1

Answer: (B)

$KE_{1}=hv_{1}-hv_{o}$
$KE_{2} =hv_{2} -hv_{o}$
As$=\frac{kE_{1}}{KE_{2}}=\frac{1}{K}$
$\frac{h_{v_1}-h_{v_o}}{h_{v_2}-h_{v_o}}=\frac{1}{K}$
$\left(Kv_{1}-v_{2}\right)= v_{o}\left(K-1\right) $
$v_{o}= \frac{Kv_{1} -v_{2}}{k-1}$