Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident…
Photoelectric emission is observed from a metallic surface for frequencies $v_1$ and $v_2$ of the incident light $\left(v_1>v_2\right)$. If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio $1: n$, then the threshold frequency of the metallic surface is
$\frac{\left(v_1-v_2\right)}{(n-1)}$
$\frac{\left(n v_1-v_2\right)}{(n-1)}$
$\frac{\left(n v_2-v_1\right)}{(n-1)}$
$\frac{\left(v_1-v_2\right)}{n}$
Solution
By using $h v-h v_0=K_{\max }$
$h\left(v_1-v_0\right)=K_1$
$h\left(v_2-v_0\right)=K_2$
$\therefore \quad \frac{v_1-v_0}{v_2-v_0}=\frac{K_1}{K_2}=\frac{1}{n}$
or $\quad v_0=\frac{n v_1-v_2}{(n-1)}$