Question

When a monochromatic light of frequency $v$ is incident on a metal, stopping potential is $V_0$. Frequency of the incident light for which stopping potential becomes double is :

• A. $v$
• B. $v+\frac{e{V}_0}{h}$
• C. $2v-\frac{e{V}_0}{h}$
• D. $v-\frac{e{V}_0}{h}$

Answer 1

The correct option is B $v+\frac{e{V}_0}{h}$

$hv=h{v}_0+K.E$

At stopping potential, $K.E=eV$

$\Rightarrow hv=h{v}_0+eV$

$V=\frac{h}{e}(v-v_0)$

Given for, $V=V_0;v=v.$

$\therefore V_0=\frac{h}{e}(v-v_0)$

For $V=2V_0;v_1=?$

$\Rightarrow 2V_0=\left(\frac{h}{e}\right(v_1-v_0\left)\right)$

$v_1=\frac{2e{V}_0}{h}+v_0$
$v_1=\frac{2e{V}_0}{h}+v-\frac{e{V}_0}{h}$

$v_1=\frac{e{V}_0}{h}+v$