Question

The stopping potential for photoelectrons from a metal surface is $V_1$ when monochromatic light of frequency ${\nu }_1$ is incident on it. The stopping potential becomes $V_2$ when monochromaitc light of another frequency is incident on the same metal surface. If $h$ be the Planck's constant and $e$ be the charge of an electron then the frequency of light in the second case is

• A. ${\nu }_1-\frac{e}{h}(V_2+V_1)$
• B. ${\nu }_1+v\frac{e}{h}(V_2+V_1)$
• C. ${\nu }_1-\frac{e}{h}(V_2-V_1)$
• D. ${\nu }_1+\frac{e}{h}(V_2-V_1)$

Answer 1

Answer: (D)

${\nu }_1+\frac{e}{h}(V_2-V_1)$

$h{\nu }_1={\varphi }_0+e{V}_1$ .......(i)
$h{\nu }_2={\varphi }_0+e{V}_2$ .......(ii)

From equations $\left(i\right)$ and $\left(ii\right)$ we have:
$\Rightarrow h({\nu }_1-{\nu }_2)=e(V_1-V_2)$

$\Rightarrow {\nu }_1-{\nu }_2=\frac{e}{h}(V_1-V_2)$

$\Rightarrow {\nu }_2=\frac{e}{h}(V_2-V_1)+{\nu }_1$