Question

A metallic surface is irradiated by a monochromatic light of frequency ${\upsilon }_1$ and stopping potential is found to be $V_1$. If the light of frequency ${\upsilon }_2$ irradiates the surface, the stopping potential will be

• A. $V_1+\frac{h}{e}({\upsilon }_1+{\upsilon }_2)$
• B. $V_1+\frac{h}{e}({\upsilon }_1-{\upsilon }_2)$
• C. $V_1+\frac{e}{h}({\upsilon }_2+{\upsilon }_1)$
• D. $V_1-\frac{h}{e}({\upsilon }_1+{\upsilon }_2)$

Answer 1

Answer: (B)

$V_1+\frac{h}{e}({\upsilon }_1-{\upsilon }_2)$

Maximum kinetic energy
$K_{(}max)=\frac{1}{2}m{v}^2=e{v}_0$
where $e{v}_0$ is the stopping potential.
According to Einstein's photoelectric equation
$h{\upsilon }_1={\varphi }_0+e{V}_1$
$h{\upsilon }_2={\varphi }_0+e{V}_2$
$\therefore h({\upsilon }_1-{\upsilon }_2)=e(V_1-V_2)$
$\frac{h}{e}({\upsilon }_1-{\upsilon }_2)=V_1-V_2$

$or{V}_2=V_1+\frac{h}{e}({\upsilon }_2-{\upsilon }_1)$