Question

Light of wavelength $2000\dot{\text{A}}$ falls on a metallic surface, for which a stopping potential of $2\text{V}$ is required. Find the wavelength of the incident radiation for which the stopping potential required is $5\text{V}$.

• A. $1348\dot{A}$
• B. $1200\dot{A}$
• C. $1648\dot{A}$
• D. $2300\dot{A}$

Answer 1

The correct option is A $1348\dot{A}$

According to Einstein's photoelectric equation,

$\Rightarrow e{V}_0=\frac{hc}{\lambda }-\varphi$

As the metal surface is the same in both cases, work function remains the same.

$\Rightarrow -e{V}_1+\frac{hc}{{\lambda }_1}=-e{V}_2+\frac{hc}{{\lambda }_2}$

$\Rightarrow \frac{hc}{{\lambda }_2}-\frac{hc}{{\lambda }_1}=e(V_2-V_1)$

$\Rightarrow 12400\text{eV}\dot{A}(\frac{1}{{\lambda }_2}-\frac{1}{2000\dot{A}})=3\text{eV}$

$\Rightarrow 12400\dot{A}(2000\dot{A}-{\lambda }_2)=3\times 2000\dot{A}{\lambda }_2$

$\Rightarrow 2000\dot{A}-{\lambda }_2=\frac{3\times 20}{124}{\lambda }_2$

$\Rightarrow 2000\dot{A}=\frac{46}{31}{\lambda }_2$

$\Rightarrow {\lambda }_2=1347.82\dot{A}$

Hence, option $\left(A\right)$ is correct.