Question

When a metallic surface is illuminate with radiation of wavelength $\lambda$, the stopping potential is $V$. If the same surface is illuminated with radiation of wavelength $2\lambda$, the stopping potential is $4V$​. The threshold wavelength for the metallic surface is:

• A. $4\lambda$
• B. $5\lambda$
• C. $\frac{5}{2}\lambda$
• D. $3\lambda$

Answer 1

The correct option is D

$3\lambda$

Step 1: Given data

For case 1, $$\begin{gathered} \mathrm{wavelength}=\lambda \\ \mathrm{stopping}\mathrm{potential}=V \end{gathered}$$

For case 2, $$\begin{gathered} \mathrm{wavelength}=2\lambda \\ \mathrm{stopping}\mathrm{potential}=4V \end{gathered}$$

Step 2: Equation for the energy

1. For the first case.

$$\begin{gathered} eV=\frac{hc}{\lambda }-\varphi \\ eV=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0} \end{gathered}$$

Here, $\lambda$ is the wavelength, $h$ is the planks constant, $c$ is the velocity of light, ${\lambda }_0$ is the threshold wavelength, and $\varphi$ is the phase difference.

2. For the second case.

$$\begin{gathered} e\frac{V}{4}=\frac{hc}{2\lambda }-\varphi \\ \frac{eV}{4}=\frac{hc}{2\lambda }-\frac{hc}{{\lambda }_0} \end{gathered}$$

.Step 3: Calculation of the threshold wavelength

Dividing the first equation by the second equation.

$$\begin{gathered} \frac{eV}{{\displaystyle \frac{eV}{4}}}=\frac{{\displaystyle \frac{hc}{\lambda }}-{\displaystyle \frac{hc}{{\lambda }_0}}}{{\displaystyle \frac{hc}{2\lambda }}-{\displaystyle \frac{hc}{{\lambda }_0}}} \\ 4\left(\frac{1}{2\lambda }-\frac{1}{{\lambda }_0}\right)=\left(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}\right) \\ \frac{3}{{\lambda }_0}=\frac{1}{\lambda } \\ {\lambda }_0=3\lambda \end{gathered}$$

Hence, the threshold wavelength for the metallic surface is $3\lambda$.

Therefore, option (D) is correct.