Question

When a metallic surface is illuminated 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 $\frac{V}{4}$ . The threshold wavelength for the metallic surface is :

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

Answer 1

The Correct Option is C

Equation Manipulation 

We are given the following two equations:

\(eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_0}\) ....(i)

\(eV/4 = \frac{hc}{2 \lambda} - \frac{hc}{\lambda_0}\) .....(ii)

From equations (i) and (ii), we can proceed as follows:

\(\Rightarrow 4 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_0}}{\frac{1}{2 \lambda} - \frac{1}{\lambda_0}}\)

On solving the above equation, we find that:

\(\lambda_0 = 3 \lambda\)

Conclusion:

The wavelength \(\lambda_0\) is three times the wavelength \(\lambda\).