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 \( \displaystyle\frac{V}{4} \), the threshold wavelength for metallic surface is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 3\lambda \)
By Einstein's photoelectric equation, the maximum kinetic energy of the photoelectrons, \( eV_0 \), equals the photon energy minus the work function \( \phi_0 \).
\( \displaystyle eV = \frac{hc}{\lambda} - \phi_0 \quad\text{(1)} \)
\( \displaystyle \frac{eV}{4} = \frac{hc}{2\lambda} - \phi_0 \quad\text{(2)} \)
Multiplying (2) by 4:
\( \displaystyle eV = \frac{2hc}{\lambda} - 4\phi_0 \)
Equating this with (1):
\( \displaystyle \frac{hc}{\lambda} - \phi_0 = \frac{2hc}{\lambda} - 4\phi_0 \;\Rightarrow\; 3\phi_0 = \frac{hc}{\lambda} \)
The work function is the energy of a photon of the threshold wavelength \( \lambda_0 \):
\( \displaystyle \phi_0 = \frac{hc}{\lambda_0} = \frac{hc}{3\lambda} \;\Rightarrow\; \lambda_0 = 3\lambda \)