In a photoelectric experiment, the wavelength of incident light is increased from \( 5\times10^{-10}\,\mathrm{m} \) to \( 9\times10^{-10}\,\mathrm{m} \), while the intensity of radiation remains the same. Choose the correct statement from the following.
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → The cut off potential will decrease
By Einstein's photoelectric equation the maximum kinetic energy of a photoelectron is the photon energy minus the work function, and the cut off (stopping) potential measures it, \( eV_0 = K_{\max} \).
\( \displaystyle eV_0 = K_{\max} = \frac{hc}{\lambda} - \phi_0 \)
The work function \( \phi_0 \) is fixed for the metal. When \( \lambda \) increases, the photon energy \( hc/\lambda \) falls, so \( K_{\max} \) and the cut off potential both decrease.
Why the other options are incorrect
Option (1) The photoelectric current will decrease — Incorrect. At the same intensity the energy arriving per second is unchanged, but each photon now carries less energy, so more photons arrive per second and the saturation current does not fall.
Option (2) The maximum kinetic energy of electron will increase — Incorrect. It decreases, as shown above.
Option (3) The intensity of radiation will increase — Incorrect. The question states that the intensity remains the same.
NCERT (2026-27):
Effect of Frequency of Incident Radiation on Stopping Potential – “… This implies that greater the frequency of incident light, greater is the maximum kinetic energy of the photoelectrons. Consequently, we need greater retarding potential to stop them completely. … The graph shows that (i) the stopping potential V0 varies linearly with the frequency of incident radiation for a given photosensitive material.”