When Ultra-violet (UV) radiations of intensity I0 are incident on a given photocell, the maximum K.E of emitted photo electrons is (K.E.)max, and stopping potential is V0. If the UV radiations are replaced by a blue light of same intensity I0 (assuming that photoelectrons are emitted), then identify the correct statement(s) among the following:
A. V0 will increase but (K.E.)max will remain same
B. (K.E.)max will remain same and V0 will decrease
C. V0 and (K.E.)max both will decrease
D. V0 will remain same and (K.E.)max will decrease
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → C only
By Einstein’s photoelectric equation, the maximum kinetic energy of the photoelectrons depends on the frequency of the light, not on its intensity, and the stopping potential is fixed by that kinetic energy.
\( \displaystyle K_{\max} = h\nu - \phi_0 \)
\( \displaystyle eV_0 = K_{\max} \)
Blue light has a lower frequency than ultraviolet radiation, so \( K_{\max} \) decreases, and \( V_0 \) decreases with it. Keeping the intensity the same does not change this.
(A) V0 will increase but (K.E.)max will remain same — Incorrect. Both quantities decrease, and they always change together because \( eV_0 = K_{\max} \).
(B) (K.E.)max will remain same and V0 will decrease — Incorrect. \( V_0 \) cannot change while \( K_{\max} \) stays the same.
(C) V0 and (K.E.)max both will decrease — Correct. The lower frequency lowers \( K_{\max} \) and hence \( V_0 \).
(D) V0 will remain same and (K.E.)max will decrease — Incorrect. A smaller \( K_{\max} \) needs a smaller stopping potential.
NCERT (2026-27):
Effect of Frequency of Incident Radiation on Stopping Potential – “… The energy of the emitted electrons depends on the frequency of the incident radiations. The stopping potential is more negative for higher frequencies of incident radiation. … 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.”