Question 341 markMCQSection D

For radiation of frequency ν>ν2\nu > \nu_2 incident on the surfaces of A and B, the maximum kinetic energy of ejected electron is

Graph of stopping potential vs frequency for metals A and B
Dual Nature of Radiation and Matter
Comparing maximum kinetic energy of photoelectrons

Options

(A)greater for metal A because it has a smaller work function.
(B)greater for metal B because it has a larger work function.
(C)greater for metal B because it has higher threshold frequency.
(D)the same for both metal A and metal B because it is independent of work functions of metals.

Official answer

Correct option: (A) greater for metal A because it has a smaller work function.


Maximum kinetic energy: Kmax=hν−ϕK_{max} = h\nu - \phi.


  • Metal A has the smaller threshold frequency (ν1<ν2)(\nu_1 < \nu_2), so ϕA=hν1<ϕB=hν2\phi_A = h\nu_1 < \phi_B = h\nu_2.
  • For the same incident frequency ν, a smaller φ leaves more energy as kinetic energy.

Hence KmaxK_{max} is greater for metal A.

Answer keywords:maximum kinetic energyK_max = hν − φsmaller work function larger KEthreshold frequency comparisonmetal A metal Bsame incident frequencyphotoelectric effect

Marking scheme

  1. 1.1 mark: correct option (A) — greater for A because it has the smaller work function.
  2. 2.Reasoning credit for Kmax=hν−ϕK_{max} = h\nu - \phi and identifying ϕA<ϕB\phi_A < \phi_B from ν1<ν2\nu_1 < \nu_2.
  3. 3.No marks for options B, C (B greater) or D (independent of φ).

Hint

Kmax=hν−ϕK_{max} = h\nu - \phi at fixed ν; smaller work function ⇒ larger KmaxK_{max}, and A has the smaller threshold frequency.

Quick oral answer

Since KmaxK_{max} equals hνh\nu minus the work function, at the same frequency the metal with the smaller work function keeps more kinetic energy; metal A has the lower threshold ν1\nu_1, so its photoelectrons are faster.

Common mistakes

  1. 1.Thinking a larger work function gives a larger kinetic energy (it does the opposite).
  2. 2.Assuming kinetic energy is independent of the metal because the graph lines are parallel — the intercept still shifts KmaxK_{max}.
  3. 3.Confusing threshold frequency with the energy of the emitted electron.

Analysis and explanation

Concept:


Einstein's equation Kmax=hν−ϕK_{max} = h\nu - \phi shows that for a fixed incident frequency the photoelectron's maximum kinetic energy depends only on the work function: the smaller the work function, the larger the leftover kinetic energy.


Reading the graph:


  • Metal A intercepts the frequency axis at the smaller value ν1\nu_1, so its threshold frequency and work function are smaller.
  • At any common ν>ν2\nu > \nu_2 (so both metals emit), Kmax(A)−Kmax(B)=ϕB−ϕA=h(ν2−ν1)>0K_{max}(A) - K_{max}(B) = \phi_B - \phi_A = h(\nu_2 - \nu_1) > 0.

Why the distractors fail:


  • (B), (C) claim B is greater — a larger work function or higher threshold means less kinetic energy, not more.
  • (D) is wrong because K_max clearly depends on φ; only the slope (h/e)(h/e), not the intercept, is metal-independent.

Exam trap:


Do not confuse 'more energy needed to escape' with 'more energy left over' — a higher work function takes away kinetic energy.

Interesting facts

Because the two lines are parallel, the vertical gap between them — the difference in stopping potential at any frequency — is constant and equals (ϕB−ϕA)/e(\phi_B - \phi_A)/e.

Metals with low work functions like sodium (2.28 eV) and potassium (2.30 eV) release faster photoelectrons than high-work-function metals like platinum (about 6 eV) under the same light.

KmaxK_{max} depends only on frequency and work function, never on intensity — a fact that sealed the case for the photon model over classical wave theory.

Question details

Q34 of 39 questions

Board
CBSE
Class
Class 12
Subject
Physics
Year
2026
Section
Section D
Marks
1
Type
MCQ
Chapter
Dual Nature of Radiation and Matter
Topic
Comparing maximum kinetic energy of photoelectrons
Bloom's level
analyze
Syllabus topic
Unit VII: Dual Nature of Radiation and Matter — Einstein's Photoelectric Equation and Maximum Kinetic Energy
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why does metal A give photoelectrons with more kinetic energy than metal B?

Maximum kinetic energy is Kmax=hν−ϕK_{max} = h\nu - \phi. Metal A has the smaller threshold frequency (ν1<ν2)(\nu_1 < \nu_2), so its work function ϕA=hν1\phi_A = h\nu_1 is smaller. At the same incident frequency the two metals absorb the same photon energy hνh\nu, but A spends less of it escaping the surface, leaving more as kinetic energy. So KmaxK_{max} is larger for A by exactly ϕB−ϕA=h(ν2−ν1)\phi_B - \phi_A = h(\nu_2 - \nu_1).

Does the maximum kinetic energy depend on the intensity of the light?

No. KmaxK_{max} depends only on the frequency of the incident light and the metal's work function. Increasing intensity increases the number of photoelectrons per second but not their maximum kinetic energy — one of the key experimental facts explained only by the photon picture.