Question 331 markMCQSection D

From the graph, the work functions of A and B are (h is Planck's constant and e value of charge on an electron)

Graph of stopping potential vs frequency for metals A and B
Dual Nature of Radiation and Matter
Work function from V_s–ν graph

Options

(A)ν1\nu_1 and ν2\nu_2
(B)V1V_1 and V2V_2
(C)hν1h\nu_1 and hν2h\nu_2
(D)hν1/eh\nu_1/e and hν2/eh\nu_2/e

Official answer

Correct option: (C) hν1h\nu_1 and hν2h\nu_2


From Einstein's photoelectric equation, eVs=hν−ϕeV_s = h\nu - \phi. Each line meets the ν-axis (Vs=0)(V_s = 0) at the threshold frequency:


  • For metal A the intercept is ν1\nu_1, so ϕA=hν1\phi_A = h\nu_1
  • For metal B the intercept is ν2\nu_2, so ϕB=hν2\phi_B = h\nu_2

The work function equals Planck's constant times the threshold frequency, giving hν1h\nu_1 and hν2h\nu_2.

Answer keywords:work functionthreshold frequencyphi = h nu_0stopping potential frequency graphEinstein photoelectric equationx-intercept thresholdhν_1 and hν_2

Marking scheme

  1. 1.1 mark: correct option (C) hν1h\nu_1 and hν2h\nu_2.
  2. 2.Reasoning credit for stating ϕ=hν0\phi = h\nu_0 where ν0\nu_0 is the x-intercept (threshold frequency).
  3. 3.No marks for options with wrong dimensions (ν1\nu_1, V1V_1 or hν/eh\nu/e).

Hint

Set Vs=0V_s = 0 in eVs=hν−ϕeV_s = h\nu - \phi: the x-intercept is the threshold frequency, and ϕ=h×\phi = h \times (that intercept).

Quick oral answer

Set the stopping potential to zero in eVs=hν−ϕeV_s = h\nu - \phi; the line then crosses the frequency axis at the threshold frequency, so the work function is Planck's constant times that intercept — hν1h\nu_1 for A and hν2h\nu_2 for B.

Common mistakes

  1. 1.Reading the y-intercept (−V1,−V2)(-V_1, -V_2) as the work function instead of using the x-intercept.
  2. 2.Quoting ν1\nu_1 and ν2\nu_2 as the work functions, ignoring that a work function must have units of energy.
  3. 3.Confusing threshold frequency with the frequency at which the two lines cross.

Analysis and explanation

Concept:


Einstein's equation eVs=hν−ϕeV_s = h\nu - \phi, written as Vs=(h/e)ν−ϕ/eV_s = (h/e)\nu - \phi/e, is a straight line of slope h/eh/e. Where the line crosses the frequency axis, Vs=0V_s = 0, so 0=hν0−ϕ0 = h\nu_0 - \phi, i.e. ϕ=hν0\phi = h\nu_0. The x-intercept of each line is therefore its threshold frequency.


Reading the graph:


  • Line A cuts the ν-axis at ν1⇒ϕA=hν1\nu_1 \Rightarrow \phi_A = h\nu_1.
  • Line B cuts the ν-axis at ν2⇒ϕB=hν2\nu_2 \Rightarrow \phi_B = h\nu_2.

Why the other options fail:


  • (A) ν1,ν2\nu_1, \nu_2 are frequencies, not energies — dimensionally wrong for a work function.
  • (B) V1,V2V_1, V_2 are potentials (volts); multiplying by e would give energy, but the plain V-values are not the work function.
  • (D) hν/eh\nu/e has units of volts, again not energy.

Exam trap:


Watch units — a work function is an energy (joule or eV), so only hν0h\nu_0 has the right dimensions.

Interesting facts

Robert Millikan, who initially doubted Einstein's photon idea, spent a decade measuring these VsV_s–ν\nu lines and in 1916 used their slope to give one of the best early values of Planck's constant.

The x-intercept (threshold frequency) is a fixed property of the metal — the two parallel lines never move sideways unless you change the metal.

Caesium, with a work function near 2.1 eV, has such a low threshold that even yellow light ejects electrons, which is why it is used in photocathodes.

Question details

Q33 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
Work function from V_s–ν graph
Bloom's level
understand
Syllabus topic
Unit VII: Dual Nature of Radiation and Matter — Photoelectric Effect and Einstein's Equation (stopping potential vs frequency graph)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

How do you read the work function off a stopping-potential vs frequency graph?

Use eVs=hν−ϕeV_s = h\nu - \phi. When the stopping potential VsV_s is zero, the incident frequency equals the threshold frequency ν0\nu_0, and the line crosses the frequency axis there. Since 0=hν0−ϕ0 = h\nu_0 - \phi, the work function is ϕ=hν0\phi = h\nu_0 — Planck's constant times the x-intercept. For metal A that intercept is ν1\nu_1 and for B it is ν2\nu_2, giving work functions hν1h\nu_1 and hν2h\nu_2.

Why are the lines for A and B parallel?

The slope of every V_s–ν line is h/eh/e, which depends only on Planck's constant and the electron charge — universal constants. So all metals give lines of the same slope; they differ only in where they cross the frequency axis, which reflects their different work functions.