Question 361 markMCQSection D

The threshold frequency for a metal surface is ν0\nu_0. If the radiation of frequency 3ν03\nu_0 illuminates the surface, the maximum kinetic energy (KE) of photoelectrons is E1E_1. If the frequency were increased to 6ν06\nu_0, the maximum KE of the photoelectrons becomes E2E_2. Then (E1/E2)(E_1/E_2) equals


OR


Let m be the slope of the graph line for metal B. If e is the value of electron charge, then Planck's constant 'h' is given by

(A) meme (B) 1/(me)1/(me) (C) m/em/e (D) e/me/m

Dual Nature of Radiation and Matter
Ratio of maximum kinetic energies at 3ν03\nu_0 and 6ν06\nu_0

Options

(A)1/31/3
(B)1/21/2
(C)2/52/5
(D)3/43/4

Official answer

Correct option: (C) 2/52/5


Primary part — KE ratio:


Using Kmax=h(ν−ν0)K_{max} = h(\nu - \nu_0):


  • E1=h(3ν0−ν0)=2hν0E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0
  • E2=h(6ν0−ν0)=5hν0E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0
  • E1/E2=2hν0/5hν0=2/5E_1/E_2 = 2h\nu_0 / 5h\nu_0 = 2/5

OR alternative — Planck's constant from slope:


Since Vs=(h/e)ν−φ/eV_s = (h/e)\nu - \varphi/e, the slope is m=h/em = h/e, so h=meh = me → option (A).

Answer keywords:K_max = h(ν − ν_0)kinetic energy ratio 2/5threshold frequency subtractionEinstein photoelectric equationslope h/ePlanck constant h = mephotoelectric effect numerical

Marking scheme

  1. 1.1 mark: correct option (C) 2/52/5 for the primary KE-ratio part.
  2. 2.Full credit for the OR alternative: option (A) meme from slope m=h/em = h/e.
  3. 3.No marks for the 1/21/2 distractor obtained by forgetting to subtract the threshold frequency.

Hint

Use Kmax=h(ν−ν0)K_{max} = h(\nu - \nu_0): subtract the threshold before taking the ratio; for the OR part slope m=h/e⇒h=mem = h/e \Rightarrow h = me.

Quick oral answer

Maximum kinetic energy is h times (ν minus ν-naught), so E1=2hν0E_1 = 2h\nu_0 at 3ν03\nu_0 and E2=5hν0E_2 = 5h\nu_0 at 6ν06\nu_0; their ratio is 2/52/5, and from the graph's slope h/eh/e we get Planck's constant h=m⋅eh = m\cdot e.

Common mistakes

  1. 1.Taking the ratio of frequencies (3ν0/6ν0=1/2)(3\nu_0/6\nu_0 = 1/2) without subtracting the threshold frequency ν0\nu_0.
  2. 2.Forgetting that Kmax=h(ν−ν0)K_{max} = h(\nu - \nu_0), not hνh\nu, so the work function/threshold term is dropped.
  3. 3.In the OR part, writing h=m/eh = m/e or e/me/m instead of h=meh = me from slope =h/e= h/e.

Analysis and explanation

Concept:


Einstein's equation in threshold form, Kmax=h(ν−ν0)K_{max} = h(\nu - \nu_0), gives the maximum kinetic energy directly as Planck's constant times the excess of frequency over threshold.


Working the primary part:


  • At 3ν03\nu_0: E1=h(3ν0−ν0)=2hν0E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0.
  • At 6ν06\nu_0: E2=h(6ν0−ν0)=5hν0E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0.
  • Ratio E1/E2=2/5E_1/E_2 = 2/5. The hν0h\nu_0 cancels, so the answer is independent of the metal.

OR part:


  • Rearranging eVs=hν−φeV_s = h\nu - \varphi gives Vs=(h/e)ν−φ/eV_s = (h/e)\nu - \varphi/e, a line of slope h/eh/e.
  • If that slope is m, then m=h/e⇒h=mem = h/e \Rightarrow h = me → option (A).

Exam trap:


Students often forget to subtract ν0\nu_0 and wrongly take the ratio 3/6=1/23/6 = 1/2 (distractor B). The threshold subtraction is essential.

Interesting facts

The relation Kmax=h(ν−ν0)K_{max} = h(\nu - \nu_0) means a photoelectric plot's slope gives Planck's constant regardless of the metal — Millikan's 1916 measurement this way matched the value from black-body radiation, a triumph for quantum theory.

Because the hν0h\nu_0 term cancels in the ratio, E1/E2=2/5E_1/E_2 = 2/5 holds for every metal illuminated at 3ν03\nu_0 and 6ν06\nu_0.

Doubling the frequency here (3ν0→6ν03\nu_0 \to 6\nu_0) more than doubles the kinetic energy (2hν0→5hν02h\nu_0 \to 5h\nu_0) precisely because the fixed threshold energy hν0h\nu_0 is subtracted first.

Question details

Q36 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
Ratio of maximum kinetic energies at 3ν03\nu_0 and 6ν06\nu_0
Bloom's level
apply
Syllabus topic
Unit VII: Dual Nature of Radiation and Matter — Einstein's Photoelectric Equation (Kmax=h(ν−ν0)K_{max} = h(\nu - \nu_0))
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Why is E1/E2=2/5E_1/E_2 = 2/5 and not 1/21/2 for frequencies 3ν03\nu_0 and 6ν06\nu_0?

Maximum kinetic energy is Kmax=h(ν−ν0)K_{max} = h(\nu - \nu_0), so you must subtract the threshold frequency before comparing. At 3ν₀, E1=h(3ν0−ν0)=2hν0E_1 = h(3\nu_0 - \nu_0) = 2h\nu_0; at 6ν₀, E2=h(6ν0−ν0)=5hν0E_2 = h(6\nu_0 - \nu_0) = 5h\nu_0. The ratio is 2hν0/5hν0=2/52h\nu_0/5h\nu_0 = 2/5. Taking 3/6=1/23/6 = 1/2 is the classic mistake of forgetting the ν₀ subtraction.

How does the OR part give Planck's constant as h=meh = me?

Rearranging Einstein's equation eVs=hν−φeV_s = h\nu - \varphi gives Vs=(h/e)ν−φ/eV_s = (h/e)\nu - \varphi/e, a straight line with slope h/e. If the measured slope of metal B's line is m, then m=h/em = h/e, so Planck's constant h=meh = me — the slope multiplied by the electron charge.