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Photoelectric Effect - JEE Main Previous Year Questions with Solutions

Photoelectric Effect questions in JEE Main predominantly test Einstein's photoelectric equation, work function calculations, stopping potential, and threshold frequency. NTA/AIEEE has asked 1–2 questions from this topic almost every year. Step-by-step solutions to all major previous year questions are listed below.
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Photoelectric Effect - JEE Main Previous Year Questions with Solutions

eSaral > JEE Main Previous Year Questions > Photoelectric Effect JEE Main Previous Year Questions with Solutions

JEE Main Previous Year Question of Physics with Solutions are available here. Practicing JEE Main Previous Year Papers Questions of Physics will help all the JEE aspirants in realizing the question pattern as well as help in analyzing their weak & strong areas. Get detailed Class 11th &12th Physics Notes to prepare for Boards as well as competitive exams like IIT JEE, NEET, etc. eSaral helps the students in clearing and understanding each topic in a better way. eSaral is providing complete chapter-wise notes of Class 11th and 12th for all subjects. Besides this, eSaral also offers NCERT Solutions, Previous year questions for JEE Main and Advanced, Practice questions, Test Series for JEE Main, JEE Advanced and NEET, Important questions of Physics, Chemistry, Math, and Biology and many more. Download the eSaral app for free study material and video tutorials.   Simulator   Previous Years AIEEE/JEE Mains Questions

Q. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is : (hc = 1240 eV-nm) (1) 1.51 eV            (2) 1.68 eV          (3) 3.09 eV              (4) 1.41 eV [AIEEE - 2009]
Ans. (4) $\mathrm{E}_{\mathrm{k}}=\frac{\mathrm{hc}}{\lambda}-\phi_{0} \Rightarrow 1.68=\frac{12400}{4000}-\phi_{0}$ By solving it $\phi_{0}=1.42 \mathrm{eV}$
Q. Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is $\mathrm{K}_{\mathrm{max}}$. When the ultraviolet light is replaced by X-rays, both $\mathrm{V}_{0}$ and $\mathrm{K}_{\mathrm{max}}$ increase. Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light (1) Statement–1 is true, Statement–2 is false (2) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement– 1 (3) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement– 1 (4) Statement–1 is false, Statement–2 is true [AIEEE - 2010]
Ans. (1) Speed of emitted electrons is independent of frequency of incident light.
Q. This question has Statememtn-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements. Statement–1 : A metallic surface is irradiated by a monochromatic light of frequency $\mathrm{v}>\mathrm{v}_{0}$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $\mathrm{K}_{\max }$ and $\mathrm{V}_{0}$ respectively. If the frequency incident on the surface is doubled, both the $\mathrm{K}_{\max }$ and $\mathrm{V}_{0}$ are also boubled. Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light. (1) Statement–1 is true, Statement–2 is true, Statement–2 is not the correct explanationof Statement– 1 (2) Statement–1 is false, Statement–2 is true (3) Statement–1 is true, Statement–2 is false (4) Statement–1 is true, Statement–2 is true, Statement–2 is the correct explanation of Statement– 1 [AIEEE-2011]
Ans. (2)
Q. The anode voltage of photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows [AIEEE-2013]
Ans. (4) For constant intensity as wavelength decreases energy of photons increases and number of photons decreases. So it may seem that current should decrease. But the probability that a photon will be successful in emitting an electron will also increase. So as wavelength decreases current increases
Q. Radiation of wavelength $\lambda$, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength of changed to $\frac{3 \lambda}{4}$, the speed of the fastest emitted electron will be :- [JEE Main-2016]
Ans. (2) $\mathrm{E}=(\mathrm{KE})_{\max }+\mathrm{f}$ $\left[\frac{\mathrm{hc}}{\lambda}=(\mathrm{KE})_{\max }+\phi\right] \ldots .$ (1) \frac{4}{3} \frac{\mathrm{hc}}{\lambda}=\left(\frac{4}{3} \mathrm{KE}_{\max }+\frac{\phi}{3}\right)+\phi

Explore more resources to strengthen your JEE Main preparation:

- NCERT Solutions for Class 12 Physics — complete chapter-wise solutions including Dual Nature of Radiation

- NCERT Solutions for Class 11 Physics — Foundation Concepts for Modern Physics

-NCERT Books for Class 12 — download the official textbook chapters

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Frequently Asked Questions

Find answers to common questions.

Why does doubling the frequency not double the maximum kinetic energy?

Because KE_max = hf − φ, not KE_max = hf. The work function φ is a constant subtracted from the photon energy. When f doubles, KE_max becomes 2hf − φ, which is not twice (hf − φ) unless φ = 0. This is a classic JEE Main trap tested directly in AIEEE 2011.

What is Einstein's photoelectric equation and how is it used in JEE?

Einstein's equation is KE_max = hf − φ, where h is Planck's constant, f is the frequency of incident light, and φ is the work function of the metal. In JEE Main, this equation is used to find the work function when KE_max and wavelength are given, or to find the speed of the fastest electron when the wavelength changes.

What is the weightage of Photoelectric Effect in JEE Main?

Photoelectric Effect typically contributes 1–2 questions (4–8 marks) in JEE Main every year. It falls under Modern Physics, which as a whole carries 3–4 questions. Based on NTA's official papers from 2009 to 2024, this topic has appeared in over 85% of JEE Main sessions, making it one of the most reliable scoring topics in Physics.

How should I prepare Photoelectric Effect for JEE Main 2025–2026?

Start with NCERT Class 12 Physics Chapter 11 to build conceptual clarity, then solve AIEEE/JEE Main previous year questions year-by-year. Focus on: (1) numerical problems using Einstein's equation, (2) graph interpretation (I vs λ, KE vs f), and (3) statement-based questions on stopping potential and intensity. Aim to solve 30+ previous year questions before your exam

Does increasing light intensity increase the stopping potential?

No. Increasing intensity increases the number of emitted electrons (and therefore photocurrent), but it does not change KE_max or stopping potential. Stopping potential is determined solely by photon frequency. This is a fundamental result of Einstein's quantum explanation and is frequently tested in JEE Main conceptual questions

What is stopping potential in the photoelectric effect?

Stopping potential (V₀) is the minimum reverse voltage applied to a photocell to stop all emitted photoelectrons. It satisfies eV₀ = KE_max. Stopping potential depends only on the frequency of incident light, not on its intensity. Higher frequency light produces a higher stopping potential.

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April 21, 2021, 1:59 p.m.
Statement type question are tougher than numerical type question
TANISHKA
April 15, 2021, 3:27 a.m.
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Dec. 26, 2020, 4:06 p.m.
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Itte Revanth Kumar
Oct. 19, 2020, 8:02 p.m.
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WQW
Sept. 8, 2020, 6:23 p.m.
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April 21, 2020, 2:54 p.m.
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March 25, 2020, 9:28 p.m.
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