Modern Physics NEET PYQ — The Photoelectric and Fission Traps NTA Repeats Every Year (2015-2026)
Modern Physics NEET PYQ analysis (2015-2026). Photoelectric effect, de Broglie, Bohr model, and nuclei traps decoded, plus the deleted syllabus, a formula sheet, and 12 exam-level PYQs with traps.
Three chapters, a handful of formulas, and the same traps every year.
If you want fast, safe marks in NEET Physics, learn this block cold. Modern Physics rewards clear concepts over heavy maths. Most questions turn on one idea and one formula. Get the idea right and the arithmetic is short.
The decade of data is blunt. Three vectors carry the block. Einstein's photoelectric equation with stopping potential leads everything. The de Broglie wavelength for charged particles is close behind. On the nuclei side, binding energy and the fission Q-value keep coming back. Learn these three and you cover most of what NTA asks.
The format has shifted, though. NEET now leans on Assertion-Reason, two-statement, and "how many of the following are correct" setups. These kill option-guessing. They punish a student who half-knows four NCERT lines. Passive reading no longer clears this block. Sharp, discriminating recall does.
We analysed every Modern Physics question NEET asked from 2015 to 2026, including NEET 2026 and Re-NEET 2026. This is part of Logic Bloom's NEET PYQ analysis series.
| 🎯 We mapped every NEET Modern Physics question of the decade. The app has them all, ready to play and practice. | |
|---|---|
| This block is formula-and-concept driven. You master it by drilling the photoelectric equation and the decay law until they are reflex, not by re-reading NCERT. Logic Bloom's Playground turns it into practice: read stopping-potential graphs, scale de Broglie wavelengths across particles, and step through fission to see where the 216 MeV comes from. Then drill every PYQ mapped by year. When the intensity-vs-stopping-potential trap or the per-nucleon fission trap catches you, TarQ teaches the fix, and your Mistake Book logs it. | Get the app → Free to start. |
The decade at a glance: where the marks actually sit.
Across 2015 to 2026 the block held a steady 5 to 8 questions per paper. Dual Nature is the biggest single source. Nuclei is the most calculation-heavy. Atoms is the smallest but the most trap-dense. Here is the split.
| Chapter | Typical questions/year | Character of the questions |
|---|---|---|
| Dual Nature of Radiation and Matter | 2 to 5 | Photoelectric equation, stopping potential, de Broglie scaling, graph reading |
| Atoms | 1 to 3 | Bohr scaling, hydrogen spectrum series, energy signs, closest approach |
| Nuclei | 1 to 3 | Binding energy, fission Q-value, decay law, half-life vs mean life |
The takeaway for a revision plan is simple. Dual Nature is where you spend the most time, because it gives the most marks and repeats the most. Nuclei is where careless students bleed marks on the fission trap. Atoms is short, so nail the sign conventions and move on.
Chapter 1: Dual Nature of Radiation and Matter
This is the highest-yield chapter in the block. It sits between two ideas. Light behaves as particles, called photons. Matter behaves as waves, described by de Broglie. Nearly every question tests one of these two.
What NEET asks, year by year
| Year | Qs | Sub-topics tested |
|---|---|---|
| Re-NEET 2026 | 4 | Photon vs electron momentum ratio; power-vs-photocurrent graph; threshold logic; de Broglie of orbital electrons |
| NEET 2026 | 5 | Ratio of de Broglie wavelengths; work function from threshold; stopping potential vs wavelength; photon flux; Einstein equation |
| NEET 2025 | 3 | de Broglie of an alpha particle; cut-off voltage; kinetic energy independent of intensity |
| NEET 2024 | 4 | Work function comparison; de Broglie vs accelerating voltage; threshold frequency; photon energy in eV |
| NEET 2023 | 4 | Stopping potential independence; matter-wave concept; Einstein equation; photoelectron KE |
| NEET 2022 | 3 | Ratio of stopping potentials; de Broglie of a thermal neutron; photon momentum |
| NEET 2021 | 2 | Photons emitted per second; wave on a near-zero work-function surface |
| NEET 2020 | 2 | Energy flux over time; de Broglie of an electron from rest |
| NEET 2019 | 2 | Work function; de Broglie vs momentum |
| NEET 2018 | 2 | Threshold wavelength; stopping-potential graph |
| NEET 2017 | 2 | Emission feasibility from work function; de Broglie of an electron |
| NEET 2016 | 1 | Ratio of de Broglie wavelengths, electron vs photon, same energy |
| NEET 2015 | 2 | Stopping potential from two wavelengths; KE change and de Broglie shift |
The highest-yield sub-topics
| Sub-topic | Decade count | What to have cold |
|---|---|---|
| Einstein equation + stopping potential | ~22 | eV₀ = hc/λ − φ₀. Often two wavelengths and two potentials, solved as a pair. |
| de Broglie wavelength scaling | ~17 | λ = h/p, λ = h/√(2mK), λ = h/√(2mqV). Ratio questions across particles are common. |
| Graph reading (V₀ vs ν) | ~8 | Slope is h/e, a universal constant. The x-intercept is the threshold frequency ν₀. |
The traps NEET repeats here
Trap 1: intensity does not change stopping potential. A question doubles the light intensity and asks for the new stopping potential. Students pick "doubled." Wrong. Intensity only adds more photons, so it raises the saturation current. It does not touch the maximum kinetic energy of a single photoelectron. The stopping potential stays the same. This is the single most-used trap in the whole physics paper.
Trap 2: the de Broglie shortcut is electron-only. The clean form λ = 1.227/√V nm works only for an electron. When the question switches to a proton or an alpha particle, that constant is wrong. Go back to λ = h/√(2mqV) and put in the right mass and charge. Students who reuse 1.227 lose the mark.
Trap 3: work function vs threshold frequency. These are linked by φ₀ = hν₀, but they are not the same quantity. Work function is an energy, in joules or eV. Threshold frequency is a frequency, in hertz. NTA plants distractors with swapped units. Also recall NCERT Table 11.1: caesium has the lowest work function at 2.14 eV, platinum the highest at 5.65 eV.
A worked NEET-style question
Q. Light of frequency ν falls on a metal of threshold frequency ν₀, with ν > ν₀. The intensity is then halved and the frequency is doubled. How do the saturation photocurrent and the stopping potential change?
(A) Halved, doubled (B) Halved, more than doubled (C) Unchanged, doubled (D) Halved, less than doubled
Solution. Saturation current tracks intensity. Half the intensity means half the current, so (C) is out. For the potential, start with eV₀₁ = hν − hν₀. The new value is eV₀₂ = 2hν − hν₀. Rewrite it: eV₀₂ = 2(hν − hν₀) + hν₀, which is 2(eV₀₁) + hν₀. Since hν₀ is positive, the new potential is more than double the old one. Choice (A) is the trap for students who assume V₀ scales straight with frequency and forget the work-function term.
Answer: (B)
Chapter 2: Atoms
This chapter joins classical mechanics to early quantum rules. NEET tests two things here. Can you handle the Bohr model formulas, and can you read the hydrogen spectrum? It is short but it is where sign errors cost marks.
What NEET asks, year by year
| Year | Qs | Sub-topics tested |
|---|---|---|
| Re-NEET 2026 | 2 | Bohr radius vs n; shortest wavelength of the Paschen series |
| NEET 2026 | 3 | Distance of closest approach; ratio of KE to PE in the nth orbit; spectral-line matching |
| NEET 2025 | 1 | Orbital speed vs principal quantum number |
| NEET 2024 | 2 | Transition levels for a given series; ground-state ionisation energy scaling |
| NEET 2023 | 2 | Energy to excite hydrogen; impact-parameter definition |
| NEET 2022 | 1 | Frequency of the photon in a Bohr transition |
| NEET 2021 | 0 | Merged into Dual Nature that year |
| NEET 2020 | 1 | Bohr orbit energy |
| NEET 2019 | 1 | Total energy in an excited state |
| NEET 2018 | 1 | Ratio of KE to total energy in a Bohr orbit |
| NEET 2017 | 1 | Ratio of wavelengths, last Balmer line vs last Lyman line |
| NEET 2016 | 2 | Wavelength for 4→3 transition; wave number at the Balmer limit |
| NEET 2015 | 2 | Hydrogen transitions; angular-momentum quantisation |
The traps NEET repeats here
Trap 1: energy signs. For a bound electron the total energy E is negative, kinetic energy K is positive, potential energy U is negative. The fixed links are K = −E and U = 2E. Ask for the potential energy in the first excited state and students stop at E = −3.4 eV. The right answer is U = 2 times −3.4, so −6.8 eV. NTA fills the options with every sign and magnitude to catch this.
Trap 2: shortest vs longest wavelength. The longest wavelength in a series is the smallest energy jump, like 3→2 in Balmer. The shortest wavelength, the series limit, is the largest jump, from n = ∞ down to the base level. Students flip these under time pressure.
Trap 3: ionisation vs excitation. Excitation energy lifts the electron from the ground state to a higher level. Ionisation energy removes it fully to n = ∞ from wherever it sits. An electron in n = 2 has energy −3.4 eV, so its ionisation energy is +3.4 eV, not 13.6 eV. The word "ionisation" makes students reach for 13.6 by reflex.
| 🧠 Bohr scaling, in one line each | |
|---|---|
| Radius | rₙ = 0.529 (n²/Z) Å. Grows as n², shrinks as Z. |
| Speed | vₙ ∝ Z/n. Faster for higher Z, slower for higher n. |
| Energy | Eₙ = −13.6 (Z²/n²) eV. Also K = +13.6 Z²/n², U = −27.2 Z²/n². |
| Series | Lyman is UV, Balmer is visible, Paschen and beyond are infrared. |
A worked NEET-style question
Q. An alpha particle of kinetic energy K heads toward a gold nucleus of atomic number Z. Its distance of closest approach is d. The kinetic energy is now doubled and a lighter nucleus of atomic number Z/2 is used. What is the new distance of closest approach?
(A) d (B) d/2 (C) d/4 (D) 2d
Solution. Set kinetic energy equal to electrostatic potential energy at the turning point. This gives d ∝ Z/K. The new atomic number is Z/2 and the new kinetic energy is 2K. So d′ ∝ (Z/2)/(2K), which is one quarter of Z/K. Both changes push in the same direction, so the distance drops by a factor of four. Choice (B) is the trap for anyone who applies only one of the two changes.
Answer: (C)
Chapter 3: Nuclei
This is the most calculation-heavy chapter in the block. It rests on two pillars. Mass turns into energy, which drives binding energy and the fission Q-value. Decay is a matter of probability over time, which drives half-life and the decay law.
What NEET asks, year by year
| Year | Qs | Sub-topics tested |
|---|---|---|
| Re-NEET 2026 | 2 | Q-value of a reaction; mass defect and constant nuclear density |
| NEET 2026 | 3 | Nuclear radius vs mass number; binding-energy gain in fission; alpha-beta cascade |
| NEET 2025 | 1 | Fraction of a sample left after several half-lives |
| NEET 2024 | 1 | Reaction products in alpha decay |
| NEET 2023 | 2 | Activity calculation; binding energy per nucleon |
| NEET 2022 | 3 | Energy in fusion; decay graph; nuclear-force properties |
| NEET 2021 | 2 | Binding-energy gain (A=240 → 120); half-life fraction |
| NEET 2020 | 1 | Isotope bombardment and neutron release |
| NEET 2019 | 1 | Composition of an alpha particle |
| NEET 2018 | 1 | Time for a set number of nuclei to decay |
| NEET 2017 | 0 | Weight shifted to Optics and current electricity |
| NEET 2016 | 0 | No direct question |
| NEET 2015 | 2 | Half-life and mean-life link; nuclear radius |
The traps NEET repeats here
Trap 1: total binding energy vs per-nucleon, the fission classic. This is the most famous number trap in the block. It appeared in 2016 and 2021. A nucleus of A = 240 has binding energy 7.6 MeV per nucleon. It splits into two fragments of A = 120, each at 8.5 MeV per nucleon. The energy released is asked. Students subtract 8.5 − 7.6 = 0.9 MeV. That is wrong. Energy comes from total binding energy. Multiply by the nucleon count: 240 × (8.5 − 7.6) = 216 MeV. The per-nucleon gain must be scaled by all 240 nucleons.
Trap 2: half-life vs mean life. They are not equal. Half-life T½ = 0.693/λ. Mean life τ = 1/λ. So T½ = 0.693 τ. Ask what fraction is left after one mean life and students say 50%. The real answer is e⁻¹, about 37% left, so 63% has decayed.
Trap 3: density is independent of mass number. Nuclear radius follows R = R₀A^(1/3). So volume grows as A, and density is mass over volume, which is A over A, a constant. Every nucleus has about the same density. NTA tests this as a concept line, and students who only memorised the radius formula miss it.
| ⚡ Last-night nuclei checklist | |
|---|---|
| Fission energy | Q = total nucleons × gain in BE per nucleon. Never subtract per-nucleon values directly. |
| Decay law | N = N₀ (½)^(t/T½). Count half-lives first, then apply. |
| After one mean life | 37% remains, 63% has decayed. Not 50%. |
| Density | Same for all nuclei. R = R₀A^(1/3), so ρ does not depend on A. |
| Nuclear force | Short-ranged, charge-independent, shows saturation. Common assertion-reason line. |
A worked NEET-style question
Q. A radioactive sample decays by alpha emission. Its initial activity is A₀. After a time equal to three half-lives, what is the activity, and what fraction of the original nuclei has decayed?
(A) A₀/8, 12.5% (B) A₀/6, 87.5% (C) A₀/8, 87.5% (D) A₀/3, 33.3%
Solution. Activity tracks the number of nuclei left. After three half-lives the fraction left is (½)³ = 1/8, so the activity is A₀/8. That removes (B) and (D). The fraction left is 12.5%, so the fraction decayed is 1 − 1/8 = 7/8, which is 87.5%. Choice (A) is the trap for anyone who reports the fraction remaining instead of the fraction decayed, which is the last clause of the question.
Answer: (C)
The formula and constant sheet you need cold
NEET gives you about a minute a question. Deriving a constant mid-paper is a waste of that minute. Have these at reflex speed.
| Quantity | Value or formula | Note |
|---|---|---|
| Photon energy shortcut | hc = 1240 eV·nm | Fastest way to energy from wavelength. |
| Einstein equation | eV₀ = hc/λ − φ₀ | Keep every term in eV before subtracting. |
| de Broglie, electron | λ = 1.227/√V nm | Electron only. Others use λ = h/√(2mqV). |
| Bohr energy | Eₙ = −13.6 Z²/n² eV | K = −E, U = 2E. |
| Bohr radius | rₙ = 0.529 n²/Z Å | Ground-state radius is 0.529 Å. |
| Rydberg formula | 1/λ = RZ²(1/n₁² − 1/n₂²) | R = 1.097 × 10⁷ m⁻¹. Shortest λ is the largest jump. |
| Mass-energy unit | 1 u = 931.5 MeV/c² | Convert mass defect straight to MeV. |
| Nuclear radius | R = R₀A^(1/3), R₀ = 1.2 fm | Density is constant across nuclei. |
| Decay law | N = N₀(½)^(t/T½) | T½ = 0.693 τ, where τ = 1/λ. |
What to expect in NEET 2027 and 2028
The pattern across the decade, plus the two 2026 papers, points to four clear moves. Prepare for these, not for deleted content.
1. More Assertion-Reason from NCERT lines. NTA is pushing Statement I and Statement II formats. Read the NCERT prose, not just the formulas. High-probability targets: the properties of nuclear forces, why classical wave theory fails to explain the photoelectric effect, and the exact alpha-scattering wording.
2. Graphs over long algebra. Expect the V₀ vs ν graph, where the slope is h/e and the x-intercept is ν₀. Expect the binding-energy curve, peaking near iron at about 8.8 MeV per nucleon, which explains why both fission and fusion release energy.
3. A mechanics step joined to a modern step. A likely 2027 shape: an electron accelerated from rest in a field. First find the speed with kinematics, then the de Broglie wavelength with λ = h/mv. One classical step, one quantum step.
4. Nothing from deleted content. No Davisson-Germer accelerating voltage. No isobar-isotone definitions. No X-ray spectra. Every minute spent there is a minute lost. Confirm the cut against your current NCERT edition.
| 🎯 Drill the whole decade of Modern Physics PYQs, sorted by year and trap. | |
|---|---|
| Reading a trap once is not the same as beating it under exam pressure. Logic Bloom's Playground has every NEET Modern Physics PYQ from 2015 to 2026, tagged by chapter and by the exact trap it hides. Miss the intensity trap or the fission trap and TarQ walks you through the fix, then your Mistake Book brings it back until it sticks. Deleted topics are stripped out, so you never waste a rep. | Get the app → Free to start. |
12 must-attempt NEET Modern Physics PYQs, with the traps named
These twelve cover the real spread of the block, from single-correct to assertion-reason to multi-statement. Each is written at NEET exam level. Try each before reading the answer.
1. Light of wavelength λ ejects photoelectrons of maximum kinetic energy K from a metal. When the wavelength is changed to λ/2, the maximum kinetic energy becomes 3K. What is the work function of the metal in terms of hc and λ?
(A) hc/λ (B) hc/2λ (C) 2hc/λ (D) hc/3λ
Work. Two equations: K = hc/λ − φ and 3K = 2hc/λ − φ. Subtract the first from the second: 2K = hc/λ, so K = hc/2λ. Put K back into the first: φ = hc/λ − hc/2λ = hc/2λ.
Answer: (B). Trap: students set φ = hc/λ by dropping the K term.
2. An electron and a proton are accelerated from rest through the same potential difference. What is the ratio of their de Broglie wavelengths, λ(electron) to λ(proton)?
(A) 1 (B) √(mₚ/mₑ) (C) mₚ/mₑ (D) √(mₑ/mₚ)
Work. For a charged particle from rest, λ = h/√(2mqV). Same q and same V, so λ ∝ 1/√m. The ratio is √(mₚ/mₑ). Since the proton is heavier, the electron has the longer wavelength, which matches a ratio above 1.
Answer: (B). Trap: choice (D) inverts the mass ratio.
3. In a photoelectric experiment the stopping potential is 1.5 V for incident light of 400 nm. The work function of the metal is closest to (take hc = 1240 eV·nm):
(A) 1.6 eV (B) 1.9 eV (C) 3.1 eV (D) 4.6 eV
Work. Photon energy = 1240/400 = 3.1 eV. Stopping potential gives K = 1.5 eV. So φ = 3.1 − 1.5 = 1.6 eV.
Answer: (A). Trap: choice (C) reports the photon energy, forgetting to subtract K.
4. The electron in a hydrogen atom jumps from n = 4 to n = 2. Which series does the emitted line belong to, and is it visible?
(A) Lyman, UV (B) Balmer, visible (C) Paschen, infrared (D) Balmer, UV
Work. Any transition ending at n = 2 is a Balmer line. The Balmer series lies in the visible range. So the line is Balmer and visible.
Answer: (B). Trap: choice (A) confuses the base level with Lyman, which ends at n = 1.
5. A nucleus of mass number 240 and binding energy 7.6 MeV per nucleon fissions into two equal fragments of 8.5 MeV per nucleon. The energy released is:
(A) 0.9 MeV (B) 108 MeV (C) 216 MeV (D) 2040 MeV
Work. Energy released = total final binding energy − total initial. Total initial = 240 × 7.6 = 1824 MeV. Total final = 2 × 120 × 8.5 = 2040 MeV. Released = 2040 − 1824 = 216 MeV. Shortcut: 240 × (8.5 − 7.6) = 216 MeV.
Answer: (C). Trap: choice (A) subtracts the per-nucleon values without scaling.
6. The potential energy of the electron in the first excited state of hydrogen is:
(A) −3.4 eV (B) −6.8 eV (C) +3.4 eV (D) +6.8 eV
Work. The first excited state is n = 2, so total energy E = −13.6/4 = −3.4 eV. For a Bohr orbit U = 2E, which gives U = −6.8 eV.
Answer: (B). Trap: choice (A) stops at the total energy and never applies U = 2E.
7. After what fraction of its mean life does a radioactive sample decay to 1/e of its initial number?
(A) 0.693 mean lives (B) 1 mean life (C) 1.44 mean lives (D) 2 mean lives
Work. N = N₀ e^(−λt) and mean life τ = 1/λ. Setting N = N₀/e means e^(−λt) = e⁻¹, so λt = 1, so t = 1/λ = τ. That is exactly one mean life.
Answer: (B). Trap: choice (A) confuses mean life with half-life, since half-life is 0.693 τ.
8. Two statements are given. Statement I: The stopping potential in a photoelectric experiment does not depend on the intensity of incident light. Statement II: Intensity changes the number of photons per second but not the energy of each photon. Choose the correct option.
(A) Both true, II explains I (B) Both true, II does not explain I (C) I true, II false (D) I false, II true
Work. Stopping potential is set by the maximum kinetic energy, which depends on photon energy, so on frequency, not intensity. Statement I is true. Intensity only changes the photon count per second, which is Statement II, and that is exactly why the stopping potential is fixed. So II explains I.
Answer: (A). Trap: choice (B) accepts both facts but misses that II is the reason for I.
9. The shortest wavelength of the Lyman series is λ. What is the shortest wavelength of the Balmer series in terms of λ?
(A) 2λ (B) 4λ (C) λ/4 (D) λ/2
Work. The series limit uses 1/λ = R(1/n₁² − 0), so 1/λ = R/n₁². For Lyman n₁ = 1, so 1/λ_L = R. For Balmer n₁ = 2, so 1/λ_B = R/4, giving λ_B = 4/R = 4λ_L. So the Balmer limit is 4λ.
Answer: (B). Trap: choice (C) inverts the ratio.
10. How many of the following statements about nuclei are correct? (i) Nuclear density is the same for all nuclei. (ii) Nuclear force is charge-independent. (iii) Binding energy per nucleon is highest for the heaviest nuclei. (iv) Nuclear force shows saturation.
(A) One (B) Two (C) Three (D) Four
Work. Statement (i) is true, since R = R₀A^(1/3) keeps density constant. Statement (ii) is true. Statement (iii) is false, because binding energy per nucleon peaks near iron, not at the heaviest nuclei. Statement (iv) is true. So three are correct.
Answer: (C). Trap: (iii) tempts students who remember only that heavy nuclei have large total binding energy.
11. An electron is accelerated from rest through 100 V. Its de Broglie wavelength is closest to (take λ = 1.227/√V nm):
(A) 0.123 nm (B) 0.0123 nm (C) 1.227 nm (D) 12.27 nm
Work. λ = 1.227/√100 = 1.227/10 = 0.1227 nm, which rounds to 0.123 nm.
Answer: (A). Trap: choice (C) forgets the √V in the denominator entirely.
12. A radioactive nucleus ⁸⁸²³⁸X emits one alpha particle and then two beta-minus particles in sequence. What are the atomic number and mass number of the final nucleus?
(A) Z = 88, A = 234 (B) Z = 86, A = 234 (C) Z = 90, A = 234 (D) Z = 88, A = 238
Work. Alpha emission drops Z by 2 and A by 4, giving Z = 86, A = 234. Each beta-minus raises Z by 1 and leaves A unchanged. Two of them raise Z back by 2, to Z = 88, with A still 234.
Answer: (A). Trap: choice (B) stops after the alpha step and forgets the two beta emissions.
How to actually clear this block
The plan is short because the block is short. Learn the three lead formulas until they are reflex: the photoelectric equation, the de Broglie forms, and the decay law. Then drill the six named traps until you spot them in the stem, not after the mistake. That is the whole game.
Reading traps in an article is step one. Beating them under a running clock is step two, and that only comes from reps. That is what Logic Bloom's Playground is built for: every PYQ mapped by year and trap, TarQ to teach the fix the moment you slip, and a Mistake Book that brings your weak spots back until they hold.
| 🚀 Turn this analysis into a real score. | |
|---|---|
| You now know where the marks sit and which traps to fear. Close the gap between knowing and scoring. Logic Bloom's Playground drills every NEET Modern Physics PYQ of the decade, and Battleground lets you test the block in 1v1 duels against other aspirants. Understand through games. Score through practice. | Get the app → Free to start. |