Why this chapter has the best marks-per-hour ratio
Compare it with Rotational Motion. There, a single question can need a moment of inertia, an energy equation and the rolling condition, and it builds on Kinematics and Laws of Motion before that. Here, a question typically needs one relation applied once, and the topics within the chapter barely depend on each other - you can learn radioactivity without having understood the photoelectric effect.
That independence is what makes it the right chapter to prioritise late. It also overlaps directly with Atomic Structure in Chemistry, so the Bohr work is done twice for the price of once.
How JEE actually asks Modern Physics
NTA does not publish chapter-wise weightage, so figures online are coaching estimates from past papers; confirm the syllabus and pattern for your session in the official NTA information bulletin.
Past papers show short, self-contained questions across five areas: the photoelectric equation, de Broglie wavelengths, Bohr energy levels and spectral transitions, radioactive decay and nuclear energy, and basic semiconductor behaviour. Conceptual questions - why intensity does not change photoelectron energy, why the Bohr formula fails for helium - appear at least as often as numerical ones.
Key concepts, compressed
- ✓Light delivers energy in quanta of E = hf. This explains the frequency threshold and the absence of any time lag in photoemission, both of which the wave model fails to predict.
- ✓The photoelectric equation hf = φ + KE(max) is energy conservation: the photon energy pays the work function first, and whatever remains becomes kinetic energy.
- ✓Matter has a wavelength λ = h/p. It is measurable for electrons and utterly negligible for everyday objects because h is so small.
- ✓Bohr quantised angular momentum, giving discrete radii and energies. Transitions between levels emit or absorb photons of exactly the energy difference.
- ✓Nuclear binding energy comes from mass defect via E = Δmc². Energy is released whenever the products have higher binding energy per nucleon than the reactants.
- ✓Radioactive decay is random per nucleus but exponential in bulk, characterised by the decay constant λ or equivalently the half-life.
Formulas you need before attempting the questions
| Quantity | Formula | Note |
|---|---|---|
| Photon energy | E = hf = hc/λ | h = 6.63 × 10⁻³⁴ J·s |
| Photoelectric equation | hf = φ + KE(max) | |
| Stopping potential | eV₀ = KE(max) | |
| Threshold frequency | f₀ = φ/h | |
| de Broglie wavelength | λ = h/p = h/mv | |
| Electron accelerated through V | λ = 12.27/√V Å | V in volts |
| Uncertainty principle | Δx · Δp ≥ h/4π | |
| Bohr energy | E(n) = −13.6 Z²/n² eV | hydrogen-like only |
| Bohr radius | r(n) = 0.529 n²/Z Å | |
| Rydberg equation | 1/λ = R Z²(1/n₁² − 1/n₂²) | |
| Radioactive decay | N = N₀e^(−λt) | |
| Half-life | t½ = 0.693/λ | |
| Mass-energy | 1 u = 931.5 MeV |
The five mistakes that cost the most marks
- ✓Using E(n) = −13.6/n² for multi-electron atoms. It applies only to hydrogen and one-electron ions such as He⁺ and Li²⁺, where the Z² factor matters.
- ✓Mixing electronvolts and joules mid-calculation. Convert once at the start and stay in one unit throughout.
- ✓Believing brighter light gives photoelectrons more energy. Intensity changes the number emitted; only frequency changes their maximum kinetic energy.
- ✓Confusing the work function with the stopping potential. The work function is a property of the metal alone; the stopping potential also depends on the incident frequency.
- ✓Treating radioactive decay as linear. After two half-lives one quarter remains, not zero - a fixed fraction is lost per interval, not a fixed amount.
Practice set 1: the photoelectric effect
1. What is the energy of a photon of frequency 10¹⁵ Hz?
6.63 × 10⁻¹⁹ J, which is about 4.14 eV. E = hf = 6.63 × 10⁻³⁴ × 10¹⁵ = 6.63 × 10⁻¹⁹ J. Dividing by 1.6 × 10⁻¹⁹ converts to electronvolts. Decide at the start which unit you will work in - most photoelectric questions are cleaner in eV.
2. What is the energy of a photon of wavelength 400 nm?
About 4.97 × 10⁻¹⁹ J, or roughly 3.11 eV. E = hc/λ = (6.63 × 10⁻³⁴ × 3 × 10⁸)/(4 × 10⁻⁷) = 1.989 × 10⁻²⁵/(4 × 10⁻⁷) ≈ 4.97 × 10⁻¹⁹ J. Converting the nanometres to metres before substituting is essential - leaving λ as 400 gives an answer wrong by nine orders of magnitude.
3. Light of photon energy 3 eV falls on a metal of work function 2 eV. What is the maximum kinetic energy of the emitted electrons?
1 eV. The photoelectric equation gives KE(max) = hf − φ = 3 − 2 = 1 eV. Working entirely in electronvolts makes this a one-line subtraction. Note that this is the maximum: electrons originating deeper in the metal lose energy on the way out and emerge with less.
4. What is the stopping potential in that case?
1 V. The stopping potential is defined by eV₀ = KE(max), so with KE(max) = 1 eV the stopping potential is exactly 1 V. This numerical coincidence is why electronvolts are convenient here: the stopping potential in volts equals the maximum kinetic energy in electronvolts.
5. What is the threshold frequency for a metal of work function 2 eV?
About 4.83 × 10¹⁴ Hz. First convert: φ = 2 × 1.6 × 10⁻¹⁹ = 3.2 × 10⁻¹⁹ J. Then f₀ = φ/h = 3.2 × 10⁻¹⁹/(6.63 × 10⁻³⁴) ≈ 4.83 × 10¹⁴ Hz. Below this frequency no electrons are emitted whatever the intensity, which is the observation that forced the photon model.
6. What happens to the photocurrent and the electron energy when the light intensity is doubled at fixed frequency?
The photocurrent doubles; the maximum kinetic energy is unchanged. Intensity determines how many photons arrive per second and therefore how many electrons are ejected, while each individual electron's energy depends only on the energy of the single photon that freed it. This separation is the central result of the photoelectric effect.
7. Why is there no measurable time lag in photoemission, even at very low intensity?
Because a single photon transfers its whole energy to one electron in a single interaction, so emission begins as soon as the first suitable photon arrives. The wave model predicts the electron must accumulate energy gradually, implying a delay at low intensity that is not observed. This absence of a lag is one of the strongest pieces of evidence for quantisation.
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Practice set 2: matter waves and uncertainty
8. What is the de Broglie wavelength of an electron accelerated through 100 V?
About 1.227 Å, or 1.227 × 10⁻¹⁰ m. Using the standard shortcut λ = 12.27/√V Å = 12.27/√100 = 12.27/10 = 1.227 Å. This is comparable to atomic spacing in crystals, which is exactly why electron diffraction works and why the Davisson-Germer experiment could confirm matter waves.
9. A particle of mass 10⁻³⁰ kg moves at 10⁶ m/s. What is its de Broglie wavelength?
6.63 × 10⁻¹⁰ m. λ = h/mv = 6.63 × 10⁻³⁴/(10⁻³⁰ × 10⁶) = 6.63 × 10⁻³⁴/10⁻²⁴ = 6.63 × 10⁻¹⁰ m. Computing the momentum first as a single number keeps the powers of ten manageable.
10. Two particles move at the same speed but one is heavier. Which has the longer de Broglie wavelength?
The lighter one. Since λ = h/mv, wavelength is inversely proportional to mass at fixed speed. This is why wave behaviour is observable for electrons and entirely undetectable for a cricket ball - the ball's wavelength is smaller than any length that could ever be measured.
11. State the uncertainty principle and what it actually asserts.
Δx · Δp ≥ h/4π: position and momentum cannot both be known to arbitrary precision simultaneously. It is a statement about nature rather than about instruments - the limitation would remain with perfect equipment, because a particle does not possess exact values of both quantities at once.
12. What is meant by wave-particle duality?
That matter and radiation both display wave and particle behaviour depending on the experiment. Light behaves as a wave in interference and diffraction but as particles in the photoelectric effect; electrons behave as particles in a cathode-ray tube but produce diffraction patterns in a crystal. No single classical picture accounts for both sets of observations.
Practice set 3: the Bohr model and atomic spectra
13. What is the energy of the electron in the n = 2 level of a hydrogen atom?
−3.4 eV. E(n) = −13.6/n² = −13.6/4 = −3.4 eV. The negative sign means the electron is bound; zero energy corresponds to a free electron at rest infinitely far away. Levels become more closely spaced as n increases, converging on zero.
14. What is the ionisation energy of hydrogen from the ground state?
13.6 eV. Ionisation means moving the electron from n = 1, where E = −13.6 eV, to n = ∞, where E = 0. The energy required is the difference, 13.6 eV. This single number anchors the whole chapter and is worth knowing without calculation.
15. What is the energy of the photon emitted in a transition from n = 3 to n = 2 in hydrogen?
About 1.89 eV. E₃ = −13.6/9 ≈ −1.51 eV and E₂ = −13.6/4 = −3.4 eV, so the emitted photon carries the difference: −1.51 − (−3.4) = 1.89 eV. This is the first line of the Balmer series and lies in the visible red, which is why hydrogen discharge tubes glow pink.
16. What is the radius of the n = 2 orbit in hydrogen?
About 2.116 Å. r(n) = 0.529 n²/Z Å = 0.529 × 4/1 = 2.116 Å. Radius grows as n² while energy grows as 1/n², so higher orbits are much larger and much less tightly bound - which is why outer electrons are the ones involved in chemistry.
17. What is the ground-state energy of a singly ionised helium atom, He⁺?
−54.4 eV. He⁺ has one electron, so the hydrogen-like formula applies with Z = 2: E = −13.6 × 2²/1² = −54.4 eV. It is four times more tightly bound than hydrogen. Note carefully that neutral helium, with two electrons, cannot be treated this way at all.
18. Which spectral series of hydrogen lies in the visible region?
The Balmer series, which consists of transitions ending at n = 2. The Lyman series ends at n = 1 and lies in the ultraviolet, while the Paschen series ends at n = 3 and lies in the infrared. Remembering which level each series terminates on is enough to place all three.
19. Why does the Bohr model fail for multi-electron atoms?
Because it ignores electron-electron repulsion and treats the electron as a particle in a definite orbit. With more than one electron, each is screened from the nucleus by the others by an amount that cannot be captured in a single Z. The model succeeds only for hydrogen and one-electron ions, and this limitation is asked about directly.
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Practice set 4: nuclear physics
20. What fraction of a radioactive sample remains after three half-lives?
One eighth. Each half-life leaves half of what was there, so three successive halvings give (1/2)³ = 1/8, which is 12.5%. Answering zero, or subtracting halves to reach a negative amount, comes from treating decay as linear - it is exponential, so a fixed fraction is lost each interval, not a fixed amount.
21. A radioactive isotope has a half-life of 10 days. What is its decay constant?
About 0.0693 per day. λ = 0.693/t½ = 0.693/10 = 0.0693 day⁻¹. The relation follows from setting N = N₀/2 in N = N₀e^(−λt), since ln2 ≈ 0.693. Keep the units consistent: a half-life in days gives a decay constant per day.
22. How much energy corresponds to a mass defect of 1 atomic mass unit?
931.5 MeV. This conversion comes from E = Δmc² with 1 u = 1.66 × 10⁻²⁷ kg, and it is used constantly in binding-energy calculations. Knowing it removes the need to carry c² through every nuclear problem.
23. Which nucleus has the highest binding energy per nucleon, and why does it matter?
Iron-56, at about 8.8 MeV per nucleon. It matters because it sets the direction of energy release: nuclei lighter than iron release energy by fusing, and nuclei heavier than iron release energy by fissioning. Both processes move products toward the peak of the binding-energy curve.
24. Why do both fission and fusion release energy despite being opposite processes?
Because both move nuclei toward higher binding energy per nucleon. Fusing very light nuclei climbs the steep left side of the curve; splitting very heavy nuclei climbs the gentle right side. The energy released is the difference in total binding energy, appearing as a mass defect between reactants and products.
25. How do the mass number and atomic number change in alpha decay?
The mass number falls by 4 and the atomic number by 2, because an alpha particle is a helium nucleus with two protons and two neutrons. So uranium-238 alpha-decaying becomes thorium-234. Conservation of both A and Z is what lets you complete any decay equation.
26. How do they change in beta-minus decay?
The mass number is unchanged and the atomic number increases by one. A neutron converts into a proton, emitting an electron and an antineutrino, so the nucleon count stays the same while the charge rises. The daughter is the next element in the periodic table.
Practice set 5: semiconductors
27. What is the difference between intrinsic and extrinsic semiconductors?
An intrinsic semiconductor is chemically pure, with electrons and holes always equal in number since each is created in a pair. An extrinsic semiconductor has been deliberately doped with impurity atoms to make one carrier type dominant, which increases its conductivity by orders of magnitude and is what makes devices possible.
28. What does doping silicon with a pentavalent element produce?
An n-type semiconductor. Four of the dopant's five valence electrons form covalent bonds and the fifth is loosely bound, so electrons become the majority carriers. Phosphorus and arsenic are the standard dopants. Crucially, the material stays electrically neutral - each donor atom contributes a free electron and an equal fixed positive charge.
29. What does doping with a trivalent element produce?
A p-type semiconductor, in which holes are the majority carriers. A trivalent atom such as boron has only three valence electrons, leaving one bond incomplete - a hole that neighbouring electrons can move into. Again the material remains neutral overall; the holes are absences of electrons, not added positive charges.
30. What happens to a p-n junction diode in forward bias?
The depletion region narrows, the potential barrier falls, and current flows readily once the applied voltage exceeds the barrier - roughly 0.7 V for silicon. In reverse bias the depletion region widens and only a tiny leakage current flows. This asymmetry is what makes the diode a rectifier.
How to study this chapter efficiently
- ✓Decide your unit before starting each question. Photoelectric and Bohr problems are almost always cleaner in electronvolts; nuclear problems usually need MeV and the 931.5 conversion.
- ✓Memorise three anchor numbers: 13.6 eV for hydrogen ionisation, 931.5 MeV per u, and 12.27/√V Å for accelerated electrons. They shortcut a large fraction of the questions.
- ✓Note explicitly where each formula stops applying. The Bohr expressions fail for multi-electron atoms, and that boundary is examined directly.
- ✓Practise the conceptual questions, not just the numerical ones. Why intensity does not change electron energy, and why there is no time lag, appear as often as any calculation.
- ✓Prepare this chapter late if you are short of time. It is nearly independent of the rest of the syllabus, so it can be secured in isolation without leaving gaps.
- ✓Study it alongside Atomic Structure in Chemistry - the Bohr model, de Broglie relation and hydrogen spectrum are shared between the two papers.
Turn this into active practice
Because the questions here are short, this chapter rewards volume more than most. You can work through a large number of items quickly, and each one tests a distinct idea rather than a long chain of reasoning - which makes weak spots unusually easy to isolate.
The JEE Modern Physics quiz on QUFF generates fresh questions across the photoelectric effect, matter waves, the Bohr model, nuclear physics and semiconductors, marks them instantly and explains each answer. Do a timed set and check whether your errors cluster in one area; if they do, that is a single evening's work to fix.