Modern physics for JEE Main

Modern physics is often called the best return on time in JEE Main Physics. The formulas are few, the questions are mostly direct substitution, and the same half-dozen ideas keep coming back. The risk is carelessness with units: electron-volts against joules, nanometres against metres. A couple of shortcuts, used consistently, remove most of it.

The official 2026 syllabus has three units here. Dual nature of matter and radiation: the photoelectric effect, Hertz's and Lenard's observations, Einstein's equation, the particle nature of light, matter waves and the de Broglie relation. Atoms: alpha-particle scattering, Rutherford's model, the Bohr model, energy levels and the hydrogen spectrum. Nuclei: composition and size, atomic masses, mass–energy relation, mass defect, binding energy per nucleon and its variation with mass number, and fission and fusion.

Two shortcuts

• E (in eV) ≈ 1240/λ (in nm) for a photon.
• ke² ≈ 1.44 MeV fm (1 fm = 10⁻¹⁵ m) for nuclear distances.

Also: 1 eV = 1.6 × 10⁻¹⁹ J, and h ≈ 6.63 × 10⁻³⁴ J s.

The photoelectric effect

Light falling on a metal can eject electrons. The experiments of Hertz and Lenard, and later careful measurements, showed four things that the wave picture cannot explain:

• Below a threshold frequency ν₀, no electrons come out, however bright the light.
• Above it, emission is instantaneous.
• The maximum kinetic energy of the electrons depends on the frequency, not the intensity.
• The photocurrent (number of electrons per second) is proportional to the intensity.

Einstein's explanation: light arrives in photons of energy hν. One photon gives all its energy to one electron, which spends at least the work function φ to escape:

hν = φ + KE_max, and KE_max = eV₀

where V₀ is the stopping potential. The threshold frequency is ν₀ = φ/h.

A photon also carries momentum p = h/λ = E/c.

Matter waves

De Broglie proposed that every moving particle has a wavelength

λ = h/p = h/√(2mK)

For an electron accelerated from rest through V volts, K = eV, which gives λ ≈ 1.227/√V nm. For particles with the same kinetic energy, λ ∝ 1/√m. For the same speed, λ ∝ 1/m.

Rutherford scattering and the Bohr model

In the alpha-scattering experiment, most alpha particles passed straight through a gold foil, but a few bounced back through large angles. Rutherford concluded that the positive charge and nearly all the mass sit in a tiny nucleus.

For an alpha particle (charge 2e) fired head-on at a nucleus of charge Ze with kinetic energy K, all of K turns into potential energy at the distance of closest approach:

r₀ = k(2e)(Ze)/K = 2kZe²/K

Rutherford's model could not explain why electrons do not spiral into the nucleus. Bohr added two postulates: electrons move in stationary orbits with angular momentum mvr = nh/(2π), and they emit or absorb a photon only when they jump between orbits, with hν = E_upper − E_lower. The quantisation condition is the same as fitting a whole number of de Broglie wavelengths round the orbit, 2πr = nλ.

For a hydrogen-like atom of atomic number Z:

Hydrogen spectrum. 1/λ = RZ²(1/n₁² − 1/n₂²), with R ≈ 1.097 × 10⁷ m⁻¹.

• Lyman series (n₁ = 1): ultraviolet
• Balmer series (n₁ = 2): visible
• Paschen (n₁ = 3), Brackett (4), Pfund (5): infrared

From level n, a group of atoms can emit n(n − 1)/2 different lines.

The nucleus

A nucleus of mass number A has Z protons and A − Z neutrons. Its radius is R = R₀∛A, with R₀ ≈ 1.2 fm. Volume is proportional to A, so nuclear density is roughly the same for all nuclei, about 2.3 × 10¹⁷ kg/m³.

Mass defect and binding energy. A nucleus weighs less than its separate nucleons:

Δm = [Z m_p + (A − Z) m_n] − M_nucleus, and BE = Δm c², with 1 u ≈ 931.5 MeV.

Using atomic masses (with the hydrogen atom's mass in place of the proton's) gives the same result, because the electron masses cancel.

Binding energy per nucleon is low for light nuclei, rises to a peak of about 8.8 MeV near A ≈ 56 (iron), and falls slowly to about 7.6 MeV for uranium. Both directions towards the peak release energy:

• Fission: a heavy nucleus such as uranium-235 splits into two middle-sized ones, releasing about 200 MeV per event.
• Fusion: light nuclei such as hydrogen isotopes join. This powers the Sun and needs very high temperatures to overcome the repulsion between nuclei.

The nuclear force is strong, short-ranged (a few femtometres) and roughly the same between any two nucleons.

Worked problems

Problem 1: photoelectric equation. Light of wavelength 310 nm falls on a metal of work function 2 eV. Find the stopping potential and the threshold wavelength.

E = 1240/310 = 4 eV. KE_max = 4 − 2 = 2 eV, so V₀ = 2 V.
Threshold: λ₀ = 1240/2 = 620 nm.

Problem 2: a Balmer line. Find the wavelength and photon energy for the hydrogen transition from n = 3 to n = 2.

1/λ = R(1/4 − 1/9) = 5R/36, so λ = 36/(5 × 1.097 × 10⁷) ≈ 6.56 × 10⁻⁷ m = 656 nm (red).
E = 13.6 × 5/36 ≈ 1.89 eV; check 1240/656 ≈ 1.89.

Problem 3: binding energy of helium-4. Atomic masses: hydrogen 1.007825 u, neutron 1.008665 u, helium-4 4.002603 u. Find the binding energy and the binding energy per nucleon.

2(1.007825) + 2(1.008665) = 4.032980 u. Δm = 4.032980 − 4.002603 = 0.030377 u.
BE = 0.030377 × 931.5 ≈ 28.3 MeV, so about 7.07 MeV per nucleon.

Problem 4 (numerical answer): closest approach. An alpha particle of kinetic energy 7.2 MeV heads straight for a tin nucleus (Z = 50). Find the distance of closest approach in fm (take ke² = 1.44 MeV fm).

r₀ = 2Z × ke²/K = (100 × 1.44)/7.2 = 20 fm.

Practice set

1. How many spectral lines can hydrogen atoms emit when electrons fall from n = 4 to the ground state?
2. A proton and an alpha particle have the same kinetic energy. Find the ratio of their de Broglie wavelengths (proton : alpha).
3. Find the ionisation energy of He⁺ in its ground state.
4. The intensity of light on a photocell is doubled at the same frequency. What happens to the stopping potential and the saturation current?
5. Find the ratio of the radii of nuclei with mass numbers 27 and 125.
6. Find the de Broglie wavelength of an electron accelerated through 100 V.
7. Find the ratio of the shortest wavelengths of the Lyman and Balmer series.
8. Find the radius of the third Bohr orbit in hydrogen.

Answers

1. 4 × 3/2 = 6.
2. λ ∝ 1/√m, so √(4/1) = 2 : 1.
3. 13.6 × 2² = 54.4 eV.
4. Stopping potential unchanged; saturation current doubles.
5. ∛27 : ∛125 = 3 : 5.
6. 1.227/√100 ≈ 0.123 nm.
7. Shortest Lyman: 1/λ = R. Shortest Balmer: 1/λ = R/4. Ratio 1 : 4.
8. 0.529 × 3² ≈ 4.76 Å.

What to do next

• Learn the two shortcuts (1240/λ and 1.44 MeV fm) and use them in every problem this week.
• Draw the three photoelectric graphs and say what shifts when intensity or frequency changes.
• Solve 20 Bohr-model problems for H, He⁺ and Li²⁺, covering radius, speed, energy and wavelength.
• Move on to semiconductors, and revise wave optics for the wave–particle contrast.