By Kumar Sir, Physics faculty at Pramana Institute (IIT Kanpur, 25 years teaching, ex-Allen)
Modern physics is the most frequent Physics chapter in JEE Main by published 10-year counts, at about 2.4 questions a year. It is also one of the easiest: almost every question is a direct formula with one or two numbers. Learn the shortcut constants below (1240 and 13.6 do most of the work) and you can answer these questions in under a minute each.
Photons and the photoelectric effect
| Quantity | Formula | Shortcut |
|---|---|---|
| Photon energy | E = hν = hc/λ | E (eV) = 1240/λ (nm) |
| Photon momentum | p = h/λ = E/c | |
| Einstein’s equation | Kₘₐₓ = hν − φ | φ = work function |
| Stopping potential | eV₀ = Kₘₐₓ | V₀ in volts equals Kₘₐₓ in eV |
| Threshold | ν₀ = φ/h, λ₀ = hc/φ | λ₀ (nm) = 1240/φ (eV) |
Matter waves
| Quantity | Formula | Shortcut |
|---|---|---|
| de Broglie wavelength | λ = h/p = h/mv = h/√(2mK) | |
| Charged particle through voltage V | λ = h/√(2mqV) | |
| Electron through V volts | λ = 1.227/√V nm | 100 V gives 0.1227 nm |
Atoms (hydrogen-like, atomic number Z)
| Quantity | Formula |
|---|---|
| Bohr’s quantisation | mvr = nh/2π |
| Orbit radius | rₙ = 0.529 n²/Z Å |
| Orbit speed | vₙ = 2.18 × 10⁶ Z/n m/s |
| Energy | Eₙ = −13.6 Z²/n² eV |
| Spectral lines | 1/λ = RZ²(1/n₁² − 1/n₂²), R = 1.097 × 10⁷ m⁻¹ |
| Number of lines from level n | n(n − 1)/2 |
| Closest approach (Rutherford) | r₀ = (1/4πε₀)(2Ze²/K), for an alpha particle of kinetic energy K |
| Series | Lower level n₁ | Region |
|---|---|---|
| Lyman | 1 | Ultraviolet |
| Balmer | 2 | Visible |
| Paschen | 3 | Infrared |
| Brackett | 4 | Infrared |
| Pfund | 5 | Infrared |
Nuclei
| Quantity | Formula |
|---|---|
| Nuclear radius | R = R₀A^(1/3), R₀ ≈ 1.2 fm |
| Nuclear density | Roughly the same for all nuclei, about 2.3 × 10¹⁷ kg/m³ |
| Mass defect | Δm = [Zmₚ + (A − Z)mₙ] − M |
| Binding energy | BE = Δm c², and 1 u = 931.5 MeV/c² |
Binding energy per nucleon peaks near A = 56 (iron). Heavy nuclei release energy by fission and light nuclei by fusion, because both move towards that peak.
Worked examples
1. Photoelectric effect. Light of 300 nm falls on a metal with φ = 2.3 eV. E = 1240/300 = 4.13 eV, so Kₘₐₓ = 4.13 − 2.3 = 1.83 eV and the stopping potential is 1.83 V.
2. Balmer line. For hydrogen, n = 3 → 2: 1/λ = R(1/4 − 1/9) = 5R/36, so λ = 36/(5 × 1.097 × 10⁷) m ≈ 656 nm, the red H-alpha line.
FAQs
Why does intensity not change Kₘₐₓ? Intensity changes the number of photons, not the energy of each. Kₘₐₓ depends only on frequency.
Is radioactive decay still in the syllabus? It was removed from the current NCERT book and the JEE Main syllabus; check the NEET notice and your board. Nuclear size, mass defect, binding energy, fission and fusion remain.
What does the 1240 shortcut mean? hc = 1240 eV nm, so a photon’s energy in eV is 1240 divided by its wavelength in nm.
Practise modern physics with weekly tests in a batch of at most 40 at Pramana, Geeta Bhawan Square, Indore. See our IIT JEE and NEET programs, or book free counselling.
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