JEE Physics — Modern Physics Deep Dive

Chapter Overview & Weightage

Modern Physics is the highest-weight Physics topic in JEE Main and Advanced. It includes photoelectric effect, atoms (Bohr model), nuclei, semiconductors, and de Broglie waves. JEE Main pulls 4 to 6 questions per paper, totalling 16 to 24 marks. JEE Advanced typically has 1 to 2 questions worth up to 8 marks. NEET pulls 3 to 4 questions.

This is a “scoring zone” chapter — the formulas are short, the concepts are limited, and most questions are direct substitutions.

Year	JEE Main Weightage	JEE Advanced
2024	20 marks	8 marks
2023	16 marks	6 marks
2022	20 marks	8 marks
2021	16 marks	4 marks
2020	24 marks	8 marks

Key Concepts You Must Know

Photoelectric effect: light is quantised; only photons above the threshold frequency $\nu_0$ eject electrons.
Einstein’s equation: $h\nu = \phi + KE_{\max}$ , where $\phi = h\nu_0$ is the work function.
Bohr model: electrons orbit in stationary states with quantised angular momentum $mvr = n\hbar$ .
Energy levels of hydrogen: $E_n = -13.6/n^2\,\text{eV}$ .
De Broglie wavelength: $\lambda = h/p$ . Every particle has wave nature.
Nuclear binding energy: $\Delta m c^2$ released when nucleons come together.
Radioactive decay: $N(t) = N_0 e^{-\lambda t}$ . Half-life $T_{1/2} = \ln 2 / \lambda$ .

Important Formulas

$h\nu = \phi + \frac{1}{2}m v_{\max}^2$

The stopping potential $V_0$ is given by $eV_0 = h\nu - \phi$ .

$E_n = -\frac{13.6}{n^2}\,\text{eV}$

For other hydrogenic atoms (single electron): $E_n = -13.6 \, Z^2/n^2\,\text{eV}$ .

$\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$

with $R = 1.097 \times 10^7\,\text{m}^{-1}$ .

$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_{\text{KE}}}}$

For an electron accelerated through potential $V$ : $\lambda = \frac{12.27}{\sqrt{V}}\,\text{Å}$ .

$E = \Delta m c^2$

$1\,\text{u} = 931.5\,\text{MeV/c}^2$ (memorise!).

$N(t) = N_0 e^{-\lambda t}, \quad T_{1/2} = \frac{\ln 2}{\lambda}, \quad T_{\text{mean}} = \frac{1}{\lambda} = 1.44 \, T_{1/2}$

Activity $A = \lambda N$ .

Solved Previous Year Questions

PYQ 1 (JEE Main 2024, 4 marks)

The work function of a metal is $2.5\,\text{eV}$ . Find the maximum kinetic energy of photoelectrons emitted by light of wavelength $400\,\text{nm}$ . (Use $hc = 1240\,\text{eV nm}$ .)

Solution:

Photon energy: $E = hc/\lambda = 1240/400 = 3.1\,\text{eV}$ .

$KE_{\max} = E - \phi = 3.1 - 2.5 = 0.6\,\text{eV}$ .

Answer: 0.6 eV.

PYQ 2 (JEE Main 2023, 4 marks)

An electron in the hydrogen atom transitions from $n = 4$ to $n = 2$ . Find the wavelength of the emitted photon.

Solution:

Energy emitted:

$\Delta E = -13.6\left(\frac{1}{16} - \frac{1}{4}\right) = -13.6 \times \left(-\frac{3}{16}\right) = 2.55\,\text{eV}$

$\lambda = \frac{hc}{\Delta E} = \frac{1240\,\text{eV nm}}{2.55\,\text{eV}} \approx 486\,\text{nm}$

This is the H-beta line in the Balmer series — visible blue-green.

PYQ 3 (JEE Advanced 2022)

Two radioactive nuclei A and B have decay constants $\lambda$ and $2\lambda$ . Initially, both samples have $N_0$ atoms. Find the time at which the ratio $N_A/N_B = e$ .

Solution:

$\frac{N_A}{N_B} = \frac{N_0 e^{-\lambda t}}{N_0 e^{-2\lambda t}} = e^{\lambda t}$

Setting this equal to $e$ : $\lambda t = 1 \implies t = 1/\lambda$ .

This is the mean lifetime of A — appears often in JEE Advanced.

Difficulty Distribution

Easy (direct formula): 8-12 marks per JEE Main paper
Medium (multi-step photoelectric or Bohr): 4-8 marks
Hard (semiconductors, multi-decay chains, de Broglie + Bohr combined): 4 marks in Main, 4-8 in Advanced

Expert Strategy

JEE Main Modern Physics is the closest thing to free marks in the entire syllabus. Drill the 6 main formulas until you can solve any PYQ in under 90 seconds.

Memorise the constants in convenient units:

$hc = 1240\,\text{eV nm}$
$13.6\,\text{eV}$ for hydrogen ground state
$1\,\text{u} = 931.5\,\text{MeV}$
$\lambda_e = 12.27/\sqrt{V}\,\text{Å}$ for electron at potential $V$

For Bohr-model questions, always identify $n_1$ and $n_2$ first. The negative sign in $E_n$ catches students out — energy is always negative for bound states.

Common Traps

Trap 1 — Forgetting the minus sign in Bohr energy. $E_n = -13.6/n^2$ , always negative for bound electrons.

Trap 2 — Mixing up frequency and wavelength. $E = h\nu = hc/\lambda$ . Make sure the formula matches what the problem gives.

Trap 3 — Assuming $\lambda = h/(mv)$ for all particles. True for non-relativistic. For relativistic particles, use $\lambda = h/p$ where $p$ is relativistic momentum.

Trap 4 — Using mass number A for nucleon mass. A nucleus of mass number A has mass approximately $A \times 1\,\text{u}$ , but the binding energy depends on the mass defect $\Delta m$ .

Trap 5 — Treating half-life and mean lifetime as the same. $T_{\text{mean}} = T_{1/2}/\ln 2 \approx 1.44 T_{1/2}$ .