Class 12 — Atoms

Chapter Overview & Weightage

The Atoms chapter introduces Bohr’s model, the hydrogen spectrum, and the foundation of quantum theory in physics. CBSE Class 12 boards typically allot $3$ – $5$ marks here, often through derivation questions (Bohr radius, energy levels) and one numerical on spectral lines.

Year	Marks	Question Pattern
2024	5	Bohr model derivation
2023	3	Hydrogen spectrum series
2022	4	Energy of nth orbit
2021	5	Rutherford scattering + Bohr
2020	3	Wavelength calculation

CBSE rewards derivations heavily. Memorise the Bohr radius and energy formula derivations cold. Numericals usually involve the Rydberg formula.

Key Concepts You Must Know

Rutherford scattering — alpha particle deflection, nuclear model, distance of closest approach.
Bohr’s postulates — quantised orbits, stationary states, frequency condition.
Bohr radius $r_n = n^2 a_0$ where $a_0 = 0.529\,\text{Å}$ .
Energy of $n$ th level $E_n = -13.6/n^2\,\text{eV}$ for hydrogen.
Spectral series — Lyman (UV), Balmer (visible), Paschen, Brackett, Pfund (IR).
Rydberg formula for hydrogen-like spectra.
Limitations of Bohr’s model — fails for multi-electron atoms.

Important Formulas

$mvr = \frac{nh}{2\pi}$

When to use: deriving orbital radii or velocities.

$r_n = \frac{n^2 h^2}{4\pi^2 m k e^2}, \quad E_n = -\frac{13.6}{n^2}\,\text{eV (hydrogen)}$

For hydrogen-like ion of charge $Z$ : $E_n = -13.6\,Z^2/n^2\,\text{eV}$ .

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

with $R = 1.097 \times 10^7\,\text{m}^{-1}$ . Choose $n_1 < n_2$ for emission.

When to use: any spectral line wavelength calculation.

Solved Previous Year Questions

PYQ 1 (CBSE 2024, 5 marks)

Derive the expression for the radius of the $n$ th Bohr orbit and the energy of the electron in that orbit.

Solution. Centripetal force balance: $kZe^2/r^2 = mv^2/r$ , so $mv^2 = kZe^2/r$ .

Bohr’s quantisation: $mvr = nh/(2\pi)$ , so $v = nh/(2\pi mr)$ .

Substituting: $m\left(\frac{nh}{2\pi mr}\right)^2 = kZe^2/r$ , giving

$r_n = \frac{n^2 h^2}{4\pi^2 m k Z e^2}$

For $Z = 1$ , $r_1 = a_0 \approx 0.529\,\text{Å}$ . So $r_n = n^2 a_0$ .

Total energy $E =$ KE + PE $= \tfrac{1}{2}mv^2 - kZe^2/r = -kZe^2/(2r)$ .

Plugging $r_n$ : $E_n = -2\pi^2 m k^2 Z^2 e^4/(n^2 h^2) = -13.6 Z^2/n^2\,\text{eV}$ .

PYQ 2 (CBSE 2023, 3 marks)

Calculate the wavelength of the first line in the Balmer series of hydrogen.

Solution. Balmer means $n_1 = 2$ . First line: $n_2 = 3$ .

$\frac{1}{\lambda} = R\left(\frac{1}{4} - \frac{1}{9}\right) = R \cdot \frac{5}{36}$

$\lambda = \frac{36}{5R} = \frac{36}{5 \times 1.097\times 10^7} \approx 6.56 \times 10^{-7}\,\text{m} = 656\,\text{nm}$

This is the famous H-alpha line, red.

PYQ 3 (CBSE 2022, 4 marks)

The energy of the electron in the second orbit of hydrogen is $-3.4\,\text{eV}$ . Find (a) the kinetic energy, (b) the potential energy, (c) the wavelength of the photon emitted when the electron jumps from $n=2$ to $n=1$ .

Solution. (a) $KE = -E_2 = +3.4\,\text{eV}$ . (b) $PE = 2E_2 = -6.8\,\text{eV}$ . (c) $E_1 = -13.6\,\text{eV}$ , transition energy $= 13.6 - 3.4 = 10.2\,\text{eV}$ . $\lambda = 1240/10.2\,\text{nm} \approx 121.6\,\text{nm}$ (Lyman alpha).

Difficulty Distribution

Easy ( $\sim 30\%$ ): definitions, postulates, simple Rydberg numericals.
Medium ( $\sim 50\%$ ): derivations of $r_n$ and $E_n$ , spectral series.
Hard ( $\sim 20\%$ ): hydrogen-like ions ( $\text{He}^+$ , $\text{Li}^{2+}$ ), photon energy chains.

Expert Strategy

Memorise the values: $a_0 = 0.529\,\text{Å}$ , $E_1 = -13.6\,\text{eV}$ , $R = 1.097\times 10^7\,\text{m}^{-1}$ . With these and the formulas, you can answer most questions in under three minutes.

Approach in order:

Master Bohr’s three postulates by heart.
Derive $r_n$ and $E_n$ once, then drill the steps.
Practice five Rydberg-formula problems on different series.
Memorise wavelength ranges: Lyman in UV, Balmer in visible, Paschen+ in IR.

Common Traps

Trap 1: Forgetting the $Z^2$ factor for hydrogen-like ions like $\text{He}^+$ . Energy scales as $Z^2/n^2$ .

Trap 2: Confusing $n_1$ and $n_2$ in Rydberg formula. $n_1$ is the lower level (final for emission, initial for absorption).

Trap 3: Forgetting that PE $= 2E_{\text{total}}$ and KE $= -E_{\text{total}}$ . Sign matters.

Trap 4: Using Bohr’s model for multi-electron atoms. It only works for hydrogen-like systems.

Trap 5: Mixing wavenumber and wavelength. Rydberg formula gives $1/\lambda$ , not $\lambda$ directly.

Quick Revision Notes

Bohr radius scales as $n^2$ .
Energy scales as $-1/n^2$ (negative — bound states).
Velocity scales as $1/n$ .
Series limit (highest energy in a series) corresponds to $n_2 \to \infty$ .
$1\,\text{eV} = 1240/\lambda(\text{nm})$ — useful conversion for photon energies.

A high-scoring chapter for students who derive Bohr’s model carefully and memorise the spectral series with their wavelength ranges.