Atoms | Class 12 Physics (NCERT Ch 12)

CBSE Class 12 › Unit VII: Dual Nature & Atoms

Atoms

NCERT Chapter 12 • Rutherford & Bohr Models, Hydrogen Spectrum

NCERT 2025–26 Unit VII • ~4 Marks JEE Main • 1 Question

1. Thomson’s Plum Pudding Model

Proposed in 1898, J.J. Thomson suggested that an atom is a uniform sphere of positive charge with electrons embedded in it — like “plums in a pudding”. This model was later disproved by Rutherford’s alpha-scattering experiment.

2. Rutherford’s Nuclear Model

The alpha-particle scattering experiment (Geiger-Marsden, 1911) led Rutherford to propose the nuclear model of the atom.

Rutherford’s gold foil experiment revolutionized our understanding of atomic structure.

Key Observations:

Most α-particles passed through undeflected → atom is mostly empty space.
A few scattered at large angles → concentrated positive charge.
About 1 in 8000 deflected by >90° → nucleus is tiny but massive.

Distance of Closest Approach ( $r_0$ ):

At this distance, kinetic energy = electrostatic potential energy:
$K = \dfrac{1}{4\pi\epsilon_0} \dfrac{(2e)(Ze)}{r_0}$ → $r_0 = \dfrac{1}{4\pi\epsilon_0} \dfrac{2Ze^2}{K}$ .

Impact Parameter ( $b$ ):
The perpendicular distance between the initial velocity vector of the α-particle and the center of the nucleus.
– Small $b$ → large deflection
– Large $b$ → small deflection

3. Failure of Rutherford’s Model

Two major contradictions:

Instability: Accelerated electrons should radiate energy and spiral into the nucleus in ~10⁻¹⁰ s. But atoms are stable.
Continuous Spectrum: Spiraling electrons would emit a continuous range of frequencies. But atoms emit only discrete spectral lines.

4. Bohr’s Model of Hydrogen Atom

To resolve these issues, Niels Bohr (1913) proposed a quantum model for hydrogen.

Diagram showing forces acting on an electron in the Bohr model. — The electrostatic force provides the necessary centripetal force.

Bohr’s Postulates:

Stable Orbits: Electrons revolve in stable circular orbits without radiating energy (“stationary states”).
Quantisation Condition: Angular momentum is quantised: $L = mvr = \dfrac{nh}{2\pi}$ ( $n = 1,2,3...$ ).
Photon Emission/Absorption: Radiation occurs only during transitions: $h\nu = E_i - E_f$ .

5. Derivations: Radius, Velocity & Energy

Derivation: Radius of $n^{th}$ Orbit ( $r_n$ )

3 Marks

Step 1: Force Balance
Centripetal force = Electrostatic force:
$\dfrac{mv^2}{r} = \dfrac{1}{4\pi\epsilon_0} \dfrac{e^2}{r^2}$ → $mv^2 = \dfrac{e^2}{4\pi\epsilon_0 r}$ … (i)

Step 2: Quantisation
From Bohr’s second postulate: $v = \dfrac{nh}{2\pi mr}$ .

Step 3: Substitute v
Substitute $v$ into (i):
$m \left(\dfrac{nh}{2\pi mr}\right)^2 = \dfrac{e^2}{4\pi\epsilon_0 r}$ .

Step 4: Solve for r
$\dfrac{n^2 h^2}{4\pi^2 m r^2} = \dfrac{e^2}{4\pi\epsilon_0 r}$ → $r_n = \dfrac{\epsilon_0 n^2 h^2}{\pi m e^2}$ .
For Hydrogen ( $n=1$ ): $r_1 = 0.529 \, \AA$ (Bohr Radius).
For Hydrogenic atoms: $r_n=0.529\dfrac{n^2}{Z}\, \AA$

Derivation: Total Energy ( $E_n$ )

3 Marks

Kinetic Energy (K):
From (i), $K = \dfrac{1}{2}mv^2 = \dfrac{e^2}{8\pi\epsilon_0 r}$ .

Potential Energy (U):
$U = -\dfrac{e^2}{4\pi\epsilon_0 r}$ (negative since force is attractive).

Total Energy (E):
$E = K + U = -\dfrac{e^2}{8\pi\epsilon_0 r}$ .

Substitute $r_n$ :
$E_n = -\dfrac{m (e^2)^2}{8 \epsilon_0^2 h^2} \cdot \dfrac{1}{n^2}$ .
For Hydrogen atom: $E_n=-\dfrac{13.6}{n^2} \, \text{eV}$
For Hydrogenic atoms: $E_n =-\dfrac{13.6 Z^2}{n^2} \, \text{eV}$ .

Ground State & Ionisation:

– Ground state ( $n=1$ ): $E_1 = -13.6 \, \text{eV}$
– Ionisation energy = $+13.6 \, \text{eV}$ (energy to remove electron to $n=\infty$ )

6. Atomic Spectra (Hydrogen Series)

When an electron jumps from higher level $n_i$ to lower level $n_f$ , a photon is emitted with:

$\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)$
where $R = 1.097 \times 10^7 \, \text{m}^{-1}$ (Rydberg constant)

Energy level diagram of Hydrogen showing spectral series. — Electron transitions between quantized energy levels produce the spectral lines.

Series	Transition ( $n_f \leftarrow n_i$ )	Region
Lyman	$1 \leftarrow 2,3,4...$	Ultraviolet
Balmer	$2 \leftarrow 3,4,5...$	Visible
Paschen	$3 \leftarrow 4,5,6...$	Infrared
Brackett	$4 \leftarrow 5,6,7...$	Infrared
Pfund	$5 \leftarrow 6,7,8...$	Far Infrared

7. de Broglie’s Explanation of Bohr’s Postulate

Louis de Broglie (1923) explained Bohr’s quantization using matter waves.

For a stable orbit, the electron wave must form a standing wave around the circumference.
Circumference = $n \times$ wavelength → $2\pi r = n\lambda$ .
Using de Broglie relation $\lambda = \dfrac{h}{mv}$ :
$2\pi r = n \left(\dfrac{h}{mv}\right)$ → $mvr = \dfrac{nh}{2\pi}$ .
This justifies Bohr’s second postulate.

8. Limitations of Bohr’s Model

Why Bohr’s model is incomplete:

Only works for hydrogenic atoms (H, He⁺, Li²⁺).
Fails for multi-electron atoms (ignores electron-electron repulsion).
Cannot explain fine structure of spectral lines.
Does not predict relative intensities of spectral lines.
Contradicts Heisenberg’s uncertainty principle (fixed orbits imply known position & momentum).
Bohr’s fixed orbits contradict Heisenberg’s uncertainty principle — electrons don’t have definite trajectories in quantum mechanics (leads to Schrödinger wave mechanics).

These limitations were resolved by **quantum mechanics**.

Practice Time!

Calculate energy levels and spectral lines: Chapter 12 Important Questions →

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