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AXIOM PHYSICS Class 12 • Open Access

CBSE Class XII • Unit 08 (Atoms & Nuclei)

Chapter 12: Atoms

Master Revision Dossier • Rutherford Scattering, Distance of Closest Approach, Bohr Postulates, Energy Quantization, Rydberg Formula & Spectral Series

1. Definitive Formula & Atomic Physics Matrix

Concept Mathematical Formulation Proportionality & Values Boundary Conditions & Notes
Distance of Closest Approach $$r_0 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K} = \frac{1}{4\pi\varepsilon_0}\frac{4Ze^2}{mv^2}$$ $$r_0 \propto \frac{Z}{K} \propto \frac{1}{v^2}$$ Head-on collision ($b = 0 \implies \theta = 180^\circ$). Entire kinetic energy turns into electrostatic potential energy. Puts upper bound on nuclear size ($R_N < r_0 \approx 10^{-14}\text{ m}$).
Impact Parameter $$b = \frac{1}{4\pi\varepsilon_0}\frac{Ze^2 \cot(\theta/2)}{K}$$ $$b \propto \cot\left(\frac{\theta}{2}\right)$$ Perpendicular distance of the initial velocity vector from central nuclear axis. If $b$ is large $\implies \theta \approx 0^\circ$ (undeviated). If $b = 0 \implies \theta = 180^\circ$.
Bohr Angular Momentum $$L = m v_n r_n = \frac{nh}{2\pi} = n\hbar$$ $n = 1, 2, 3\dots$
$\hbar = h/2\pi$		 Non-radiating stationary circular orbits. Proven by de Broglie through circular standing waves ($2\pi r_n = n\lambda$).
Bohr Orbital Radius $$r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2} = 0.529 \frac{n^2}{Z}\text{ \AA}$$ $$r_n \propto \frac{n^2}{Z}$$ Bohr radius for Hydrogen ground state ($n=1, Z=1$): $a_0 = r_1 \approx 0.529\text{ \AA} = 5.29 \times 10^{-11}\text{ m}$. Orbit area scales as $n^4$.
Orbital Velocity & Period $$v_n = \frac{Z e^2}{2 \varepsilon_0 n h} = \left(\frac{c}{137}\right)\frac{Z}{n}, \quad T_n = \frac{2\pi r_n}{v_n}$$ $$v_n \propto \frac{Z}{n}, \quad T_n \propto n^3$$ Ground state speed in Hydrogen: $v_1 \approx 2.18 \times 10^6\text{ m/s} \approx c/137$. Orbital frequency scales as $f_n = 1/T_n \propto 1/n^3$.
Energy Levels (Hydrogen-like) $$E_n = -\frac{m Z^2 e^4}{8 \varepsilon_0^2 h^2 n^2} = -13.6 \frac{Z^2}{n^2}\text{ eV}$$ $$E_n \propto -\frac{Z^2}{n^2}$$ Energy Partition Triangle: $\mathbf{K = -E_n}$, $\mathbf{U = 2E_n}$, and $\mathbf{E_n = K + U = -K}$. Negative sign indicates bound electrostatic state.
Rydberg Transition Formula $$\frac{1}{\lambda} = \bar{\nu} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$$ $$R = 1.097 \times 10^7\text{ m}^{-1}$$ Useful constant: $\mathbf{1/R \approx 912\text{ \AA}}$. Number of spectral lines emitted during cascade from level $n$: $\mathbf{N = \frac{n(n-1)}{2}}$.

2. Core Derivations Blueprint (Step-by-Step Board Proofs)

Blueprint 12.1: Deduction of the Distance of Closest Approach ($r_0$) in Head-On Alpha Scattering

1. Consider an $\alpha$-particle ($q_1 = +2e$, mass $m$) directed straight towards a target nucleus of charge $q_2 = +Ze$ in a head-on collision ($b = 0$).

2. Initially at infinity, the $\alpha$-particle has kinetic energy $K = \frac{1}{2}mv^2$ and zero electrostatic potential energy.

3. As it approaches the positively charged nucleus, repulsive Coulomb force decelerates it until at distance $r_0$, its velocity momentarily drops to zero ($v = 0$).

4. By the Law of Conservation of Energy, all initial kinetic energy $K$ is transformed into electrostatic potential energy $U$ at $r_0$:

$$K = U \implies K = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r_0} = \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r_0}$$
$$r_0 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K} = \frac{1}{4\pi\varepsilon_0}\frac{4Ze^2}{mv^2}$$

Blueprint 12.2: Derivations of Orbital Radius ($r_n$), Orbital Velocity ($v_n$), and Quantized Energy ($E_n$)

1. For an electron of mass $m$, charge $-e$ revolving in orbit of radius $r_n$ with speed $v_n$ around nucleus $+Ze$, centripetal force is provided by Coulomb force:

$$\frac{m v_n^2}{r_n} = \frac{1}{4\pi\varepsilon_0}\frac{Z e^2}{r_n^2} \implies m v_n^2 r_n = \frac{Z e^2}{4\pi\varepsilon_0} \quad \text{--- [Eq. 1]}$$

2. From Bohr's angular momentum quantization: $m v_n r_n = \frac{nh}{2\pi} \implies v_n = \frac{nh}{2\pi m r_n}$ [Eq. 2].

3. Substituting Eq. 2 into Eq. 1: $m \left(\frac{nh}{2\pi m r_n}\right)^2 r_n = \frac{Ze^2}{4\pi\varepsilon_0} \implies \frac{n^2 h^2}{4\pi^2 m r_n} = \frac{Ze^2}{4\pi\varepsilon_0}$:

$$r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m Z e^2} \implies r_n \propto \frac{n^2}{Z} \quad \left(r_1 = 0.529\text{ \AA for } H\right)$$

4. Substituting $r_n$ into Eq. 2 gives orbital velocity:

$$v_n = \frac{Z e^2}{2 \varepsilon_0 n h} \implies v_n \propto \frac{Z}{n} \quad \left(v_1 = \frac{c}{137} \approx 2.18 \times 10^6\text{ m/s for } H\right)$$

5. Kinetic energy $K = \frac{1}{2}mv_n^2 = \frac{Ze^2}{8\pi\varepsilon_0 r_n}$; Potential energy $U = -\frac{Ze^2}{4\pi\varepsilon_0 r_n}$. Total Energy $E_n = K + U = -\frac{Ze^2}{8\pi\varepsilon_0 r_n}$:

$$E_n = -\frac{m Z^2 e^4}{8 \varepsilon_0^2 h^2 n^2} = -13.6 \frac{Z^2}{n^2}\text{ eV}$$

Blueprint 12.3: de Broglie's Standing Wave Explanation of Bohr's Angular Momentum Quantization

1. According to de Broglie, an electron orbiting in a stationary state behaves as a matter wave running along a closed circular loop of radius $r_n$.

2. To maintain a stationary, non-radiating orbit without destructive self-cancellation, the circumference of the orbit must fit an integral number of de Broglie wavelengths:

$$2\pi r_n = n \lambda \quad (n = 1, 2, 3\dots)$$

3. Substituting the de Broglie wavelength relation $\lambda = \frac{h}{p} = \frac{h}{m v_n}$:

$$2\pi r_n = n \left(\frac{h}{m v_n}\right) \implies m v_n r_n = \frac{nh}{2\pi} = n\hbar$$
$$L = m v_n r_n = \frac{nh}{2\pi} \quad (\text{Bohr's 2nd Postulate Proven from Wave Mechanics!})$$

3. High-Yield "Red Alert" Exam Trap Matrix

Master Trap 1: Comprehensive Spectral Series Spectrum & Wavelength Bounds ($Z = 1$)

Series Name Ground Level ($n_1$) Upper Levels ($n_2$) Spectral Region Longest Line ($\lambda_{\max}$, $n_2=n_1+1$) Series Limit ($\lambda_{\min}$, $n_2=\infty$)
1. Lyman Series $n_1 = 1$ $2, 3, 4\dots$ Ultraviolet (UV) $\frac{4}{3R} \approx 1216\text{ \AA}$ $\frac{1}{R} \approx 912\text{ \AA}$
2. Balmer Series $n_1 = 2$ $3, 4, 5\dots$ Visible Region $\frac{36}{5R} \approx 6563\text{ \AA}$ ($H_\alpha$) $\frac{4}{R} \approx 3646\text{ \AA}$
3. Paschen Series $n_1 = 3$ $4, 5, 6\dots$ Near-Infrared (IR) $\frac{144}{7R} \approx 18751\text{ \AA}$ $\frac{9}{R} \approx 8208\text{ \AA}$
4. Brackett Series $n_1 = 4$ $5, 6, 7\dots$ Mid-Infrared (IR) $\frac{400}{9R} \approx 40518\text{ \AA}$ $\frac{16}{R} \approx 14585\text{ \AA}$
5. Pfund Series $n_1 = 5$ $6, 7, 8\dots$ Far-Infrared (IR) $\frac{900}{11R} \approx 74578\text{ \AA}$ $\frac{25}{R} \approx 22789\text{ \AA}$

⚠️ Trap 2: Energy Partition Signs & Relations

If total energy is $E_n = -3.4\text{ eV}$ (for $n = 2$ in H):
• Kinetic Energy is strictly positive: $\mathbf{K = -E_n = +3.4\text{ eV}}$.
• Potential Energy is negative and twice total: $\mathbf{U = 2E_n = -6.8\text{ eV}}$.
Never reverse these signs in objective or numerical questions!

⚠️ Trap 3: Classical Collapse Paradox of Rutherford

Accelerating electrons radiate continuous EM waves according to Maxwell's electrodynamics.
• In Rutherford's model, the orbit should collapse into the nucleus within $\mathbf{\approx 10^{-8}\text{ s}}$.
• The emitted spectrum should be continuous, contradicting the discrete line spectrum observed experimentally.

4. Handcrafted Exemplars (Analytical Numerical Solutions)

Problem 1: Closest Approach in Head-On Alpha Collision Exemplar 01

In a Geiger-Marsden experiment, what is the distance of closest approach to the nucleus of a $7.7\text{ MeV}$ $\alpha$-particle before it comes momentarily to rest and reverses its direction? ($Z = 79$ for gold).

Solution: $K = 7.7\text{ MeV} = 7.7 \times 1.6 \times 10^{-13}\text{ J} \approx 1.232 \times 10^{-12}\text{ J}$, $Z = 79$.

$$r_0 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K} = \frac{(9 \times 10^9) \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{1.232 \times 10^{-12}} = \frac{3640.32 \times 10^{-29}}{1.232 \times 10^{-12}}$$
$$r_0 \approx 2.95 \times 10^{-14}\text{ m} = 29.5\text{ fm} \approx 30\text{ fm}$$

Since the $\alpha$-particle turns back before touching nuclear matter, nuclear radius $R_N < 30\text{ fm}$.

Problem 2: Spectral Line Emission Cascade Exemplar 02

A hydrogen atom initially in the ground state absorbs a photon which excites it to the $n = 4$ state. (a) Calculate the wavelength of the absorbed photon. (b) Find the total number of spectral lines that can be emitted when the atom de-excites to the ground state.

Solution: Transition from $n_1 = 1$ to $n_2 = 4$:

$$\frac{1}{\lambda_{\text{abs}}} = R \left(\frac{1}{1^2} - \frac{1}{4^2}\right) = R \left(1 - \frac{1}{16}\right) = \frac{15R}{16} \implies \lambda_{\text{abs}} = \frac{16}{15R} = \frac{16}{15 \times 1.097 \times 10^7} \approx 9.72 \times 10^{-8}\text{ m} = 97.2\text{ nm}$$

Total number of emitted spectral lines in de-excitation cascade from $n = 4$:

$$N = \frac{n(n - 1)}{2} = \frac{4 \times (4 - 1)}{2} = \frac{12}{2} = \mathbf{6\text{ spectral lines}}$$

Breakdown: 3 lines in Lyman ($4\to 1, 3\to 1, 2\to 1$), 2 lines in Balmer ($4\to 2, 3\to 2$), and 1 line in Paschen ($4\to 3$).

5. Printable Pocket Flashcard Grid

Closest Approach

$r_0 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K}$

Bohr Radius

$r_n = 0.529\frac{n^2}{Z}\text{ \AA}$

Energy Quantization

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

Cascade Lines

$N = \frac{n(n-1)}{2}$