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CBSE • Class XII • Physics • Ch 12

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Atoms

In Class 12 Physics, "Atoms" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

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⚛️ Have You Ever Wondered?

How did Ernest Rutherford discover that almost the entire mass of an atom is concentrated in a microscopic atomic nucleus 100,000 times smaller than the atom itself, proving that all solid matter is 99.9999999% empty space? The Alpha Particle Scattering Experiment and Bohr's quantized orbits unlocked atomic structure.

Why This Chapter Matters

In Class 12 Physics, "Atoms" provides an authoritative, curriculum-verified master resource aligned with the 2026–27 NCERT syllabus.

Before You Begin (Prerequisites)

Rutherford model from Class 9.
Coulomb's Law from Chapter 1.
Photons from Chapter 11.

What You Will Learn (Core Objectives)

Explain Geiger-Marsden Alpha Particle Scattering Experiment and Rutherford's Planetary Nuclear Model.
Calculate Distance of Closest Approach ($d = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K}$) and Impact Parameter ($b$).
State Bohr's Postulates: Quantization of angular momentum ($mvr = \frac{nh}{2\pi}$) and energy transition ($E_2 - E_1 = h\nu$).
Derive orbital radius ($r_n = n^2 a_0, a_0 = 0.529\text{ Å}$) and energy levels of Hydrogen: $E_n = -\frac{13.6}{n^2}\text{ eV}$.
Explain the Hydrogen Spectral Series: Lyman, Balmer, Paschen, Brackett, Pfund (Rydberg formula $\frac{1}{\lambda} = R_H(\frac{1}{n_1^2} - \frac{1}{n_2^2})$).

Chapter Roadmap & Progression

1 1. Rutherford Scattering & Nuclear...

2 2. Bohr's Quantized Hydrogen Model

3 3. Hydrogen Spectral Emission Serie...

Complete Concept Guide (100% Curriculum Coverage)

1. Rutherford Scattering & Nuclear Atom

Geiger and Marsden fired alpha particles ($^4\text{He}^{2+}$) at a thin gold foil ($10^{-7}\text{ m}$):
• Most passed undeviated ($99.86\%$), proving matter is mostly empty space.
• 1 in 8000 rebounded by $>90^\circ$, discovering the massive positive Nucleus!
• Distance of Closest Approach ($d$): Conservation of energy: $\frac{1}{2}m v^2 = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{d}$.

2. Bohr's Quantized Hydrogen Model

Bohr resolved orbital collapse by postulating that electrons orbit only in non-radiating stationary states: $$\mathbf{L = m v r = \frac{n h}{2\pi}} \quad (n = 1, 2, 3\dots)$$
• Bohr Radius: $\mathbf{r_n = n^2 \left(\frac{h^2 \varepsilon_0}{\pi m e^2}\right) = 0.529 n^2\text{ Å}}$.
• Quantized Energy: $\mathbf{E_n = -\frac{13.6}{n^2}\text{ eV}}$ (Negative sign proves electron is bound to nucleus; $E = 0$ means free ionized electron!).

3. Hydrogen Spectral Emission Series

Photon emitted during transitions: $\mathbf{\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)}$ ($R_H = 1.097 \times 10^7\text{ m}^{-1}$).
• Lyman ($n_1=1$): Ultraviolet region.
• Balmer ($n_1=2$): Visible region (H-alpha red, H-beta cyan).
• Paschen ($n_1=3$), Brackett ($n_1=4$), Pfund ($n_1=5$): Infrared region.

Atoms - Key Conceptual & Analytical Model

Chapter Summary & 10 Key Takeaways

Takeaway 1

Rutherford Alpha Experiment: Discovery of dense positively charged atomic nucleus.

Takeaway 2

Bohr Angular Momentum Quantization: $mvr = nh/2\pi$ stabilizing stationary non-radiating orbits.

Takeaway 3

Ground State Energy: $E_1 = -13.6\text{ eV}$ for hydrogen requiring $13.6\text{ eV}$ ionization energy.

Takeaway 4

Balmer Visible Series: Only hydrogen spectral transitions falling into the visible optical window ($n_1 = 2$).

Takeaway 5

Rydberg Constant: $R_H = 1.097 \times 10^7\text{ m}^{-1}$ predicting all atomic emission wavelengths.

Check Your Understanding (Diagnostic Practice Questions)

Diagnostic questions testing core conceptual clarity. Answers are hidden initially — solve each problem first, then click to reveal the step-by-step verified solution.

State the three postulates of Bohr's model of the hydrogen atom.

Reveal Answer & Explanation

Answer: (1) Electrons revolve around the nucleus in certain stable non-radiating orbits called stationary orbits. (2) Quantization condition: An electron can only revolve in those orbits for which its orbital angular momentum is an integral multiple of $h/2\pi$: $mvr = rac{nh}{2\pi}$. (3) Frequency condition: Radiation is emitted or absorbed only when an electron jumps from one stationary orbit to another: $h u = E_2 - E_1$.
Stationary orbits, angular momentum quantization mvr = nh/2π, and hν = E2 - E1.

The ground state energy of hydrogen atom is $-13.6\text{ eV}$. (a) What is the kinetic energy of the electron in the 2nd excited state? (b) What is its potential energy in this state?

Reveal Answer & Explanation

Answer: 2nd excited state corresponds to $n = 3$. Total energy $E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} \approx -1.51\text{ eV}$. (a) Kinetic energy $K = -E_3 = +1.51\text{ eV}$. (b) Potential energy $U = 2E_3 = 2(-1.51) = -3.02\text{ eV}$.
(a) K = +1.51 eV, (b) U = -3.02 eV.

Calculate the shortest and longest wavelengths in the Balmer series of hydrogen spectrum ($R_H = 1.097 \times 10^7\text{ m}^{-1}$).

Reveal Answer & Explanation

Answer: In Balmer series, $n_1 = 2$. Longest wavelength ($n_2 = 3$): $\frac{1}{\lambda_{\text{max}}} = R_H(\frac{1}{4} - \frac{1}{9}) = \frac{5R_H}{36} \implies \lambda_{\text{max}} = \frac{36}{5 \times 1.097 \times 10^7} \approx 656.3\text{ nm}$. Shortest wavelength ($n_2 = \infty$): $\frac{1}{\lambda_{\text{min}}} = R_H(\frac{1}{4} - 0) = \frac{R_H}{4} \implies \lambda_{\text{min}} = \frac{4}{1.097 \times 10^7} \approx 364.6\text{ nm}$.
Longest = 656.3 nm; Shortest = 364.6 nm.

Why did Rutherford's planetary model of the atom fail theoretically?

Reveal Answer & Explanation

Answer: According to classical electromagnetic theory, an orbiting electron accelerates continuously and must radiate electromagnetic energy, losing speed and spiraling into the nucleus in $10^{-8}\text{ seconds}$, destroying atomic stability. It also could not explain discrete line spectra.
Accelerating electron should radiate energy and collapse into nucleus.

What is the physical significance of the negative total energy of an electron in an atom?

Reveal Answer & Explanation

Answer: The negative sign indicates that the electron is bound to the positive nucleus by attractive electrostatic forces. Energy must be supplied from outside to liberate the electron to infinity ($E = 0$).
Indicates the electron is bound to the nucleus by attractive forces.

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