Welcome to HSLC Guru! This comprehensive guide covers ASSEB Class 12 Physics Chapter 12 — Atoms in the English medium. The chapter traces the historical evolution of atomic models from J.J. Thomson’s plum pudding picture, through Rutherford’s nuclear discovery via alpha-particle scattering, to Niels Bohr’s quantum postulates that finally explained the hydrogen spectrum. You will master the distance of closest approach, impact parameter, Bohr radius $r_1 = 0.53$ Å, the famous energy formula $E_n = -13.6/n^2$ eV, spectral series (Lyman, Balmer, Paschen, Brackett, Pfund), and the Rydberg formula. Each concept includes derivations, NCERT exercise solutions, additional ASSEB-pattern problems, scalable SVG illustrations, and a glossary aligned with the Assam State School Education Board syllabus.

Chapter Summary

Atoms are the smallest units of an element that retain its chemical identity. Throughout the late 19th and early 20th centuries, physicists struggled to reconcile experimental data on atomic spectra and matter-radiation interaction with classical electromagnetism. Thomson’s 1898 model imagined the atom as a uniform sphere of positive charge with electrons embedded like raisins in a pudding. While it accounted for atomic neutrality, it predicted no large-angle scattering of fast charged projectiles. In 1911, Ernest Rutherford’s gold-foil experiment with alpha particles overturned the picture: a tiny, dense, positive nucleus carries virtually all the atomic mass, while electrons orbit at relatively vast distances. Rutherford’s planetary atom, however, suffered a fatal classical defect — accelerating electrons should radiate continuously and spiral into the nucleus within $10^{-8}$ s. Niels Bohr (1913) rescued the atom with three bold postulates: electrons reside only in stationary orbits, angular momentum is quantized as $L = n\hbar$, and radiation occurs only during transitions with $h\nu = E_i – E_f$. Bohr’s theory perfectly reproduced the hydrogen spectrum and yielded the Rydberg constant $R = 1.097 \times 10^7$ m$^{-1}$ from first principles. This chapter is a cornerstone for understanding modern quantum mechanics and prepares you for Chapter 13 (Nuclei) and the broader physics of matter.

Key Concepts at a Glance

Concept	Symbol / Formula	Value / Meaning
Bohr radius (1st orbit)	$r_1$	$0.529 \times 10^{-10}$ m $\approx 0.53$ Å
Ground state energy (H)	$E_1$	$-13.6$ eV
Rydberg constant	$R$	$1.097 \times 10^7$ m$^{-1}$
Ionisation energy of H	$E_\infty – E_1$	$13.6$ eV
Speed in 1st orbit	$v_1$	$c/137 \approx 2.19 \times 10^6$ m/s
Angular momentum quantum	$L = n\hbar$	$n = 1, 2, 3, \dots$
Lyman series (UV)	$n_f = 1$	91.2 nm – 121.6 nm
Balmer series (visible)	$n_f = 2$	364.6 nm – 656.3 nm
Paschen series (IR)	$n_f = 3$	820 nm – 1875 nm

12.1 Thomson’s Model and Its Failures

J. J. Thomson, fresh from discovering the electron in 1897, proposed in 1898 that an atom is a sphere of uniform positive charge of radius $\sim 10^{-10}$ m, with electrons embedded inside it — the famous plum-pudding model. The negative electrons balance the positive jelly, making the atom electrically neutral. Electrons could oscillate about equilibrium positions and emit light of definite frequencies. Although elegant for its time, Thomson’s model failed on several counts:

• It predicted only small-angle deflections of alpha particles, contrary to Rutherford’s observations.
• It could not explain the line spectra of hydrogen and other elements.
• It offered no mechanism for the stability of complex multi-electron atoms.
• The predicted oscillation frequencies did not match the observed spectral frequencies.

12.2 Rutherford’s Alpha-Scattering Experiment (1911)

Hans Geiger and Ernest Marsden, under Rutherford’s guidance, fired a collimated beam of alpha particles ($^4_2$He nuclei, energy $\approx$ 5.5 MeV from radium-bismuth source) at a thin gold foil ($\sim 10^{-7}$ m thick). A movable zinc-sulphide screen detected scintillations as the alphas struck it. The startling results were:

• About 99% of alpha particles passed straight through with negligible deflection — atoms are mostly empty space.
• A few (about 1 in 8000) suffered large-angle deflections (greater than 90°).
• Some particles were even scattered backward (180°).

Rutherford famously remarked: “It was almost as incredible as if you fired a 15-inch shell at a piece of tissue paper and it came back and hit you.” He concluded that the entire positive charge and nearly all the mass of the atom is concentrated in a tiny central region — the nucleus — of radius $\sim 10^{-15}$ m, while electrons orbit at distances $\sim 10^{-10}$ m. The atom is therefore $10^5$ times wider than its nucleus.

12.3 Distance of Closest Approach

Consider an alpha particle (charge $+2e$, kinetic energy $E_K$) heading directly toward a gold nucleus (charge $+Ze$, $Z = 79$). At the closest distance $r_0$, all the kinetic energy is converted into electrostatic potential energy:

$$r_0 = \frac{1}{4\pi\epsilon_0}\cdot\frac{2Ze^2}{E_K}$$

For a 7.7 MeV alpha particle and gold ($Z = 79$):

$$r_0 = \frac{(9 \times 10^9)(2)(79)(1.6\times 10^{-19})^2}{(7.7 \times 10^6)(1.6\times 10^{-19})} \approx 3.0 \times 10^{-14} \text{ m}$$

This sets an upper bound on the nuclear radius. Modern measurements give $R_\text{nucleus} \sim 10^{-15}$ m, consistent with $r_0$.

12.4 Impact Parameter

The impact parameter $b$ is the perpendicular distance from the initial straight-line path of the alpha particle to the centre of the nucleus, had there been no force. Rutherford derived:

$$b = \frac{Ze^2 \cot(\theta/2)}{4\pi\epsilon_0 \cdot E_K}$$

Smaller $b$ means a closer encounter and hence larger scattering angle $\theta$. A head-on collision ($b = 0$) gives $\theta = 180°$ (back-scattering).

12.5 Failures of Rutherford’s Model

• Stability problem: An electron in circular orbit is centripetally accelerated; classical electromagnetism predicts continuous radiation, energy loss, shrinking orbit, and collapse in $\sim 10^{-8}$ s.
• Continuous spectrum predicted: As radius decreases, frequency would change continuously, contradicting the observed line spectra.
• No quantization: Cannot explain why specific wavelengths (Balmer, Lyman lines) are emitted.
• Limited applicability: Cannot account for fine structure, Zeeman effect, or chemical bonding.

12.6 Bohr’s Postulates (1913)

1. Stationary orbits: Electrons revolve in certain non-radiating, allowed circular orbits called stationary states.
2. Quantization of angular momentum: Only those orbits are permitted in which the angular momentum is an integral multiple of $\hbar = h/2\pi$:

$$L = mvr = \frac{nh}{2\pi}, \quad n = 1, 2, 3, \dots$$

3. Frequency condition: Radiation is emitted/absorbed only when the electron jumps between two stationary states:

$$h\nu = E_i – E_f$$

12.7 Bohr Radius Derivation

Coulomb attraction provides the centripetal force for an electron of mass $m$ orbiting a nucleus of charge $+Ze$:

$$\frac{1}{4\pi\epsilon_0}\cdot\frac{Ze^2}{r^2} = \frac{mv^2}{r}$$

Combining with $mvr = nh/2\pi$, eliminating $v$ gives the radius of the $n^\text{th}$ orbit:

$$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2 Z}$$

For hydrogen ($Z = 1$, $n = 1$): $r_1 = 0.529 \times 10^{-10}$ m $= 0.53$ Å, the celebrated Bohr radius $a_0$. Higher orbits scale as $r_n = n^2 a_0$, so $r_2 = 2.12$ Å, $r_3 = 4.77$ Å.

12.8 Energy of Stationary States

Total energy $E = K + U$ where $K = \tfrac{1}{2}mv^2$ and $U = -Ze^2/(4\pi\epsilon_0 r)$. Substituting:

$$E_n = -\frac{m e^4 Z^2}{8\epsilon_0^2 h^2 n^2} = -\frac{13.6 \, Z^2}{n^2}\text{ eV}$$

For hydrogen $Z = 1$: $E_1 = -13.6$ eV (ground state), $E_2 = -3.4$ eV, $E_3 = -1.51$ eV, $E_\infty = 0$. The negative sign indicates the electron is bound. Ionisation energy = $|E_1| = 13.6$ eV. Note that $K = -E$ (always positive) and $U = 2E$ (twice as negative as the total energy).

12.9 Hydrogen Spectral Series

When an electron jumps from a higher orbit $n_i$ to a lower orbit $n_f$, a photon is emitted whose wavelength is given by the Rydberg formula:

$$\frac{1}{\lambda} = R\left(\frac{1}{n_f^2} – \frac{1}{n_i^2}\right), \quad R = 1.097 \times 10^7 \text{ m}^{-1}$$

Series	$n_f$	$n_i$	Region	Discovered	First line $\lambda$
Lyman	1	2, 3, 4, …	Ultraviolet	1906	121.6 nm
Balmer	2	3, 4, 5, …	Visible	1885	656.3 nm (H$_\alpha$, red)
Paschen	3	4, 5, 6, …	Infrared	1908	1875 nm
Brackett	4	5, 6, 7, …	Far IR	1922	4050 nm
Pfund	5	6, 7, 8, …	Far IR	1924	7460 nm

NCERT Exercise — Solved

Q1. Choose the correct alternative from the clues given:

(a) The size of the atom in Thomson’s model is __________ the atomic size in Rutherford’s model.

Answer: not different from (both are about $10^{-10}$ m).

(b) In the ground state of __________ electrons are in stable equilibrium, while in __________ electrons always experience a net force.

Answer: Thomson’s model; Rutherford’s model.

(c) A classical atom based on __________ model is doomed to collapse.

Answer: Rutherford’s model — accelerating electrons radiate, lose energy and spiral inward.

(d) An atom has a nearly continuous mass distribution in __________ but a highly non-uniform mass distribution in __________.

Answer: Thomson’s model; Rutherford’s model.

(e) The positively charged part of the atom possesses most of the mass in __________.

Answer: Both Thomson’s and Rutherford’s models.

Q2. Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen instead of gold. (Hydrogen is solid at temperatures below 14 K.) What results do you expect?

Answer: The mass of a hydrogen nucleus (proton) is about one-quarter the mass of an alpha particle. By conservation of momentum, scattering by such a light target cannot produce large-angle deflections of the heavier projectile — large-angle scattering events would be virtually absent. Hence Rutherford’s discovery of the nucleus would not have been possible with hydrogen.

Q3. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

Answer: $\Delta E = h\nu \Rightarrow \nu = \Delta E / h$. With $\Delta E = 2.3 \times 1.6 \times 10^{-19}$ J $= 3.68 \times 10^{-19}$ J and $h = 6.626 \times 10^{-34}$ J·s:

$$\nu = \frac{3.68 \times 10^{-19}}{6.626 \times 10^{-34}} = 5.55 \times 10^{14} \text{ Hz}$$

Q4. The ground state energy of hydrogen atom is $-13.6$ eV. What are the kinetic and potential energies of the electron in this state?

Answer: For a Coulomb-bound electron, $K = -E$ and $U = 2E$. Hence $K = +13.6$ eV and $U = -27.2$ eV. Sum: $E = K + U = -13.6$ eV. ✓

Q5. A hydrogen atom initially in the ground state absorbs a photon, exciting it to $n = 4$ level. Determine the wavelength and frequency of the photon.

Answer: $\Delta E = E_4 – E_1 = -13.6/16 – (-13.6) = 12.75$ eV $= 2.04 \times 10^{-18}$ J.

$$\nu = \frac{\Delta E}{h} = \frac{2.04 \times 10^{-18}}{6.626 \times 10^{-34}} = 3.08 \times 10^{15} \text{ Hz}$$

$$\lambda = \frac{c}{\nu} = \frac{3 \times 10^8}{3.08 \times 10^{15}} = 97.5 \text{ nm}$$

Q6. (a) Using the Bohr model, calculate the speed of the electron in a hydrogen atom in the $n = 1, 2, 3$ levels. (b) Calculate the orbital period in each level.

Answer: Speed in the $n^\text{th}$ orbit is $v_n = e^2/(2\epsilon_0 n h) = c\alpha/n$ where $\alpha \approx 1/137$. Thus:

• $v_1 = 2.19 \times 10^6$ m/s
• $v_2 = 1.09 \times 10^6$ m/s
• $v_3 = 7.28 \times 10^5$ m/s

Period $T_n = 2\pi r_n / v_n \propto n^3$:

• $T_1 = 1.52 \times 10^{-16}$ s
• $T_2 = 1.22 \times 10^{-15}$ s
• $T_3 = 4.12 \times 10^{-15}$ s

Q7. The radius of the innermost electron orbit of a hydrogen atom is $5.3 \times 10^{-11}$ m. What are the radii of the $n = 2$ and $n = 3$ orbits?

Answer: Since $r_n = n^2 r_1$:

• $r_2 = 4 \times 5.3 \times 10^{-11} = 2.12 \times 10^{-10}$ m
• $r_3 = 9 \times 5.3 \times 10^{-11} = 4.77 \times 10^{-10}$ m

Q8. A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Answer: The energy 12.5 eV is enough to excite the electron from $n = 1$ ($-13.6$ eV) up to $n = 3$ ($-1.51$ eV; $\Delta E = 12.09$ eV) but not to $n = 4$ ($\Delta E = 12.75$ eV). De-excitation gives Lyman ($3 \to 1$, $2 \to 1$) and Balmer ($3 \to 2$) series wavelengths.

Q9. In accordance with the Bohr model, find the quantum number that characterizes the Earth’s revolution around the Sun in an orbit of radius $1.5 \times 10^{11}$ m with orbital speed $3 \times 10^4$ m/s. (Mass of Earth = $6.0 \times 10^{24}$ kg.)

Answer: $L = mvr = 2.7 \times 10^{40}$ kg·m²/s. Setting $L = nh/2\pi$:

$$n = \frac{2\pi L}{h} = \frac{2\pi (2.7 \times 10^{40})}{6.626 \times 10^{-34}} \approx 2.6 \times 10^{74}$$

This astronomically large $n$ shows that the correspondence principle is satisfied — quantum effects are imperceptible for macroscopic systems.

Additional ASSEB-Pattern Questions

Q10. Define impact parameter and write its expression for alpha-particle scattering.

Answer: Impact parameter $b$ is the perpendicular distance from the centre of the nucleus to the initial line of motion of the alpha particle. $b = (Ze^2 \cot(\theta/2))/(4\pi\epsilon_0 E_K)$. Smaller $b$ produces larger scattering angle.

Q11. Show that $r_n \propto n^2$ in Bohr’s hydrogen atom.

Answer: From Coulomb-centripetal balance and angular momentum quantization, $r_n = n^2 h^2 \epsilon_0/(\pi m e^2 Z)$. For fixed $Z$, $r_n \propto n^2$.

Q12. Calculate the wavelength of the first line of the Balmer series.

Answer: $1/\lambda = R(1/4 – 1/9) = R(5/36)$. With $R = 1.097 \times 10^7$ m$^{-1}$, $\lambda = 656.3$ nm (red H-alpha line).

Q13. State two limitations of Bohr’s model.

Answer: (i) Applies only to hydrogen-like one-electron systems; fails for multi-electron atoms. (ii) Cannot explain fine structure of spectral lines, the Zeeman effect, or relative intensities. (iii) Mixes classical orbits with ad-hoc quantum rules — superseded by Schrödinger’s wave mechanics.

Q14. Why is the energy of an electron in a Bohr orbit negative?

Answer: The electron is bound by the attractive Coulomb force; the zero of potential energy is fixed at infinite separation, so any bound state has $E < 0$. Energy must be supplied to ionize the atom.

Q15. Determine the ratio of the longest to the shortest wavelength in the Lyman series.

Answer: Longest: $n_i=2 \to n_f=1$, $1/\lambda_L = R(1 – 1/4) = 3R/4$. Shortest: $n_i=\infty$, $1/\lambda_S = R$. Ratio $\lambda_L/\lambda_S = 4/3 \approx 1.33$.

Q16. Calculate the distance of closest approach for an alpha particle of energy 5 MeV scattering from a gold nucleus ($Z = 79$).

Answer: $r_0 = (1/4\pi\epsilon_0)(2Ze^2/E_K)$. Plugging values: $r_0 = (9 \times 10^9)(2 \times 79 \times (1.6 \times 10^{-19})^2)/(5 \times 10^6 \times 1.6 \times 10^{-19}) = 4.55 \times 10^{-14}$ m.

Q17. The wavelength of $H_\alpha$ line of hydrogen is 6563 Å. Calculate the Rydberg constant.

Answer: $1/\lambda = R(1/4 – 1/9) = 5R/36 \Rightarrow R = 36/(5\lambda) = 36/(5 \times 6.563 \times 10^{-7}) = 1.097 \times 10^7$ m$^{-1}$.

Q18. What is the de Broglie justification of Bohr’s quantization?

Answer: An electron behaves like a wave of wavelength $\lambda = h/(mv)$. A stable orbit requires the circumference to contain an integer number of wavelengths: $2\pi r = n\lambda \Rightarrow mvr = nh/2\pi$.

Q19. Why did Thomson’s model fail to account for the alpha-scattering result?

Answer: A diffuse positive cloud cannot produce the strong electric field needed to deflect a fast alpha particle through large angles. Backward scattering requires the entire positive charge concentrated in a tiny nucleus.

Q20. The total energy of an electron in the first excited state of hydrogen is $-3.4$ eV. What are its kinetic and potential energies?

Answer: $K = -E = +3.4$ eV; $U = 2E = -6.8$ eV.

Worked Numerical Examples

Example A. Compute the energy required to excite a hydrogen atom from $n = 2$ to $n = 4$.

$$\Delta E = -13.6\left(\frac{1}{16} – \frac{1}{4}\right) = -13.6 \times (-0.1875) = 2.55 \text{ eV}$$

Example B. Find the wavelength of the second line of the Lyman series.

$$\frac{1}{\lambda} = R\left(\frac{1}{1} – \frac{1}{9}\right) = \frac{8R}{9}$$

$$\lambda = \frac{9}{8 \times 1.097 \times 10^7} = 1.025 \times 10^{-7} \text{ m} = 102.5 \text{ nm}$$

Example C. An electron jumps from the $n = 5$ to $n = 3$ orbit. To which series does the emitted line belong, and what is its wavelength?

$$\frac{1}{\lambda} = R\left(\frac{1}{9} – \frac{1}{25}\right) = R \cdot \frac{16}{225}$$

$$\lambda = \frac{225}{16 \times 1.097 \times 10^7} = 1282 \text{ nm (Paschen series, IR)}$$

Example D. Calculate the orbital frequency of the electron in the ground state of hydrogen.

$$f = \frac{v_1}{2\pi r_1} = \frac{2.19 \times 10^6}{2\pi \times 5.3 \times 10^{-11}} \approx 6.58 \times 10^{15} \text{ Hz}$$

Glossary

Term	Meaning
Alpha particle	Helium-4 nucleus, $^4_2$He, charge $+2e$, used as projectile in scattering experiments.
Angstrom (Å)	Unit of length, $1$ Å $= 10^{-10}$ m, convenient for atomic dimensions.
Bohr radius	Radius of the smallest orbit in hydrogen, $a_0 = 0.529$ Å.
Distance of closest approach	Minimum separation in head-on collision; all kinetic energy turns into potential energy.
Excited state	Any allowed energy state above the ground state.
Ground state	Lowest energy state of an atom ($n = 1$ for hydrogen).
Hydrogenic atom	Single-electron ion (He⁺, Li²⁺) — Bohr formulas apply with $Z^2$ scaling.
Impact parameter	Perpendicular distance from nucleus centre to undeflected projectile path.
Ionisation energy	Energy needed to free the electron from the ground state to infinity ($13.6$ eV for H).
Lyman series	Spectral lines from transitions to $n_f = 1$, in ultraviolet.
Nucleus	Tiny dense positive core of an atom containing protons and neutrons.
Quantum number	Integer $n$ labelling allowed Bohr orbits; specifies energy and angular momentum.
Rydberg constant	$R = 1.097 \times 10^7$ m$^{-1}$; fundamental constant of hydrogen spectroscopy.
Stationary state	Allowed orbit in which an electron does not radiate energy.
Spectral series	Family of lines with common lower level $n_f$.

Detailed Conceptual Discussions

A. Why Did Thomson’s Model Persist for a Decade?

Despite its limitations, Thomson’s plum-pudding model dominated atomic physics from 1898 until 1911 because it offered the first plausible structural picture incorporating the newly discovered electron. The model could explain electrical neutrality of atoms, the existence of positive and negative ions formed during ionization, and qualitatively the emission of light by oscillating electrons. Mathematically, oscillation frequencies of electrons embedded in a uniform positive sphere of radius $R$ satisfy $\omega^2 = e^2/(4\pi\epsilon_0 m R^3)$. For $R = 10^{-10}$ m this yields $\omega \sim 10^{16}$ rad/s — comfortably in the visible/UV range. The model fell only when more energetic alpha particles became available as probes, allowing physicists to “see” the atom’s interior.

B. Geometry and Statistics of Rutherford Scattering

Rutherford derived the differential cross-section for Coulomb scattering of a charged particle by a heavy nucleus:

$$\frac{d\sigma}{d\Omega} = \left(\frac{Ze^2}{4 E_K \sin^2(\theta/2)}\cdot\frac{1}{4\pi\epsilon_0}\right)^2$$

The number of alpha particles scattered into a small solid angle $d\Omega$ at angle $\theta$ from the beam direction follows the inverse fourth power of $\sin(\theta/2)$. Geiger and Marsden’s data from 1909–1913 confirmed this dependence over four orders of magnitude in scattering angle, brilliantly vindicating the point-nucleus picture. The minute size deduced for the gold nucleus, $\sim 10^{-14}$ m, set the stage for nuclear physics and ultimately led to the discovery of the proton (1919) and the neutron (1932).

C. Energy of the Bohr Orbits — Step-by-Step Derivation

Step 1 — Equate Coulomb force to centripetal force:

$$\frac{Z e^2}{4\pi\epsilon_0 r^2} = \frac{m v^2}{r} \quad\Rightarrow\quad m v^2 = \frac{Z e^2}{4\pi\epsilon_0 r}$$

Step 2 — Use Bohr quantization $m v r = n h/(2\pi)$ to solve for $v$:

$$v = \frac{n h}{2\pi m r}$$

Step 3 — Substitute back to get the orbital radius:

$$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2 Z}$$

Step 4 — Compute the kinetic and potential energies:

$$K = \frac{1}{2} m v^2 = \frac{Z e^2}{8\pi\epsilon_0 r_n}, \quad U = -\frac{Z e^2}{4\pi\epsilon_0 r_n}$$

Step 5 — Add to obtain total energy:

$$E_n = K + U = -\frac{Z e^2}{8\pi\epsilon_0 r_n} = -\frac{m e^4 Z^2}{8 \epsilon_0^2 h^2 n^2}$$

Numerically, $E_n = -13.6 \, Z^2/n^2$ eV. The relationship $|U| = 2K$ is a manifestation of the virial theorem for inverse-square attractive forces.

D. From Bohr’s Frequency Condition to the Rydberg Formula

Combining $h\nu = E_i – E_f$ with $E_n = -13.6/n^2$ eV and $\nu = c/\lambda$:

$$\frac{1}{\lambda} = \frac{m e^4}{8 \epsilon_0^2 h^3 c}\left(\frac{1}{n_f^2} – \frac{1}{n_i^2}\right)$$

The leading constant $R = m e^4/(8\epsilon_0^2 h^3 c)$ is precisely the Rydberg constant. Plugging fundamental constants:

$$R = \frac{(9.11 \times 10^{-31})(1.6 \times 10^{-19})^4}{8 \times (8.85 \times 10^{-12})^2 \times (6.626 \times 10^{-34})^3 \times (3 \times 10^8)} = 1.097 \times 10^7 \text{ m}^{-1}$$

This was a triumph of theoretical physics — a constant previously fitted from spectral data was now derived from microscopic constants of nature.

E. Velocity, Frequency and Period in Bohr Orbits

Speed: $v_n = Z e^2 /(2 \epsilon_0 n h) = (Z\alpha c)/n$, where $\alpha = e^2/(2\epsilon_0 h c) \approx 1/137$ is the fine-structure constant. For hydrogen ground state, $v_1 \approx c/137 = 2.19 \times 10^6$ m/s, well below relativistic speeds — Bohr’s non-relativistic treatment is valid.

Orbital frequency $f_n = v_n/(2\pi r_n) \propto 1/n^3$. Period $T_n = 1/f_n \propto n^3$. Numbers for hydrogen:

$n$	$r_n$ (Å)	$v_n$ (m/s)	$T_n$ (s)	$E_n$ (eV)
1	0.53	$2.19 \times 10^6$	$1.5 \times 10^{-16}$	−13.60
2	2.12	$1.09 \times 10^6$	$1.2 \times 10^{-15}$	−3.40
3	4.77	$7.28 \times 10^5$	$4.1 \times 10^{-15}$	−1.51
4	8.48	$5.47 \times 10^5$	$9.7 \times 10^{-15}$	−0.85
5	13.25	$4.38 \times 10^5$	$1.9 \times 10^{-14}$	−0.54

F. Hydrogenic Ions: $\text{He}^+$, $\text{Li}^{2+}$, $\text{Be}^{3+}$

The Bohr theory generalises directly to any single-electron ion of nuclear charge $+Ze$: the orbit shrinks by $1/Z$ and the energy deepens by $Z^2$:

$$r_n = \frac{n^2 a_0}{Z}, \qquad E_n = -\frac{13.6 Z^2}{n^2} \text{ eV}$$

Examples:

• $\text{He}^+$ ($Z = 2$): ground-state energy $-54.4$ eV; first orbit radius $0.265$ Å.
• $\text{Li}^{2+}$ ($Z = 3$): $-122.4$ eV; $0.176$ Å.
• $\text{Be}^{3+}$ ($Z = 4$): $-217.6$ eV; $0.132$ Å.

For multi-electron neutral atoms, electron-electron repulsion forbids the simple Bohr picture; one must use Hartree-Fock or density-functional methods.

G. Excitation, De-excitation and Ionization

Excitation energy is the energy needed to lift the electron from the ground state to a higher allowed state. For hydrogen:

• First excitation: $E_2 – E_1 = 10.2$ eV
• Second excitation: $E_3 – E_1 = 12.09$ eV
• Third excitation: $E_4 – E_1 = 12.75$ eV

Ionization energy is the energy required to remove the electron entirely: $E_\infty – E_1 = 13.6$ eV. Photons of energy $\geq 13.6$ eV (wavelength $\leq 91.2$ nm) ionize the hydrogen atom from its ground state.

De-excitation occurs when the excited atom spontaneously drops to a lower level emitting a photon of frequency $\nu = (E_i – E_f)/h$. The mean lifetime of an excited state is typically $10^{-8}$ s.

H. Continuous and Line Spectra

Hot solids (incandescent filament) emit continuous spectra spanning all wavelengths — energy levels in solids form bands. Hot gases at low pressure emit discrete line spectra with sharp wavelengths characteristic of the element. Each element’s spectrum is its fingerprint: Helium was first detected in the Sun’s chromosphere through unidentified yellow lines (1868) before being isolated on Earth (1895). Bohr’s quantum theory finally explained why isolated atoms emit only discrete frequencies — only certain electron transitions are permitted.

More Practice Problems with Solutions

P1. The radius of the first Bohr orbit of hydrogen is $0.53$ Å. Find the radius of the first Bohr orbit of $\text{He}^+$.

Answer: $r_1(\text{He}^+) = a_0/Z = 0.53/2 = 0.265$ Å.

P2. Calculate the wavelength of the photon emitted when the electron in $\text{He}^+$ jumps from $n = 2$ to $n = 1$.

Answer: $1/\lambda = R Z^2 (1/1 – 1/4) = 4R(3/4) = 3R$. So $\lambda = 1/(3 \times 1.097 \times 10^7) = 3.04 \times 10^{-8}$ m $= 30.4$ nm.

P3. The shortest wavelength in the Lyman series is 91.2 nm. Find the shortest wavelength in the Balmer series.

Answer: Series limits: $\lambda_\text{Lyman}^\text{min} = 1/R$, $\lambda_\text{Balmer}^\text{min} = 4/R$. Ratio = 4. So $\lambda = 4 \times 91.2 = 364.8$ nm.

P4. The angular momentum of an electron in the second orbit of hydrogen is __________.

Answer: $L_2 = 2 h/(2\pi) = h/\pi = 2.11 \times 10^{-34}$ J·s.

P5. Compute the wave number of the first line of the Paschen series.

Answer: $\bar\nu = 1/\lambda = R(1/9 – 1/16) = R(7/144) = 5.33 \times 10^5$ m$^{-1}$.

P6. Find the value of $n$ for which the radius of the Bohr orbit equals $4.77$ Å.

Answer: $4.77 = n^2 \times 0.53 \Rightarrow n^2 = 9 \Rightarrow n = 3$.

P7. The energy of an electron in the $n^\text{th}$ orbit of hydrogen is $-1.51$ eV. Identify $n$.

Answer: $-13.6/n^2 = -1.51 \Rightarrow n^2 = 9 \Rightarrow n = 3$.

P8. Determine the recoil speed of a hydrogen atom (mass $1.67 \times 10^{-27}$ kg) when an electron jumps from $n = 4$ to $n = 1$.

Answer: Photon energy $E = 12.75$ eV $= 2.04 \times 10^{-18}$ J. Photon momentum $p = E/c = 6.8 \times 10^{-27}$ kg\cdotm/s. Recoil speed $v = p/M = 4.07$ m/s.

P9. State the postulates of Bohr’s atomic model.

Answer: (i) Electrons revolve in stationary, non-radiating circular orbits. (ii) Angular momentum is quantized: $L = n h/(2\pi)$. (iii) Photons are emitted/absorbed only during transitions: $h\nu = E_i – E_f$.

P10. The Lyman series of hydrogen lies in the ultraviolet region. Why?

Answer: Transitions to $n_f = 1$ release energies between $10.2$ eV and $13.6$ eV, corresponding to wavelengths $91.2 – 121.6$ nm — the ultraviolet region of the electromagnetic spectrum.

Common Misconceptions Clarified

• Negative energy does not mean less than zero in absolute terms — it merely indicates a bound state relative to a free electron at infinity (which is set to zero by convention).
• Bohr orbits are not literal “tracks” in space — modern quantum mechanics replaces them with probability clouds (orbitals). Bohr’s circles are useful semi-classical visualisations.
• The fine-structure constant $\alpha \approx 1/137$ is dimensionless, controls the strength of electromagnetic interaction, and appears naturally in $v_1 = \alpha c$.
• Rutherford did not “disprove” Thomson directly; he supplemented it. Both share that the positive charge carries most of the mass — they differ in how it is distributed.
• Distance of closest approach is not the nuclear radius — it is an upper bound. The actual nuclear radius is smaller and is determined by other techniques (electron scattering).

Quick Recap — Memory Aids

• $r_n \propto n^2$ — radii expand as $1, 4, 9, 16$.
• $E_n \propto -1/n^2$ — energies $-13.6, -3.4, -1.51, -0.85$ eV.
• $v_n \propto 1/n$ — speeds drop $1, 1/2, 1/3, 1/4$.
• $T_n \propto n^3$ — periods grow as $1, 8, 27, 64$.
• $L_n = n\hbar$ — angular momentum increments in units of $\hbar$.
• Mnemonic for series: Lovely Boys Play Basketball in Paris (Lyman, Balmer, Paschen, Brackett, Pfund) — descending in energy, ascending in $n_f$.

Long-Answer Theory Questions (ASSEB Pattern)

L1. Describe Rutherford’s alpha-scattering experiment with a neat diagram and write its three main observations and conclusions.

Answer: A radioactive radium-bismuth source emits 5.5 MeV alpha particles, collimated by a lead slit and directed onto a thin (~10$^{-7}$ m) gold foil placed in an evacuated chamber. A movable zinc-sulphide screen with microscope detects scintillations. Observations: (i) Most alphas pass through undeflected. (ii) A small fraction is deflected through angles greater than 1°. (iii) About 1 in 8000 is scattered through more than 90°, and a few even rebound. Conclusions: (i) Atoms are mostly empty space. (ii) Almost all the positive charge and mass is concentrated in a tiny nucleus of size ~$10^{-15}$ m. (iii) Electrons orbit at distances ~$10^{-10}$ m. The nuclear atom replaced the plum-pudding model.

L2. Derive an expression for the radius of the $n^\text{th}$ Bohr orbit and the energy of the electron in that orbit.

Answer: Equating Coulomb force and centripetal force: $Ze^2/(4\pi\epsilon_0 r^2) = mv^2/r$. Using Bohr quantization $mvr = nh/(2\pi)$, eliminate $v$ to get $r_n = n^2 h^2 \epsilon_0/(\pi m e^2 Z)$. Total energy $E = K + U = -Ze^2/(8\pi\epsilon_0 r_n) = -13.6 Z^2/n^2$ eV. The negative sign indicates a bound state.

L3. Explain the origin of the spectral series of hydrogen and write the Rydberg formula.

Answer: When an excited hydrogen atom de-excites from a higher orbit $n_i$ to a lower orbit $n_f$, it emits a photon whose energy equals the energy difference between the two states. The wavelengths follow $1/\lambda = R(1/n_f^2 – 1/n_i^2)$. Each fixed $n_f$ defines a spectral series: Lyman ($n_f = 1$, UV), Balmer ($n_f = 2$, visible), Paschen ($n_f = 3$, IR), Brackett ($n_f = 4$), Pfund ($n_f = 5$).

L4. State Bohr’s postulates and use them to explain hydrogen line spectrum.

Answer: Bohr postulated stationary non-radiating orbits, angular-momentum quantization $L = nh/2\pi$, and the frequency condition $h\nu = E_i – E_f$. Combined, these give discrete energy levels $E_n = -13.6/n^2$ eV. Transitions between these levels emit photons with wavelengths fitting the Rydberg formula, reproducing the observed Lyman, Balmer, etc. series.

L5. Show that the orbital frequency of the electron in the $n^\text{th}$ Bohr orbit varies as $1/n^3$.

Answer: $f_n = v_n/(2\pi r_n)$. Since $v_n \propto 1/n$ and $r_n \propto n^2$, we get $f_n \propto 1/n^3$. Numerically $f_1 = 6.58 \times 10^{15}$ Hz; for large $n$ this matches Bohr’s correspondence-principle limit.

L6. Derive an expression for the impact parameter $b$ in alpha-particle scattering.

Answer: Conservation of energy and angular momentum during Coulomb scattering yields $b = (Ze^2/(4\pi\epsilon_0 E_K))\cot(\theta/2)$. Head-on collision ($b = 0$) gives $\theta = 180°$; $\theta = 90°$ corresponds to $b = Ze^2/(4\pi\epsilon_0 E_K)$.

Short Definition Questions (1-Mark)

S1. What is meant by ‘stationary orbit’ in Bohr’s atomic model?

Answer: An allowed orbit in which an electron revolves without radiating energy, although it is accelerating.

S2. State the Bohr quantization rule for angular momentum.

Answer: $L = mvr = nh/(2\pi)$, where $n = 1, 2, 3, \dots$ is the principal quantum number.

S3. Define ionization energy of hydrogen.

Answer: The minimum energy needed to remove the electron from the ground state to infinity, equal to $13.6$ eV.

S4. What is the SI unit of the Rydberg constant?

Answer: $\text{m}^{-1}$ (per metre, or reciprocal metre).

S5. Which series of hydrogen lies entirely in the ultraviolet?

Answer: The Lyman series ($n_f = 1$), with wavelengths between 91.2 and 121.6 nm.

S6. Why are alpha particles preferred over beta particles in scattering experiments?

Answer: Alphas are massive ($\sim 7000 \times m_e$) and doubly charged, giving them strong electrostatic interaction with the nucleus and stable, easily-tracked trajectories.

S7. State the value of the Bohr radius.

Answer: $a_0 = 0.529 \times 10^{-10}$ m $\approx 0.53$ Å.

S8. What does a positive value of total energy of an electron imply?

Answer: The electron is unbound — it has been ionized and behaves as a free particle with kinetic energy.

S9. Why does Bohr’s model fail for multi-electron atoms?

Answer: It does not include electron-electron repulsion, screening effects, or quantum spin — all of which become significant when more than one electron is present.

S10. What is the relation between potential and kinetic energy of an electron in a Bohr orbit?

Answer: $U = -2K$ (virial theorem); the total energy $E = K + U = -K$.

Multiple-Choice Questions (Self-Test)

MCQ 1. The size of an atom is approximately: (a) $10^{-15}$ m (b) $10^{-10}$ m (c) $10^{-6}$ m (d) $10^{-3}$ m

Answer: (b) $10^{-10}$ m.

MCQ 2. According to Bohr’s postulate, angular momentum is an integral multiple of: (a) $h$ (b) $h/2\pi$ (c) $h/\pi$ (d) $2\pi h$

Answer: (b) $h/2\pi$.

MCQ 3. The energy of the electron in the second excited state of hydrogen is: (a) $-13.6$ eV (b) $-3.4$ eV (c) $-1.51$ eV (d) $-0.85$ eV

Answer: (c) $-1.51$ eV (third level, $n = 3$).

MCQ 4. Which of the following is NOT a postulate of Bohr’s model? (a) Stationary orbits (b) Quantized angular momentum (c) Continuous emission (d) Photon emission on transition

Answer: (c) Continuous emission — Bohr’s atom radiates only during quantum jumps.

MCQ 5. The shortest wavelength in the Balmer series corresponds to a transition: (a) $n = 3 \to 2$ (b) $n = \infty \to 2$ (c) $n = 2 \to 1$ (d) $n = \infty \to 1$

Answer: (b) $n = \infty \to 2$, $\lambda = 364.6$ nm.

MCQ 6. The de Broglie wavelength of an electron in the first Bohr orbit equals: (a) $r_1$ (b) $2 r_1$ (c) $2\pi r_1$ (d) $4\pi r_1$

Answer: (c) $2\pi r_1$ — orbit circumference contains exactly one wavelength.

MCQ 7. Rutherford concluded the existence of nucleus mainly because of: (a) Most alphas passed through (b) Some alphas were back-scattered (c) Alphas were absorbed (d) None

Answer: (b) Back-scattered alphas — only a concentrated positive nucleus could deflect them so strongly.

MCQ 8. The ratio of the radii of orbits $n = 2$ and $n = 1$ in hydrogen is: (a) 2 : 1 (b) 4 : 1 (c) 1 : 2 (d) 1 : 4

Answer: (b) 4 : 1, since $r_n \propto n^2$.

MCQ 9. Which series of hydrogen is in the visible region? (a) Lyman (b) Balmer (c) Paschen (d) Brackett

Answer: (b) Balmer.

MCQ 10. The product of mass, velocity and radius for an electron in the second Bohr orbit is: (a) $h/2\pi$ (b) $h/\pi$ (c) $3h/2\pi$ (d) $2h/\pi$

Answer: (b) $h/\pi$ ($n = 2$, so $L = 2 \cdot h/(2\pi) = h/\pi$).

Historical Context and Modern Relevance

Niels Bohr was a 28-year-old postdoctoral researcher when he proposed his trilogy of papers (July, September, November 1913) in Philosophical Magazine. His breakthrough married Planck’s quantum hypothesis (1900) with Rutherford’s nucleus (1911) into a working atomic theory. Although superseded a decade later by Heisenberg’s matrix mechanics (1925) and Schrödinger’s wave mechanics (1926), Bohr’s model remains pedagogically vital. It supplied the energy formulae, principal quantum number, the correspondence principle, and a template for thinking about quantum systems that physicists still use today.

The Rydberg constant is now measured to thirteen decimal places ($1.0973731568160 \times 10^7$ m$^{-1}$) — one of the most precise numbers in physics — and underpins atomic clocks, frequency comb metrology and tests of quantum electrodynamics. The hydrogen $1S$-$2S$ transition has been measured to $10^{-15}$ relative precision, providing stringent limits on time-variation of fundamental constants. Bohr’s stationary states have grown into modern quantum theory’s energy eigenstates, and his frequency rule $h\nu = E_i – E_f$ remains exactly correct.

Frequently Asked Examination Questions

F1. Why does the Rutherford model fail to explain the stability of an atom?

Answer: An accelerating charged particle radiates electromagnetic energy according to classical electromagnetism. An orbiting electron in Rutherford’s model is always accelerating (centripetally), so it should continuously radiate, lose energy and spiral into the nucleus within $\sim 10^{-8}$ s — yet atoms are stable. The model also cannot explain the discrete spectrum of hydrogen.

F2. Define wavenumber. What are its units?

Answer: Wavenumber $\bar\nu = 1/\lambda$ is the number of waves per unit length. SI unit is $\text{m}^{-1}$. Spectroscopists often use cm$^{-1}$.

F3. What is the limiting line of Lyman series?

Answer: Transition $n_i = \infty \to n_f = 1$ gives $1/\lambda = R$, hence $\lambda = 91.2$ nm. This represents the series limit beyond which the spectrum becomes a continuum (ionization).

F4. Compute the recoil energy of the hydrogen atom when the electron drops from $n=2$ to $n=1$.

Answer: Photon momentum $p = E/c = 10.2 \times 1.6 \times 10^{-19}/3 \times 10^8 = 5.44 \times 10^{-27}$ kg·m/s. Recoil energy $E_R = p^2/(2M) = 8.84 \times 10^{-27}$ J $= 5.5 \times 10^{-8}$ eV — much smaller than the photon energy.

F5. Two energy levels of an atom differ by $4.14$ eV. Find the wavelength of the photon emitted in the transition from upper to lower level.

Answer: $\lambda = hc/E = (1240 \text{eV}\cdot\text{nm})/4.14 \text{ eV} = 299.5$ nm — ultraviolet.

F6. The first member of the Balmer series of hydrogen has wavelength $6563$ Å. Calculate the wavelength of the second member.

Answer: First: $1/\lambda_1 = R(1/4 – 1/9) = 5R/36$. Second: $1/\lambda_2 = R(1/4 – 1/16) = 3R/16$. Ratio $\lambda_2/\lambda_1 = (5R/36)/(3R/16) = 80/108 = 20/27$. So $\lambda_2 = 6563 \times 20/27 = 4862$ Å.

F7. Why does the Bohr model fail to predict relative intensities of spectral lines?

Answer: Bohr’s model gives only allowed energies and frequencies, not transition probabilities. Predicting intensities requires the full quantum mechanical treatment with matrix elements between initial and final wavefunctions (Schrödinger equation).

F8. What is the significance of the negative sign in $E_n$?

Answer: It indicates that the electron is bound to the nucleus; energy must be supplied to free it. The reference (zero) of energy is taken at infinite separation where the electron has no kinetic energy.

Conclusion: Chapter 12 captures one of the most exciting transitions in modern physics — from classical orbital pictures to quantized atoms. With Rutherford’s nucleus, Bohr’s stationary states, and the Rydberg formula, you now possess the conceptual toolkit to interpret atomic spectra and to appreciate why classical electromagnetism alone cannot describe the microscopic world. Practice the worked examples and additional questions to gain fluency, and consult NCERT diagrams for the energy-level chart. Stay tuned to HSLC Guru for Chapter 13 — Nuclei, where we apply similar quantum logic to nuclear binding and radioactive decay.