Class 12 Physics NCERT REPRINT 2026-27

Chapter 12: Atoms

Complete concise study notes, alpha particle scattering, Rutherford nuclear model, distance of closest approach, Bohr's postulates, hydrogen energy levels and spectrum, de Broglie's explanation, solved examples 12.1–12.4, and exercises 12.1–12.9.

01 / Exam-Focused Notes

Chapter 12: Atoms

Complete official NCERT textbook coverage with high-precision vector diagrams: Geiger-Marsden Alpha Scattering Experiment, Impact Parameter ($b$) & Distance of Closest Approach ($d$), Rutherford Planetary Model vs Maxwell Radiation Collapse, Bohr's 3 Quantum Postulates ($L = n\hbar$), Quantised Orbital Radii ($r_n = n^2 a_0$) & Energy Levels ($E_n = -13.6\text{ eV}/n^2$), Hydrogen Spectral Series (Lyman, Balmer, Paschen, Brackett, Pfund), and de Broglie Standing Matter Wave interpretation ($2\pi r_n = n\lambda$).

100% SYLLABUS
Chapter Navigation • NCERT Sections 12.1 – 12.6
12.1-12.2Rutherford Model 12.2.2-12.3Electron Orbits 12.4-12.5Bohr Model & Spectra 12.6de Broglie Standing Waves
12.1 & 12.2
NCERT Sections

Introduction, Alpha-Particle Scattering & Rutherford's Nuclear Model

Geiger-Marsden Experiment

12.2.1 Alpha-Particle Scattering & Discovery of the Nucleus

  • Experimental Setup: $5.5\text{ MeV}$ $\alpha$-particles ($^4_2\text{He}^{2+}$) from a $^{214}_{\ 83}\text{Bi}$ radioactive source collimated by lead bricks incident on a thin gold foil (thickness $2.1 \times 10^{-7}\text{ m} \approx 210\text{ nm}$). Scintillations counted via a rotatable $\text{ZnS}$ detector screen and microscope.
  • Observations:
    • Most $\alpha$-particles ($> 99.86\%$) passed straight through without deflection $\implies$ most of the atom is empty space.
    • Only about $0.14\%$ scattered by $\theta > 1^\circ$.
    • Only about $1$ in $8000$ bounced back by $\theta > 90^\circ$ $\implies$ entire positive charge and almost all mass is concentrated in a tiny central core called the nucleus (size $\sim 10^{-15}\text{ m}$ vs atomic size $\sim 10^{-10}\text{ m}$).
  • Coulomb Scattering Force ($Z=79$ for Gold): $$F = \frac{1}{4\pi\varepsilon_0} \frac{(2e)(Ze)}{r^2}$$
  • Impact Parameter ($b$): The perpendicular distance of the initial velocity vector of the $\alpha$-particle from the centre of the nucleus.
    • Small $b \implies$ large angle deflection $\theta$ (head-on collision $b=0 \implies \theta = 180^\circ$).
    • Large $b \implies$ small deflection $\theta \approx 0^\circ$.
Fig 12.3: Alpha-Particle Trajectories & Impact Parameter (b) near Gold Nucleus
+Ze Gold Nucleus (Z=79) b=0 (d ≈ 30 fm, θ=180°) b (small) → θ > 90° Large b (θ ≈ 0°) Deflected beam Incident α-particles (5.5 MeV)
Example 12.1

Solar System vs Atomic Scale Analogy

If the solar system had the same proportions as an atom (electron orbit $10^{-10}\text{ m}$, nucleus $10^{-15}\text{ m}$), would Earth be closer to or farther from the Sun than actual? (Sun radius $7 \times 10^8\text{ m}$, Earth orbit $1.5 \times 10^{11}\text{ m}$).

Atomic ratio $\frac{r_{\text{orbit}}}{r_{\text{nucleus}}} = \frac{10^{-10}\text{ m}}{10^{-15}\text{ m}} = 10^5$.
If Earth's orbital radius were $10^5$ times the Sun's radius:
$R_{\text{Earth}} = 10^5 \times (7 \times 10^8\text{ m}) = 7 \times 10^{13}\text{ m}$.
This is $> 100$ times greater than the actual orbital radius ($1.5 \times 10^{11}\text{ m}$).
Earth would be much farther away, demonstrating that an atom contains a vastly greater fraction of empty space than our solar system.
Example 12.2

Distance of Closest Approach ($d$)

In a Geiger-Marsden experiment, calculate the distance of closest approach to a gold nucleus ($Z = 79$) for a $7.7\text{ MeV}$ $\alpha$-particle.

By conservation of mechanical energy: initial kinetic energy $K = 7.7\text{ MeV} = 1.2 \times 10^{-12}\text{ J}$ converts entirely into electrostatic potential energy $U$ at turning point $d$: $$K = \frac{1}{4\pi\varepsilon_0} \frac{2 Z e^2}{d} \implies d = \frac{1}{4\pi\varepsilon_0} \frac{2 Z e^2}{K}$$
$$d = \frac{(9.0 \times 10^9) \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{1.2 \times 10^{-12}\text{ J}} = 3.0 \times 10^{-14}\text{ m} = 30\text{ fm}$$
Distance of closest approach $d = 30\text{ fm}$ (sets an upper limit on gold nucleus radius $< 30\text{ fm}$).
12.2.2 & 12.3
NCERT Sections

Classical Electron Orbits, Energy & Limitations

Classical Orbit Mechanics

12.2.2 Kinetic, Potential and Total Energy of Bound Electron

  • Centripetal Force Balance: $$\frac{m v^2}{r} = \frac{1}{4\pi\varepsilon_0} \frac{e^2}{r^2} \implies r = \frac{e^2}{4\pi\varepsilon_0 m v^2}$$
  • Kinetic Energy: $K = \frac{1}{2} m v^2 = \frac{e^2}{8\pi\varepsilon_0 r}$
  • Electrostatic Potential Energy: $U = -\frac{e^2}{4\pi\varepsilon_0 r}$
  • Total Energy: $$E = K + U = -\frac{e^2}{8\pi\varepsilon_0 r} = -K = \frac{U}{2}$$ (Negative sign confirms the electron is gravitationally/electrostatically bound to the nucleus).
  • Two Critical Failures of Rutherford's Model:
    1. Nuclear Collapse (Instability): An accelerated orbiting electron must radiate EM waves continuously according to Maxwell's electrodynamics, losing energy and spiraling into the nucleus in $\sim 10^{-8}\text{ s}$.
    2. Continuous vs Line Spectrum: Spiraling electron frequency would change continuously, predicting a continuous emission spectrum rather than discrete sharp spectral lines.
Example 12.3

Orbital Radius and Electron Velocity in Hydrogen

If $13.6\text{ eV}$ separates a hydrogen atom into a proton and electron ($E = -13.6\text{ eV}$), compute the orbital radius and electron velocity.

Total energy $E = -13.6\text{ eV} = -2.18 \times 10^{-18}\text{ J} = -\frac{e^2}{8\pi\varepsilon_0 r}$.
$$r = \frac{e^2}{8\pi\varepsilon_0 |E|} = \frac{(9.0 \times 10^9) \times (1.6 \times 10^{-19})^2}{2 \times (2.18 \times 10^{-18})} = 5.3 \times 10^{-11}\text{ m} = 0.53\text{ \AA}$$
$$v = \frac{e}{\sqrt{4\pi\varepsilon_0 m r}} = \frac{1.6 \times 10^{-19}}{\sqrt{\frac{9.11 \times 10^{-31} \times 5.3 \times 10^{-11}}{9 \times 10^9}}} = 2.18 \times 10^6\text{ m s}^{-1} \approx 2.2 \times 10^6\text{ m/s}$$
Radius $r = 5.3 \times 10^{-11}\text{ m}$ • Speed $v = 2.2 \times 10^6\text{ m/s}$.
Example 12.4

Classical Frequency of Emitted Light

According to classical EM theory, calculate the initial frequency of light emitted by the orbiting electron in hydrogen.

Classical orbital frequency $\nu_{\text{orb}} = \frac{v}{2\pi r}$: $$\nu = \frac{2.2 \times 10^6\text{ m s}^{-1}}{2\pi \times (5.3 \times 10^{-11}\text{ m})} \approx 6.6 \times 10^{15}\text{ Hz}$$
Classical emission frequency $\nu = 6.6 \times 10^{15}\text{ Hz}$.
12.4 & 12.5
NCERT Sections

Bohr's Quantum Model of Hydrogen & Spectral Series

Bohr's 3 Postulates

12.4.1 Bohr's Postulates of the Hydrogen Atom

  1. Postulate I (Stationary Orbits): Electrons revolve only in certain non-radiating, dynamically stable orbits called stationary states with definite discrete energies.
  2. Postulate II (Quantisation of Angular Momentum): An electron can only revolve in orbits where its orbital angular momentum $L$ is an integral multiple of $h/2\pi = \hbar$: $$L = m v_n r_n = \frac{n h}{2\pi} \quad (n = 1, 2, 3, \dots)$$
  3. Postulate III (Frequency Condition): Radiation is emitted or absorbed only when an electron jumps from one stationary orbit ($E_i$) to another ($E_f$): $$h\nu = E_i - E_f \implies \nu = \frac{E_i - E_f}{h}$$
Formulae for $r_n, v_n, E_n$

12.4.2 Bohr Orbit Radius, Velocity, and Quantised Energy Levels

  • Radius of $n^{\text{th}}$ Bohr Orbit: $$r_n = \frac{n^2 h^2 \varepsilon_0}{\pi m e^2} = n^2 a_0 = n^2 \times (0.529\text{ \AA}) = n^2 \times (5.3 \times 10^{-11}\text{ m})$$
  • Speed of Electron in $n^{\text{th}}$ Orbit: $$v_n = \frac{e^2}{2 \varepsilon_0 n h} = \frac{c}{137 n} = \frac{2.18 \times 10^6}{n}\text{ m/s}$$
  • Total Energy of $n^{\text{th}}$ Orbit: $$E_n = -\frac{m e^4}{8 \varepsilon_0^2 n^2 h^2} = -\frac{2.18 \times 10^{-18}\text{ J}}{n^2} = -\frac{13.6\text{ eV}}{n^2}$$
  • Ground State & Excitation Energies:
    • Ground State ($n = 1$): $E_1 = -13.6\text{ eV}$ (Ionisation energy $= +13.6\text{ eV}$).
    • First Excited State ($n = 2$): $E_2 = -3.40\text{ eV} \implies \Delta E_{1\to 2} = 10.2\text{ eV}$.
    • Second Excited State ($n = 3$): $E_3 = -1.51\text{ eV} \implies \Delta E_{1\to 3} = 12.09\text{ eV}$.
    • Third Excited State ($n = 4$): $E_4 = -0.85\text{ eV} \implies \Delta E_{1\to 4} = 12.75\text{ eV}$.
Fig 12.8: Energy Level Diagram of Hydrogen & Spectral Series Transitions
n = ∞ (0.00 eV) n = 4 (-0.85 eV) n = 3 (-1.51 eV) n = 2 (-3.40 eV) n = 1 (-13.60 eV) Lyman (UV) Balmer (Visible) Paschen (IR) Brackett (IR)
Spectral Series of Hydrogen

12.5 Hydrogen Emission Spectral Series

Rydberg Formula: $\frac{1}{\lambda} = R \left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ where $R = 1.097 \times 10^7\text{ m}^{-1}$.

Series Final State ($n_f$) Initial States ($n_i$) Spectral Region Shortest Line ($\lambda_{\text{min}}$ for $n_i=\infty$)
Lyman Series n_f = 1 2, 3, 4, ... Ultraviolet (UV) $91.2\text{ nm}$ ($912\text{ \AA}$)
Balmer Series n_f = 2 3, 4, 5, ... Visible $364.6\text{ nm}$ ($3646\text{ \AA}$)
Paschen Series n_f = 3 4, 5, 6, ... Infra-red (IR) $820.4\text{ nm}$
Brackett Series n_f = 4 5, 6, 7, ... Infra-red (IR) $1458.5\text{ nm}$
Pfund Series n_f = 5 6, 7, 8, ... Far Infra-red $2279.0\text{ nm}$
12.6
NCERT Section

de Broglie's Explanation of Bohr's Quantisation & Limitations

Standing Waves on Orbits

12.6.1 de Broglie's Justification of Angular Momentum Quantisation

  • Standing Matter Waves on Circular Orbits:
    An orbiting electron forms a stationary standing matter wave along its circular circumference. Resonant persistence requires an integral number of de Broglie wavelengths to fit around the circumference: $$2\pi r_n = n \lambda \quad (n = 1, 2, 3, \dots)$$ Substitute de Broglie wavelength $\lambda = \frac{h}{p} = \frac{h}{m v_n}$: $$2\pi r_n = n \left(\frac{h}{m v_n}\right) \implies m v_n r_n = \frac{n h}{2\pi} = n \hbar$$ (This provides the quantum-mechanical proof for Bohr's empirical 2nd postulate).
  • Limitations of Bohr's Atomic Model:
    1. Applicable only to single-electron (hydrogenic) species ($\text{H}, \text{He}^+, \text{Li}^{2+}, \text{Be}^{3+}$); fails for multi-electron atoms due to inter-electronic repulsive interactions.
    2. Fails to explain relative intensities and transition probabilities of spectral lines.
    3. Unable to explain fine structure splitting (Zeeman effect in magnetic fields, Stark effect in electric fields).
    4. Violates Heisenberg's Uncertainty Principle by assuming well-defined circular orbits with precise position and momentum.
Fig 12.10: Standing Matter Wave for n = 4 Orbit (2πr = 4λ)
+e 2πr₄ = 4λ ⟹ L = 4(h / 2π)
02 / Interactive Assessment

Concept Check & Board Mastery Quiz

15 targeted MCQs covering alpha scattering, Rutherford's model, distance of closest approach, Bohr's postulates, energy level transitions, and spectral series with in-depth explanations.

15 MCQS
0 / 15 Answered
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03 / Textbook Solutions

NCERT Exercises 12.1 – 12.9

Complete stepwise solutions for every exercise question in NCERT Class 12 Physics Chapter 12 Atoms Reprint 2026-27.

9 QUESTIONS
Ex 12.1Thomson vs Rutherford Model Comparison

12.1 Choose the correct alternative:
(a) The size of the atom in Thomson’s model is .......... the atomic size in Rutherford’s model. (much greater than/no different from/much less than.)
(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson’s model/ Rutherford’s model.)
(c) A classical atom based on .......... is doomed to collapse. (Thomson’s model/ Rutherford’s model.)
(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson’s model/ Rutherford’s model.)
(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford’s model/both the models.)

(a) no different from • (b) Thomson's model; Rutherford's model • (c) Rutherford's model
(d) Thomson's model; Rutherford's model • (e) both the models.
(a) no different from: Both models assume an atomic size of order $\sim 10^{-10}\text{ m}$.
(b) In Thomson’s model electrons are at electrostatic equilibrium, while in Rutherford’s model revolving electrons experience continuous centripetal acceleration and net Coulomb force.
(c) Rutherford's model: Accelerated charges continuously radiate electromagnetic energy and spiral into the nucleus.
(d) Thomson’s model; Rutherford’s model: Uniform spherical mass cloud in Thomson vs concentrated tiny nucleus ($10^{-15}\text{ m}$) in Rutherford.
(e) both the models: Since electrons are extremely light ($m_e \approx m_p / 1836$), the positively charged portion contains almost the entire atomic mass in both descriptions.
Ex 12.2Alpha Scattering with Solid Hydrogen Foil

12.2 Suppose you repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil (below $14\text{ K}$). What results do you expect?

No large-angle scattering or backward rebound will be observed ($\alpha$-particles will knock proton targets forward without deflection).
An $\alpha$-particle ($m_\alpha = 4\text{ u}$) is 4 times heavier than a hydrogen nucleus/proton ($m_p = 1\text{ u}$). In elastic collision with a much lighter target, the heavy projectile carries straight through and can never be deflected through large angles ($> 90^\circ$).
Ex 12.3Transition Frequency between Energy Levels

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

Frequency $\nu = 5.55 \times 10^{14}\text{ Hz}$.
$$\Delta E = h\nu \implies \nu = \frac{\Delta E}{h} = \frac{2.3 \times 1.602 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = 5.55 \times 10^{14}\text{ Hz}$$
Ex 12.4Kinetic and Potential Energy of Hydrogen Ground State

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

Kinetic energy $K = +13.6\text{ eV}$ • Potential energy $U = -27.2\text{ eV}$.
For Coulomb potential: $$K = -E = -(-13.6\text{ eV}) = +13.6\text{ eV}$$ $$U = 2E = 2(-13.6\text{ eV}) = -27.2\text{ eV}$$ $$E = K + U = 13.6 - 27.2 = -13.6\text{ eV}$$
Ex 12.5Excitation to n = 4 Photon Parameters

12.5 A hydrogen atom initially in ground state ($n=1$) absorbs a photon which excites it to $n = 4$. Determine wavelength and frequency of the photon.

Wavelength $\lambda = 97.3\text{ nm} = 973\text{ \AA}$ • Frequency $\nu = 3.08 \times 10^{15}\text{ Hz}$.
$$E_1 = -13.6\text{ eV}, \quad E_4 = \frac{-13.6}{4^2} = -0.85\text{ eV}$$ $$\Delta E = E_4 - E_1 = -0.85 - (-13.6) = 12.75\text{ eV} = 12.75 \times 1.602 \times 10^{-19}\text{ J} = 2.043 \times 10^{-18}\text{ J}$$
$$\nu = \frac{\Delta E}{h} = \frac{2.043 \times 10^{-18}\text{ J}}{6.63 \times 10^{-34}\text{ J s}} = 3.08 \times 10^{15}\text{ Hz}$$ $$\lambda = \frac{c}{\nu} = \frac{3.0 \times 10^8\text{ m s}^{-1}}{3.08 \times 10^{15}\text{ s}^{-1}} = 9.73 \times 10^{-8}\text{ m} = 97.3\text{ nm}$$
Ex 12.6Speed and Orbital Period in n = 1, 2, 3

12.6 (a) Calculate speed of electron in hydrogen atom for $n = 1, 2, 3$.
(b) Calculate orbital period ($T$) in each of these levels.

(a) $v_1 = 2.18 \times 10^6\text{ m/s}, v_2 = 1.09 \times 10^6\text{ m/s}, v_3 = 7.27 \times 10^5\text{ m/s}$
(b) $T_1 = 1.53 \times 10^{-16}\text{ s}, T_2 = 1.22 \times 10^{-15}\text{ s}, T_3 = 4.13 \times 10^{-15}\text{ s}$.
(a) Speeds ($v_n = v_1 / n$):
$v_1 = \frac{e^2}{2\varepsilon_0 h} = 2.18 \times 10^6\text{ m s}^{-1}$
$v_2 = \frac{2.18 \times 10^6}{2} = 1.09 \times 10^6\text{ m s}^{-1}$
$v_3 = \frac{2.18 \times 10^6}{3} = 7.27 \times 10^5\text{ m s}^{-1}$
(b) Orbital Periods ($T_n = \frac{2\pi r_n}{v_n} \propto n^3$):
$r_1 = 5.3 \times 10^{-11}\text{ m} \implies T_1 = \frac{2\pi (5.3 \times 10^{-11})}{2.18 \times 10^6} = 1.53 \times 10^{-16}\text{ s}$
$T_2 = 2^3 \times T_1 = 8 \times (1.53 \times 10^{-16}) = 1.22 \times 10^{-15}\text{ s}$
$T_3 = 3^3 \times T_1 = 27 \times (1.53 \times 10^{-16}) = 4.13 \times 10^{-15}\text{ s}$
Ex 12.7Radii of n = 2 and n = 3 Orbits

12.7 Innermost orbit radius $r_1 = 5.3 \times 10^{-11}\text{ m}$. What are radii of $n = 2$ and $n = 3$ orbits?

$r_2 = 2.12 \times 10^{-10}\text{ m} = 2.12\text{ \AA}$ • $r_3 = 4.77 \times 10^{-10}\text{ m} = 4.77\text{ \AA}$.
Since $r_n = n^2 r_1$: $$r_2 = 2^2 \times (5.3 \times 10^{-11}\text{ m}) = 4 \times 5.3 \times 10^{-11} = 2.12 \times 10^{-10}\text{ m}$$ $$r_3 = 3^2 \times (5.3 \times 10^{-11}\text{ m}) = 9 \times 5.3 \times 10^{-11} = 4.77 \times 10^{-10}\text{ m}$$
Ex 12.8Wavelengths Emitted on 12.5 eV Bombardment

12.8 A $12.5\text{ eV}$ electron beam bombards gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Excites up to $n = 3$. Three wavelengths emitted: Lyman ($\lambda_{3\to 1} = 102.5\text{ nm}, \lambda_{2\to 1} = 121.5\text{ nm}$) and Balmer ($\lambda_{3\to 2} = 656.3\text{ nm}$).
Max energy of atom: $E = -13.6 + 12.5 = -1.1\text{ eV}$.
Since $E_3 = -1.51\text{ eV}$ and $E_4 = -0.85\text{ eV}$, atoms can only be excited up to $n = 3$.
Possible transitions ($N = \frac{n(n-1)}{2} = \frac{3 \times 2}{2} = 3$):
  • $3 \to 1$: $\Delta E = -1.51 - (-13.6) = 12.09\text{ eV} \implies \lambda = \frac{1242}{12.09} = 102.5\text{ nm}$ (Lyman series, UV)
  • $2 \to 1$: $\Delta E = -3.4 - (-13.6) = 10.2\text{ eV} \implies \lambda = \frac{1242}{10.2} = 121.6\text{ nm}$ (Lyman series, UV)
  • $3 \to 2$: $\Delta E = -1.51 - (-3.4) = 1.89\text{ eV} \implies \lambda = \frac{1242}{1.89} = 656.3\text{ nm}$ (Balmer $H_\alpha$ line, Red visible)
Ex 12.9Quantum Number of Earth Orbit

12.9 Find the quantum number $n$ that characterises Earth's revolution around Sun ($r = 1.5 \times 10^{11}\text{ m}, v = 3.0 \times 10^4\text{ m/s}, M_E = 6.0 \times 10^{24}\text{ kg}$).

Quantum number $n = 2.6 \times 10^{74}$ (so vast that quantum discrete states appear completely continuous in classical mechanics).
$$m v r = \frac{n h}{2\pi} \implies n = \frac{2\pi m v r}{h}$$ $$n = \frac{2\pi \times (6.0 \times 10^{24}\text{ kg}) \times (3.0 \times 10^4\text{ m s}^{-1}) \times (1.5 \times 10^{11}\text{ m})}{6.63 \times 10^{-34}\text{ J s}} = 2.56 \times 10^{74} \approx 2.6 \times 10^{74}$$
04 / Rapid Reference

Chapter Summary & Points to Ponder

Core atomic formulae, radius and energy scalings, spectral series parameters, and official NCERT Points to Ponder for Chapter 12.

100% SYLLABUS

Rutherford Nuclear Atom

r_{\text{nucleus}} \sim 10^{-15}\text{ m} \ll r_{\text{atom}} \sim 10^{-10}\text{ m}

Most of the atom is empty space. Alpha scattering distance of closest approach: $d = \frac{1}{4\pi\varepsilon_0}\frac{2Ze^2}{K}$.

Bohr Radius & Velocity

r_n \propto n^2 / Z • v_n \propto Z / n

Bohr radius $a_0 = 0.529\text{ \AA} = 5.3 \times 10^{-11}\text{ m}$, ground velocity $v_1 = c/137 \approx 2.18 \times 10^6\text{ m/s}$.

Energy Levels of Hydrogen

E_n = −13.6 / n² \text{ eV}

$K = -E = +13.6/n^2\text{ eV}$, $U = 2E = -27.2/n^2\text{ eV}$. Total energy is negative, confirming bound state.

de Broglie Standing Waves

2\pi r_n = n\lambda \implies m v r = n h / 2\pi

Justifies Bohr's 2nd postulate: quantised orbits correspond to resonant stationary matter waves on a circle.

Rydberg Spectral Formula

1 / λ = R (1/n_f² − 1/n_i²)

Lyman (UV, $n_f=1$), Balmer (Visible, $n_f=2$), Paschen (IR, $n_f=3$), Brackett ($n_f=4$), Pfund ($n_f=5$).

Excitation Transitions

Total Lines = n (n − 1) / 2

For electrons excited to level $n$, the total number of distinct emission spectral lines emitted upon de-excitation is $n(n-1)/2$.

NCERT Official

Points to Ponder

  1. Instability of Classical Models: Thomson's model is electrostatically unstable; Rutherford's model is electrodynamically unstable due to continuous radiation by accelerated orbiting charges.
  2. Natural Choice of Angular Momentum: Planck's constant $h$ has dimensions of angular momentum ($[\text{M L}^2 \text{T}^{-1}]$), making $L = n(h/2\pi)$ the most fundamental and natural quantisation condition for circular orbital motion.
  3. Quantum vs Classical Orbital Picture: Bohr's well-defined circular orbits violate the Heisenberg uncertainty principle ($\Delta x \Delta p \ge \hbar/2$). Modern wave mechanics replaces deterministic orbits with 3D probability density clouds (orbitals).
  4. Failure in Multi-Electron Systems: In multi-electron atoms, electron-electron repulsive forces are comparable in magnitude to the nuclear attractive force, causing the simple central-field Bohr model to break down.
  5. Bohr Correspondence Principle: For extremely large quantum numbers ($n \to \infty$), the quantum transition frequency matches the classical orbital revolution frequency ($\nu_{\text{quantum}} \approx \nu_{\text{classical}}$).
05 / Practice Tests

3-Tier Practice Tests

Level 1 (Foundation), Level 2 (Application), and Level 3 (Challenge). Select options and submit to evaluate your preparation.

24 QUESTIONS
Level 1 Active • 8 Questions