Atomic Models — Concepts, Formulas & Examples

Atomic models trace the evolution of our understanding of the atom. CBSE Class 9 and 11 and NEET test the main models, their predictions and their limitations. This is a high-yield conceptual topic.

Core Concepts

Dalton’s model (1808)

Atoms are indivisible, indestructible spheres. All atoms of an element are identical. Atoms of different elements combine in fixed ratios. Explained laws of chemical combination but could not explain sub-atomic particles.

What Dalton got right: Law of conservation of mass, law of definite proportions, law of multiple proportions. What Dalton got wrong: Atoms are not indivisible (electrons, protons, neutrons exist). Atoms of the same element can differ in mass (isotopes).

Thomson’s model (1904)

Plum pudding model. Atom is a positively charged sphere with electrons embedded in it. Discovered the electron via cathode ray experiments.

Thomson’s key experiment: Cathode rays (streams of electrons) in a discharge tube. He measured the charge-to-mass ratio ( $e/m = 1.76 \times 10^{11}$ C/kg). Robert Millikan later measured the charge itself ( $e = 1.6 \times 10^{-19}$ C) in the oil drop experiment.

Limitation: Could not explain Rutherford’s gold foil results — the positive charge is not spread uniformly but concentrated in a tiny nucleus.

Rutherford’s model (1911)

Gold foil experiment showed atoms are mostly empty space with a tiny dense positive nucleus. Electrons orbit the nucleus. Limitation — classical physics said orbiting electrons should radiate energy and collapse into nucleus.

Key observations from the gold foil experiment:

Most alpha particles passed straight through → atom is mostly empty space
Some deflected at small angles → positive charge causes repulsion
Very few (<1 in 8000) bounced back at large angles → positive charge is concentrated in a tiny, dense nucleus

Numerical perspective: If the atom were the size of a football stadium, the nucleus would be a marble at the centre. The nucleus contains 99.97% of the atom’s mass but only $10^{-15}$ of its volume.

Limitation: An electron moving in a circle is constantly accelerating (changing direction). Classical electrodynamics says accelerating charges radiate electromagnetic energy. An orbiting electron should continuously lose energy, spiral inward, and collapse into the nucleus in about $10^{-8}$ seconds. Atoms would be unstable — but they clearly are not.

Bohr’s model (1913)

Electrons occupy fixed circular orbits with quantised energy. They jump between orbits by absorbing or emitting specific quanta of light. Explained hydrogen spectrum but failed for multi-electron atoms.

Bohr’s postulates:

Electrons revolve in fixed circular orbits without radiating energy
Only orbits where angular momentum $L = n\hbar = n(h/2\pi)$ are allowed ( $n = 1, 2, 3...$ )
Energy is emitted/absorbed only when an electron jumps between orbits: $\Delta E = hf$

Radius: $r_n = \frac{n^2 a_0}{Z}$ where $a_0 = 0.529$ angstrom (Bohr radius)

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

Velocity: $v_n = \frac{2.18 \times 10^6 Z}{n}$ m/s

Spectral series of hydrogen:

Series	Transition to $n$	Region	Formula
Lyman	1	Ultraviolet	$1/\lambda = R(1/1^2 - 1/n^2)$ , $n \geq 2$
Balmer	2	Visible	$1/\lambda = R(1/2^2 - 1/n^2)$ , $n \geq 3$
Paschen	3	Infrared	$1/\lambda = R(1/3^2 - 1/n^2)$ , $n \geq 4$
Brackett	4	Far infrared	$1/\lambda = R(1/4^2 - 1/n^2)$ , $n \geq 5$
Pfund	5	Far infrared	$1/\lambda = R(1/5^2 - 1/n^2)$ , $n \geq 6$

Where $R = 1.097 \times 10^7$ m $^{-1}$ (Rydberg constant).

Limitation of Bohr’s model: Works perfectly for hydrogen and hydrogen-like ions (He $^+$ , Li $^{2+}$ ). Fails for multi-electron atoms because it ignores electron-electron repulsion. Cannot explain Zeeman effect or fine structure.

Quantum mechanical model

Electrons are described by wave functions (orbitals), not defined orbits. Heisenberg uncertainty principle — cannot know position and momentum exactly simultaneously. Schrodinger equation gives orbital shapes — s, p, d, f.

\Delta x \cdot \Delta p \geq \frac{h}{4\pi}

The more precisely you know position ( $\Delta x$ ), the less precisely you know momentum ( $\Delta p$ ), and vice versa.

Four quantum numbers describe each electron:

Quantum number	Symbol	Values	Determines
Principal	$n$	1, 2, 3…	Shell, energy
Azimuthal	$l$	0 to $n-1$	Subshell, shape
Magnetic	$m_l$	$-l$ to $+l$	Orbital orientation
Spin	$m_s$	$+1/2$ or $-1/2$	Spin direction

Orbital shapes: $l = 0$ → s (spherical), $l = 1$ → p (dumbbell), $l = 2$ → d (clover), $l = 3$ → f (complex).

Filling order (Aufbau principle): 1s → 2s → 2p → 3s → 3p → 4s → 3d → 4p → 5s → 4d → 5p → 6s → 4f…

Key Formulas

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

For hydrogen. Ground state at $n=1$ is $-13.6$ eV.

\lambda = \frac{h}{mv}

Every moving particle has a wave character. Smaller mass and slower speed give longer wavelength.

Worked Examples

An accelerating charge radiates electromagnetic energy classically. Orbiting electrons accelerate (changing direction), so they should radiate and spiral into the nucleus. Atoms would be unstable — but they are not. Bohr’s quantisation fixed this.

$E_n = -13.6/n^2$ eV. The ground state is $n=1$ at $-13.6$ eV. Higher $n$ gives less negative (less bound) energies.

Find the wavelength of the first line of the Balmer series ( $n=3$ to $n=2$ ).

\frac{1}{\lambda} = R\left(\frac{1}{2^2} - \frac{1}{3^2}\right) = 1.097 \times 10^7\left(\frac{1}{4} - \frac{1}{9}\right) = 1.097 \times 10^7 \times \frac{5}{36}

\frac{1}{\lambda} = 1.524 \times 10^6 \text{ m}^{-1}

$\lambda = 656 \text{ nm (red light --- this is the H-alpha line)}$

Ionisation means removing the electron completely ( $n = \infty$ , $E = 0$ ).

Energy needed = $E_\infty - E_1 = 0 - (-13.6) = 13.6$ eV = $2.18 \times 10^{-18}$ J.

For one mole: $13.6 \times 96.5 = 1312$ kJ/mol. This matches the experimental ionisation energy of hydrogen.

An electron ( $m = 9.1 \times 10^{-31}$ kg) moves at $10^6$ m/s.

\lambda = \frac{h}{mv} = \frac{6.626 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^6} = 7.28 \times 10^{-10}\;\text{m} = 0.728\;\text{nm}

This is comparable to atomic dimensions — confirming that wave behaviour matters for electrons.

Common Mistakes

Saying Dalton discovered electrons. Thomson did.

Writing that Bohr’s model works for all atoms. It works only for hydrogen-like single-electron systems.

Confusing orbit (Bohr) and orbital (quantum). Orbit is a fixed path; orbital is a probability distribution.

Assuming that the Bohr radius formula gives the actual size of the electron. In quantum mechanics, the electron does not have a definite position — $r_n$ gives the most probable distance from the nucleus.

Using the Rydberg formula with wrong $n$ values. The lower $n$ goes into $1/n_1^2$ and the higher into $1/n_2^2$ . Swapping them gives a negative wavelength, which is physically meaningless.

Exam Weightage and Revision

JEE Main 2024 tested the Rydberg formula for spectral lines. NEET 2023 asked about Bohr’s postulates. CBSE Class 9 asks about Rutherford’s experiment. CBSE Class 11 and JEE test quantum numbers and orbital shapes. This topic spans multiple classes and exams.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Draw a timeline with five models and one key feature of each. That timeline is your revision.

Practice Questions

Q1. What fraction of alpha particles were deflected in Rutherford’s experiment?

Most alpha particles passed straight through. About 1 in 20,000 were deflected by more than 1°, and about 1 in 8000 bounced back at angles greater than 90°. The tiny fraction of large deflections proved that the positive charge is concentrated in a very small nucleus.

Q2. Calculate the energy of the photon emitted when hydrogen transitions from $n=4$ to $n=2$ .

$\Delta E = 13.6\left(\frac{1}{2^2} - \frac{1}{4^2}\right) = 13.6\left(\frac{1}{4} - \frac{1}{16}\right) = 13.6 \times \frac{3}{16} = 2.55$ eV.

This corresponds to $\lambda = hc/E = 486$ nm — a blue-green line (H-beta in the Balmer series).

Q3. Why cannot we apply Bohr’s model to helium?

Helium has two electrons. Bohr’s model accounts only for the nucleus-electron interaction, not electron-electron repulsion. With two electrons, their mutual repulsion significantly changes the energy levels, and Bohr’s simple formula no longer gives correct results. The Schrodinger equation must be used instead.

Q4. What are the four quantum numbers for the outermost electron of sodium?

Sodium: $1s^2 2s^2 2p^6 3s^1$ . The outermost electron is in 3s: $n = 3$ , $l = 0$ (s subshell), $m_l = 0$ (only one orientation for s), $m_s = +1/2$ or $-1/2$ .

Q5. What is the significance of the Heisenberg uncertainty principle for atomic structure?

It means we cannot simultaneously know the exact position and momentum of an electron. Therefore, the concept of a fixed orbit (Bohr) is meaningless. Instead, we use orbitals — regions of space where the probability of finding the electron is high. The uncertainty principle is the fundamental reason why quantum mechanics uses probability distributions rather than definite paths.

FAQs

Why are atomic models important if most are wrong? Each model was the best explanation available at the time and correctly predicted some observations. Dalton explained stoichiometry, Thomson explained electrical neutrality, Rutherford explained scattering, Bohr explained hydrogen spectra. Science progresses by building better models, not by being right the first time.

What is the Schrodinger equation? It is the fundamental equation of quantum mechanics: $\hat{H}\psi = E\psi$ . Solving it for a system gives the wave function ( $\psi$ ), from which we get the probability distribution ( $|\psi|^2$ ), energy levels, and orbital shapes. For hydrogen, it gives exact solutions. For multi-electron atoms, approximate methods are used.

Why is the ground state energy of hydrogen negative? The zero of energy is defined as the electron at infinite distance (completely separated from the nucleus). Any bound state has less energy than this, hence negative. The more negative the energy, the more tightly bound the electron. $E_1 = -13.6$ eV means you need to supply 13.6 eV to ionise hydrogen from its ground state.

Atomic models show how science corrects itself. Each model fixed a flaw in the previous one, and that scientific process is itself testable.