Atoms

2.1 Core concepts

Atomic theory at the close of the nineteenth century stood as follows. Once J. J. Thomson had discovered the electron in 1897, the question became: how are the electrons arranged inside a neutral atom? Thomson proposed the plum-pudding model in 1898 (NCERT §12.1, p. 290): a uniform sphere of positive charge throughout the volume of the atom, with electrons embedded in it like seeds in a watermelon, just dense enough to make the atom neutral overall. The model could roughly account for the size of atoms (~1 Å) and was consistent with the existence of electrons, but it had no real explanation for the most striking experimental fact about atoms — their line spectra. An excited rarefied atomic gas does not emit a continuous rainbow of colours but a sharp set of discrete wavelengths characteristic of the element. Solid or condensed matter, by contrast, gives a continuous spectrum (NCERT §12.1, p. 290–291). The spectrum of an element is its fingerprint, and any successful atomic model has to predict it.

The decisive experimental input came from the Geiger–Marsden α-scattering experiment designed by Rutherford and carried out by his students Hans Geiger and Ernest Marsden between 1908 and 1911 (NCERT §12.2, p. 291–292). The setup used 5.5 MeV α-particles emitted by a radioactive source of ²¹⁴₈₃Bi. A pair of lead bricks collimated the α-beam onto a very thin (~2.1 × 10⁻⁷ m) gold foil suspended in vacuum. Scattered α-particles produced flashes of light (scintillations) on a ZnS screen viewed through a microscope mounted on an arm that could rotate around the foil to cover all scattering angles θ from 0° to 180°.

The numbers Geiger and Marsden recorded surprised everyone. Most α-particles (>99.86%) passed straight through with little or no deflection; only ~0.14% scattered through angles greater than 1°; and remarkably, about 1 in 8000 was scattered through more than 90°, with a few even bouncing straight back along the direction of incidence. Rutherford famously remarked that the result was "as if you fired a fifteen-inch shell at a piece of tissue paper and it came back and hit you." Thomson's plum-pudding picture could not account for these violent back-scatterings — a diffuse cloud of positive charge would deflect α-particles only very gradually.

Rutherford's interpretation was decisive (NCERT §12.2, p. 293). The α-particle is a helium nucleus with charge +2e and mass ~7300 times that of an electron. The electrons in the atom can scarcely deflect such a massive projectile. Large-angle scattering must come from a single close encounter with a much heavier object — a concentrated positive nucleus containing nearly all the atomic mass. The frequency of large-angle events told Rutherford that this nucleus was small: of order 10⁻¹⁵ to 10⁻¹⁴ m across, roughly 10⁴–10⁵ times smaller than the atom (~10⁻¹⁰ m). Hence the atom is mostly empty space, with electrons orbiting at distances vastly larger than the nuclear size — the nuclear (planetary) model.

The geometry of α-scattering depends on the impact parameter b — the perpendicular distance of the α-particle's incoming velocity vector from the centre of the target nucleus (NCERT §12.2.1, p. 293–294). Small b ⇒ very close encounter ⇒ large scattering angle θ. The limiting case b = 0 (head-on collision) gives back-scattering θ ≈ π, where the α-particle decelerates to a momentary halt at the distance of closest approach d before being reversed by the Coulomb repulsion. Setting kinetic energy equal to electrostatic potential energy at the closest point, K = (1/4πε₀)(2e)(Ze)/d, gives d = 2Ze²/(4πε₀ K). For a 7.7 MeV α-particle striking a gold nucleus (Z = 79), d ≈ 3 × 10⁻¹⁴ m = 30 fm, providing an upper bound on the gold-nuclear radius (NCERT Example 12.2, p. 295).

Electron orbits in the Rutherford atom (NCERT §12.2.2, p. 295–296). For a circular orbit of an electron around a nucleus of charge +Ze, the Coulomb attraction supplies the centripetal force:

(1/4πε₀) Ze² / r² = m v² / r ⇒ v² = Ze²/(4πε₀ m r).

KE = (1/2) m v² = Ze²/(8πε₀ r), PE = −Ze²/(4πε₀ r), and total energy

E = KE + PE = −Ze²/(8πε₀ r) (NCERT Eq. 12.4, p. 296).

The total energy is negative, reflecting that the electron is bound; it would need to be supplied with |E| of energy to escape the atom. For hydrogen (Z = 1) and the smallest stable orbit one would calculate, E comes out to about −13.6 eV — quite close to the experimentally measured ionisation energy of hydrogen.

But Rutherford's atom has a fatal flaw (NCERT §12.4, p. 297). According to classical electromagnetic theory, an accelerating charge — and an electron in circular orbit is centripetally accelerating — must radiate electromagnetic waves continuously, losing energy. The electron would therefore spiral inward in about 10⁻¹⁰ s, and the atom would collapse into the nucleus. Worse, the continuously changing orbital frequency would produce a continuous spectrum, contradicting the discrete line spectra observed for atoms.

Neither of these contradictions could be resolved within classical physics. Niels Bohr in 1913 proposed three radical postulates (NCERT §12.4, p. 298–299) that broke from classical EM in just the right way to save the planetary model.

Postulate 1 — Stationary states. An electron in an atom occupies certain stable orbits in which it does not radiate, contrary to classical EM. Each such orbit has a definite total energy. The classical instability problem is simply postulated away in these specific orbits.

Postulate 2 — Angular-momentum quantisation. The stationary orbits are those for which the orbital angular momentum is an integer multiple of ℏ = h/(2π):

L = m v r = n h/(2π) = nℏ, n = 1, 2, 3, …

where n is the principal quantum number.

Postulate 3 — Frequency condition for transitions. Photons are emitted or absorbed when the electron jumps between two stationary states of energies Eᵢ and E_f, with the photon frequency given by

h ν = Eᵢ − E_f

(emission if Eᵢ > E_f, absorption if Eᵢ < E_f).

These three postulates, combined with the Rutherford-orbit dynamics, completely specify the hydrogen atom. Solving Coulomb's law for circular orbit, mv²/r = e²/(4πε₀ r²), together with quantisation mvr = nℏ, gives the quantised radii

r_n = (n²/m)(h/2π)² (4πε₀/e²) (NCERT Eq. 12.7, p. 299).

For n = 1 this evaluates to the famous Bohr radius

a₀ = 5.29 × 10⁻¹¹ m ≈ 0.53 Å (NCERT p. 300).

Higher orbits scale as r_n = n² a₀: r₂ = 4 a₀, r₃ = 9 a₀, and so on — the orbits get rapidly larger as n increases.

The corresponding quantised energies are

E_n = − m e⁴ / (8 n² ε₀² h²) = −13.6/n² eV (NCERT Eq. 12.10, p. 299).

So E₁ = −13.6 eV (ground state), E₂ = −3.40 eV, E₃ = −1.51 eV, E₄ = −0.85 eV, … E_∞ = 0 (NCERT Fig. 12.7, p. 300). Several immediate consequences follow:

• The ionisation energy of hydrogen from its ground state is |E₁| = 13.6 eV — in excellent agreement with measurement.
• The excitation energy from n = 1 to n = 2 is E₂ − E₁ = 10.2 eV; from n = 1 to n = 3 it is 12.09 eV.
• The energy levels get closer together as n grows (the spacing E_{n+1} − E_n shrinks).
• The orbital speed v_n ∝ 1/n; v₁ ≈ c/137 (where 1/137 is the fine-structure constant), so electrons in higher orbits move more slowly. Line spectra explained (NCERT §12.5, p. 300–301). When the atom is excited (by heat, light, or collisions) into a state n_i and then drops to a lower state n_f, the emitted photon has h ν = E_{n_i} − E_{n_f}. Because the energies are discrete, so are the frequencies — only certain definite wavelengths appear. Conversely, when white light is sent through a cold gas, the same set of wavelengths is absorbed, producing dark absorption lines at exactly the same wavelengths the gas would emit. Bohr's frequency condition unifies emission and absorption in a single energy-conservation statement. de Broglie's interpretation (NCERT §12.6, p. 301–302). Why should angular momentum be quantised in units of h/(2π)? Louis de Broglie in 1923 offered a striking explanation. Every moving particle has an associated matter wave of wavelength λ = h/(mv). For a circular orbit to be stable, the wave must close on itself — the orbit must contain an integer number of de Broglie wavelengths: 2π r_n = n λ = n h/(m v_n). Rearranging gives m v_n r_n = n h/(2π) — Bohr's quantisation condition! Bohr's stationary orbits are simply standing matter waves on circular orbits (NCERT Fig. 12.8, p. 301, shows the n = 4 standing wave). Limitations of Bohr's model (NCERT §12.6, p. 302). Spectacularly successful as it is for hydrogen, Bohr's model fails the moment we add a second electron. It works only for hydrogenic atoms — single-electron systems like H, He⁺, Li²⁺ — because it ignores electron–electron Coulomb repulsion. It also fails to predict the relative intensities of spectral lines, a quantitative weakness. These shortcomings would eventually motivate the full quantum-mechanical treatment by Schrödinger, Heisenberg and others in 1925–26.

2.2 Definitions to memorise

2.3 Diagrams / processes to remember

• Fig. 12.1 (p. 292): Geiger–Marsden scattering apparatus — α-source, collimating lead bricks, thin gold foil, rotatable ZnS screen and microscope, all in a vacuum chamber.
• Fig. 12.2 (p. 292): Schematic arrangement of the experiment, showing the lead-brick collimator and the angular distribution of scattered α-particles.
• Fig. 12.3 (p. 293): Plot of N(θ) — number of α-particles scattered per unit angle — vs scattering angle θ. Experimental points (dots) match Rutherford's nuclear-model curve over many orders of magnitude.
• Fig. 12.4 (p. 294): α-particle trajectories in the Coulomb field of the nucleus for several different impact parameters b, showing how θ grows as b shrinks.
• Fig. 12.5 (p. 297): Emission lines in the visible (Balmer) series of hydrogen.
• Fig. 12.6 (p. 298): Spiral inward path predicted by classical electromagnetism for an accelerating orbiting electron — the instability of Rutherford's model.
• Fig. 12.7 (p. 300): Energy-level diagram for hydrogen: horizontal lines at −13.6 eV (n = 1), −3.4 eV (n = 2), −1.51 eV (n = 3), … converging to 0 eV at n = ∞, with vertical arrows showing emission transitions.
• Fig. 12.8 (p. 301): Standing de Broglie wave on a circular orbit with n = 4 (2πr_n = 4λ).

2.4 Common confusions / NTA trap points

• Confusing the two failures of Rutherford's model — instability of the atom (electron spirals in) vs failure to explain line spectra. Both arise from classical EM; NTA likes to test which postulate of Bohr's plugs which hole (p. 297).
• Mixing up the n-dependence: r_n ∝ n² (radius grows), v_n ∝ 1/n, E_n ∝ −1/n² (becomes less negative as n grows). As n increases, energy levels get closer together.
• Sign of total energy: E_n is negative because the electron is bound; "energy required to ionise" is the magnitude (13.6 eV for ground state). Do not confuse the kinetic energy (positive) with total energy (negative).
• The energy difference between ground (n = 1) and first excited (n = 2) state is 10.2 eV, NOT 3.4 eV (which is E₂ alone). Similarly E₃ − E₁ = 12.09 eV.
• Bohr's model is valid only for hydrogenic (one-electron) atoms — H, He⁺, Li²⁺, etc. It breaks for any atom with two or more electrons.
• The α-particle in Geiger–Marsden does NOT physically touch the nucleus; the distance of closest approach (e.g. 30 fm for 7.7 MeV α on gold) is larger than the sum of the radii.
• The fraction "1 in 8000" refers specifically to deflections greater than 90°. "0.14%" is the fraction deflected greater than 1°.
• Bohr's angular-momentum quantum is h/(2π) — not h, and not h/π.
• The frequency condition uses |E_i − E_f|; if E_i < E_f the atom absorbs a photon, not emits.
• Distance of closest approach d depends on K (the kinetic energy of the α-particle) — doubling the kinetic energy halves d.
• For an emission line, the wavelength is the same whether you label the transition as n_i → n_f or n_f → n_i; absorption produces the same line.
• Bohr orbits are not electron clouds; they are circular orbits in a semi-classical picture. The full quantum-mechanical picture replaces them with probability densities.

2.5 Key formulas table