Two chapters of Class 12 collapse into this single NEET topic, and together they reliably supply 3–4 questions: one or two on the Bohr atom and hydrogen spectral lines, and one or two on binding energy, radioactive decay or mass defect. The mathematics is light — mostly substitution into standard formulas — so the marks come from knowing which formula applies and remembering the small numerical constants.
From Thomson's Blob to Rutherford's Nucleus
Before 1911 the atom was pictured as a uniform sphere of positive charge with electrons embedded in it. Geiger and Marsden tested this by firing a narrow beam of alpha particles (helium nuclei, charge +2e, from a radioactive source) at a thin gold foil and counting the scattered particles at various angles on a fluorescent screen.
Most alphas passed almost straight through, but roughly 1 in 8000 turned through more than 90°. A diffuse cloud of positive charge could never deflect a fast, heavy alpha backwards — so the positive charge and nearly all the mass must be squeezed into a tiny central core. Rutherford's conclusions:
- The nucleus carries charge +Ze and essentially all the atomic mass; its radius is ~10⁻¹⁵ m against an atomic radius of ~10⁻¹⁰ m (the nucleus occupies about 10⁻¹² of the atom's volume).
- Electrons circulate around it, held by the Coulomb attraction. For a circular orbit, $\dfrac{mv^2}{r}=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Ze^2}{r^2}$, so the kinetic energy is $K = \dfrac{Ze^2}{8\pi\varepsilon_0 r}$ and the total energy $E = -\dfrac{Ze^2}{8\pi\varepsilon_0 r}$ — negative, i.e. bound.
- The number of alphas scattered at angle θ falls off as $1/\sin^4(\theta/2)$.
The distance of closest approach for a head-on collision comes from equating the initial kinetic energy to the electrostatic potential energy: $d = \dfrac{1}{4\pi\varepsilon_0}\dfrac{2Ze^2}{K}$. This is a favourite one-liner. Also note the impact parameter b: small b gives large-angle scattering, and b = 0 means a head-on hit.
Rutherford's model has a fatal flaw. An electron in a circular orbit accelerates continuously, so classical electromagnetism says it must radiate energy, spiral inward, and crash into the nucleus within about 10⁻⁸ s. It would also emit a continuous smear of frequencies, whereas real atoms emit sharp lines.
Bohr's Postulates and the Energy Levels of Hydrogen
Bohr patched the model with three assumptions for hydrogen-like atoms (one electron, nuclear charge Ze):
- Electrons move only in certain stationary orbits in which they do not radiate.
- Angular momentum is quantised: $mvr = \dfrac{nh}{2\pi}$, with n = 1, 2, 3, …
- Radiation is emitted or absorbed only when the electron jumps between levels, with $h\nu = E_i - E_f$.
Combining the quantisation condition with the Coulomb force balance gives the standard results (Z = 1 for hydrogen):
- Radius: $r_n = \dfrac{n^2}{Z}(0.53\ \text{Å})$, so the ground-state Bohr radius is 0.53 Å.
- Energy: $E_n = -13.6,\dfrac{Z^2}{n^2}$ eV. Ground state of hydrogen: −13.6 eV; hence the ionisation energy of hydrogen is 13.6 eV.
- Speed: $v_n = \dfrac{Ze^2}{2\varepsilon_0 nh}$, i.e. $v_n \propto Z/n$; for hydrogen's ground state $v \approx c/137$.
- Useful proportionalities: $K = -E$, $U = 2E = -2K$, $T \propto n^3/Z^2$, and the number of possible spectral lines from level n down to the ground state is $n(n-1)/2$.
The de Broglie interpretation makes the second postulate less arbitrary: the orbit circumference must hold a whole number of electron wavelengths, $2\pi r = n\lambda$, which with $\lambda = h/mv$ reproduces $mvr = nh/2\pi$ exactly.
Bohr's theory works beautifully for H, He⁺, Li²⁺ and other single-electron systems, but fails for multi-electron atoms, cannot explain relative line intensities, and gives no account of fine structure or the Zeeman splitting of lines.
The Hydrogen Spectrum
The photon wavelength for a transition from level $n_i$ to $n_f$ follows from the third postulate:
$$\frac{1}{\lambda} = R Z^2\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right), \qquad R = 1.097\times10^7\ \text{m}^{-1}$$
The named series are classified by the lower level:
- Lyman ($n_f = 1$): ultraviolet; series limit 91.2 nm, longest line (2→1) 121.6 nm.
- Balmer ($n_f = 2$): the only series with lines in the visible region; Hα (3→2) is red at 656.3 nm.
- Paschen ($n_f = 3$), Brackett ($n_f = 4$), Pfund ($n_f = 5$): all infrared.
Two shortcuts worth memorising: the longest wavelength (smallest energy) line of a series is the transition from the immediately higher level, while the series limit (shortest wavelength) corresponds to $n_i = \infty$. For quick numerical work, $\lambda(\text{in nm}) = 1240/E(\text{in eV})$.
Absorption spectra show the reverse process — cool hydrogen gas absorbs exactly those photons it would emit, producing dark lines. In practice, gas at room temperature is almost entirely in n = 1, so hydrogen's absorption spectrum shows only the Lyman series.
Inside the Nucleus: Size, Mass Defect and Binding Energy
A nucleus $^A_Z X$ contains Z protons and (A − Z) neutrons, A being the mass number and Z the atomic number. Nuclides with the same Z are isotopes, the same A are isobars, and the same neutron number are isotones.
Nuclear radii obey $R = R_0 A^{1/3}$ with $R_0 \approx 1.2$ fm. Since volume ∝ A, nuclear density is essentially the same for all nuclei, about 2.3 × 10¹⁷ kg m⁻³ — some 10¹⁴ times denser than ordinary matter.
Masses at this scale are quoted in unified mass units: 1 u = 1.66 × 10⁻²⁷ kg, equivalent to 931.5 MeV of energy. The measured mass of any nucleus is less than the sum of its free constituents; the difference is the mass defect
$$\Delta m = \big[Z m_p + (A - Z) m_n\big] - M_{\text{nucleus}}$$
and the energy equivalent $E_b = \Delta m c^2$ is the binding energy — the work needed to pull the nucleus apart into free nucleons. Dividing by A gives the binding energy per nucleon, the true measure of stability.
The shape of the BE/A versus A curve is examined repeatedly:
- It rises steeply for light nuclei, with sharp local peaks at $^4$He, $^{12}$C and $^{16}$O.
- It is nearly flat at about 8.5 MeV between A ≈ 30 and A ≈ 120, peaking near $^{56}$Fe (≈ 8.8 MeV) — the most stable region.
- It declines gently to about 7.6 MeV at $^{238}$U.
The flatness tells us the nuclear force is short-ranged and saturated: each nucleon interacts only with its immediate neighbours, not with the whole nucleus. The decline at large A explains why heavy nuclei release energy on splitting, and the steep rise at small A explains why light nuclei release energy on merging. In both cases the products lie closer to the peak, so energy is liberated.
Fission: $^{235}\text{U} + n \rightarrow ^{144}\text{Ba} + ^{89}\text{Kr} + 3n$ releases roughly 200 MeV per event, and the extra neutrons make a chain reaction possible (controlled by cadmium or boron rods in a reactor, with a moderator such as heavy water to slow neutrons down).
Fusion: four protons effectively combine into helium in the Sun, releasing about 26.7 MeV; fusion requires temperatures of ~10⁷ K so that nuclei have enough kinetic energy to overcome Coulomb repulsion.
Radioactivity and the Decay Law
Unstable nuclei emit radiation spontaneously in three modes:
| Decay | Emission | Change |
|---|---|---|
| Alpha | $^4_2$He nucleus | Z → Z − 2, A → A − 4 |
| Beta-minus | electron + antineutrino | Z → Z + 1, A unchanged (n → p) |
| Beta-plus | positron + neutrino | Z → Z − 1, A unchanged (p → n) |
| Gamma | high-energy photon | no change in Z or A |
Alpha particles are the least penetrating but most strongly ionising; gamma rays penetrate deepest. Gamma emission usually follows α or β decay, as the daughter nucleus drops from an excited state to its ground state. In alpha decay the Q-value $Q = (m_X - m_Y - m_{He})c^2$ is shared as kinetic energy between the daughter and the alpha, with the light alpha taking almost all of it.
Decay is a statistical process: the number of disintegrations per second is proportional to the number of undecayed nuclei present.
$$-\frac{dN}{dt} = \lambda N \quad\Longrightarrow\quad N = N_0 e^{-\lambda t}$$
Key derived quantities:
- Half-life: $T_{1/2} = \dfrac{\ln 2}{\lambda} = \dfrac{0.693}{\lambda}$. After n half-lives, $N = N_0/2^n$.
- Mean (average) life: $\tau = 1/\lambda = T_{1/2}/0.693$, so τ is always longer than the half-life.
- Activity: $R = \lambda N = R_0 e^{-\lambda t}$, measured in becquerel (1 Bq = 1 decay s⁻¹) or curie (1 Ci = 3.7 × 10¹⁰ Bq).
Because activity is directly proportional to N, activity, number of nuclei and mass of the sample all decay with the same half-life — you can apply the $1/2^n$ rule to any of them. Note also that λ is fixed for a given nuclide and is unaffected by temperature, pressure or chemical form.
Common Mistakes and Exam Traps
Dropping the Z² and the sign in Bohr energies. For He⁺ (Z = 2), the ground state is not −13.6 eV but $E_1 = -13.6\times 2^2/1^2 = -54.4$ eV — it is easy to forget to square Z, or to forget the energy is negative and treat 13.6 eV directly as a transition's photon energy. Always compute $E_n$ with its sign first, then take $E_i - E_f$ for the photon energy, which should come out positive for emission.
Confusing the series limit with the line nearest the series head. The series limit ($n_i \to \infty$) gives the shortest wavelength and highest energy of a series; the transition from the very next level up gives the longest wavelength of that series. It is easy to reverse these under time pressure when a question asks for 'the maximum wavelength of the Lyman series'.
Treating mean life and half-life as interchangeable. $\tau = T_{1/2}/0.693$, so the mean life is always about 1.44 times the half-life, never equal to it — substituting one for the other silently corrupts an otherwise correct activity or remaining-nuclei calculation.
Assuming binding energy per nucleon keeps rising with mass number. It peaks near iron-56 and then decreases for heavier nuclei, which is exactly why both fission of very heavy nuclei and fusion of very light nuclei release energy — a point NEET often tests by asking which process is energetically favourable for a nucleus at a given position on the curve.