AtomsNEET Physics · Class 12 · NCERT Chapter 12

You can expect 1 to 2 questions in NEET 2027. Common asks: Bohr radius and velocity, energy of n-th orbit, Rydberg formula for spectral lines, ionisation energy of hydrogen, and the angular momentum quantisation L = n h-bar.

Most alpha particles passed straight through a thin gold foil with no deflection. A small fraction were deflected by large angles, and roughly 1 in 8000 was deflected backwards. This was impossible if charge were spread out (Thomson model). Rutherford concluded the atom's positive charge and almost all its mass are concentrated in a tiny dense nucleus, with electrons orbiting at large distance.

The minimum distance an alpha particle gets to a nucleus when fired head-on. At this point all KE has been converted to PE: r_0 = (1 / 4 pi epsilon_0) × 2 Z e^2 / E_k, where E_k is the alpha's initial KE.

Three: (1) Electrons move in certain stationary circular orbits and do not radiate, despite being accelerated. (2) The angular momentum of an electron is quantised: L = m v r = n h-bar, n = 1, 2, 3 ... (3) When an electron jumps from a higher orbit to a lower one, it emits a photon of energy h f equal to the difference in orbit energies.

a_0 = 0.529 angstrom, the radius of the first orbit (n = 1) of a hydrogen atom (Z = 1). For any hydrogen-like atom: r_n = n^2 a_0 / Z.

E_n = -13.6 Z^2 / n^2 eV. n = 1 is the ground state with E = -13.6 eV (for H, Z = 1). The negative sign indicates a bound state. Higher n gives less negative (smaller magnitude), n = ∞ gives E = 0 (electron just freed).

1 / lambda = R Z^2 (1 / n_1^2 - 1 / n_2^2), where n_2 > n_1. R = 1.097 × 10^7 per metre. The formula gives the wavelength of light emitted (or absorbed) when an electron jumps between two levels.

Lyman (n_1 = 1, UV), Balmer (n_1 = 2, visible: H-alpha 656 nm red, H-beta 486 nm blue-green, H-gamma, H-delta), Paschen (n_1 = 3, IR), Brackett (n_1 = 4, IR), Pfund (n_1 = 5, far IR). Higher n_1 means lower energy → longer wavelength.

Excitation energy: energy needed to lift an electron from the ground state to an excited level. For H: ground (n = 1) to first excited (n = 2) requires 10.2 eV. Ionisation energy: energy to remove the electron entirely (n = 1 to n = ∞). For H: 13.6 eV.

If an electron is a standing wave around the nucleus, the orbit circumference must fit a whole number of wavelengths: 2 pi r = n lambda. Substituting lambda = h / m v gives m v r = n h-bar, exactly Bohr's second postulate. This made the random-looking quantisation rule a wave-resonance condition.