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Atomic Physics revision notes

Every idea in the Atomic Physics chapter for JEE Main, with the usual mistakes and the fast methods. Topic 1 is open to everyone; sign in free to read all 21 and download the PDF.

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Atom models and α-scattering

Thomson and Rutherford models
1× asked, last 2024
  • Rutherford's model fails classically: the orbiting electron accelerates, so it should radiate, spiral into the nucleus and give a continuous spectrum. Real atoms are stable and give line spectra.
  • Not a special case of Rutherford's model. Thomson spreads the positive charge over the whole atom with electrons embedded in it, which allows only tiny deflections. Rutherford puts it all in a tiny nucleus.
α-scattering results
3× asked, last 2026
N(θ)∝1sin⁡4(θ/2)N(\theta)\propto\frac{1}{\sin^4(\theta/2)}
Number of α scattered at angle θ\theta from a thin foil. It falls steeply as θ\theta grows; no rise at large angles.
θ\thetalog⁡10N\log_{10}NO30∘30^\circ90∘90^\circ180∘180^\circ24
N∝1/sin⁡4(θ/2)N\propto1/\sin^4(\theta/2) (log scale): ten thousand times fewer at 180∘180^\circ than at 10∘10^\circ
  • Most α go straight through and only about 1 in 8000 turn back beyond 90∘90^\circ. So the atom is mostly empty, with all positive charge and nearly all mass in a nucleus of about 10−1410^{-14} m (atom 10−1010^{-10} m).
  • Not because some α are very fast. A faster α is deflected less at the same aim. Rebound needs a near head-on aim (b≈0b\approx0), rare because the nucleus is tiny.
Distance of closest approach
3× asked, last 2026
r0=14πε0 2Ze2Kr_0=\frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^2}{K}
Head-on α of KE KK at a fixed (heavy) nucleus stops where all KE is PE. r0∝Z/Kr_0\propto Z/K; an upper limit on nuclear radius.
rrUUOr0r_0KKU=k(2e)(Ze)/rU=k(2e)(Ze)/rKE of the α\alpha
Head-on, the α\alpha stops where U=KU=K: r0=2kZe2/Kr_0=2kZe^2/K
  • Quick method. Put KK in eV so one ee cancels: r0=(9×109)(2Z)(1.6×10−19)/KeVr_0=(9\times10^9)(2Z)(1.6\times10^{-19})/K_{\text{eV}} m. A diameter is 2r02r_0. Given r0r_0, get the speed from mv2/2=k(2e)(Ze)/r0mv^2/2=k(2e)(Ze)/r_0.
Impact parameter
2× asked, last 2026
b=14πε0 Ze2Kcot⁡θ2b=\frac{1}{4\pi\varepsilon_0}\,\frac{Ze^2}{K}\cot\frac{\theta}{2}
bb = perpendicular distance of the α's line of approach from a fixed nucleus. Smaller bb, larger θ\theta; b=0b=0 is head-on.
nucleusb=0.15b=0.15b=0.5b=0.5b=1.0b=1.0b=1.6b=1.6b=2.3b=2.3
Paths traced under Coulomb repulsion (same energy): the smaller bb, the larger the turn; each obeys b∝cot⁡(θ/2)b\propto\cot(\theta/2)

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17 more concepts in 4 topics

Bohr's model and hydrogen-like ions; Scaling with n and Z; Excitation, ionisation and line count; and Hydrogen spectrum. Read the first part free. Sign in free with Google to read the rest and download the PDF.

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Bohr's model and hydrogen-like ions

  • Bohr's postulates12×
  • Orbit radius9×
  • Orbit speed4×
  • Energy levels15×
  • Kinetic, potential and total energy5×
  • Matching levels across ions11×

Scaling with n and Z

  • Scaling table12×
  • Orbital current, field and moment4×
  • Bohr's rule for other forces3×

Excitation, ionisation and line count

  • Excitation and ionisation energy13×
  • Exciting by photons or electrons7×
  • Number of spectral lines6×

Hydrogen spectrum

  • Rydberg formula16×
  • Spectral series of hydrogen4×
  • Longest and shortest line of a series11×
  • Ratios between lines17×
  • Recoil of the emitting atom3×
  • Also remember1

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