MixedJEE Physics · Original learning card10 original chapter questions

Rutherford Scattering and the Nuclear Atom

Rutherford scattering treats the alpha particle and nucleus as positive point charges interacting through Coulomb repulsion. The observed predominance of small deflections and rarity of backward scattering imply that nearly all positive charge and mass occupy a region far smaller than the atom.

Why this shows up in the exam

Interpreting alpha-scattering paths · Comparing Thomson and Rutherford models · Inferring nuclear compactness from scattering statistics

Learn the idea

Rare large alpha-particle deflections reveal a tiny, massive, positively charged nucleus. Most alpha particles cross a foil almost straight because atoms are mostly empty space. Only particles aimed very near the compact positive centre feel enough repulsion to turn sharply.

🧠 Memory hook: Mostly straight means mostly empty; rare rebounds mean a tiny positive core.

Get this one clearly and it pays off every single time it shows up in the paper. 🎯

Formulas & facts to keep ready

  • F = (1/(4 pi epsilon₀)) (2e)(Ze)/r² — repulsive Coulomb force on an alpha particle near a nucleus of charge Ze
  • b proportional to cot(theta/2) — impact parameter decreases as scattering angle theta increases for fixed energy

How to approach it

  1. 1Identify the sign and concentration of charge
  2. 2Relate smaller impact parameter to larger deflection
  3. 3Use the observation to test the proposed atomic model

Common slip-ups that cost marks

  • •Saying every alpha particle passes through the nucleus
  • •Attributing large deflection to electron collisions
  • •Confusing impact parameter with distance of closest approach

🌟 That's the whole idea — you've got this. Try the practice set below; every question you attempt makes it stick a little harder.

Original chapter practice

Original questions for this chapter, not past-paper questions or an exact mapping to this individual concept.

Question 1 of 10

In hydrogen, an electron transitions from n = 2 to n = 1. Using E_n = -13.6/n^2 eV, find the emitted photon energy.

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