PYQ Vault

JEE Mains Physics · Atoms

Rutherford Scattering and Bohr's Postulates

Rutherford's alpha scattering showed a tiny, heavy, positive nucleus; Bohr then allowed only the orbits where the angular momentum is nh/2π and made light come out only in a jump between them.

All Atoms notesJEE Mains Physics strategy

Why this matters

Nineteen PYQs, sixteen of them multiple choice, and five from 2026. Seven are about Rutherford's experiment: three use the distance of closest approach, one the impact parameter, one asks why so few alphas bounce back, and two compare Thomson's model with Rutherford's. Twelve are about Bohr's postulates: three test the statements themselves, four apply L = nh/2π directly, four first find n from an energy or turn L into an energy, and one applies the quantisation rule to a different force.

Concept 1 of 2: Rutherford scattering and the distance of closest approach

An alpha particle fired straight at a nucleus slows down as it climbs the electric hill, and stops where all its kinetic energy has become potential energy. That turning point is the distance of closest approach. Most alphas never come near a nucleus, because the atom is almost all empty space, so they pass straight through. Only the few aimed almost dead at a nucleus are turned back.

Definition

  • Thomson's model: the positive charge and the mass are spread through the whole atom, with electrons embedded in it. It cannot turn an alpha back.
  • Rutherford's model: all the positive charge and nearly all the mass sit in a tiny nucleus, about 10−1410^{-14} m across, inside an atom about 10−1010^{-10} m across. Electrons move around it.
  • Every alpha in the beam has the same energy. A few rebound only because the nucleus is tiny and only a few come nearly head-on.
  • Closest approach (head-on): K=14πϵ0(2e)(Ze)r0K = \dfrac{1}{4\pi\epsilon_0}\dfrac{(2e)(Ze)}{r_0}, so r0∝ZKr_0 \propto \dfrac{Z}{K}. The nucleus must be smaller than r0r_0, so r0r_0 is an upper limit on its radius.
  • A handy constant: e24πϵ0=1.44 MeV fm\dfrac{e^2}{4\pi\epsilon_0} = 1.44\ \text{MeV fm}, with 1 fm=10−151\ \text{fm} = 10^{-15} m.
  • Impact parameter b is the sideways miss distance of the alpha's line of approach: b=r02cot⁡θ2b = \dfrac{r_0}{2}\cot\dfrac{\theta}{2}. b = 0 means θ = 180°, a straight bounce back.
  • A classical Rutherford atom collapses: the orbiting electron accelerates, radiates energy and spirals into the nucleus. Bohr's postulates were written to stop this.

Closest approach and impact parameter

r0=14πϵ0 2Ze2K,b=r02cot⁡θ2r_0 = \frac{1}{4\pi\epsilon_0}\,\frac{2Ze^2}{K}, \qquad b = \frac{r_0}{2}\cot\frac{\theta}{2}

Worked example

An alpha particle of kinetic energy 4 MeV heads straight for a silver nucleus (Z = 47). Find (a) the distance of closest approach, (b) that distance if the energy were 8 MeV, and (c) the impact parameter for scattering through 90° at 4 MeV. Use e2/4πϵ0=1.44e^2/4\pi\epsilon_0 = 1.44 MeV fm.
Practice this conceptself-check · 5 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 21 Jan 2026 Shift 1 · Q13Moderate

Example 1 · Atoms · Rutherford Scattering and Bohr's Postulates

If an alpha particle with energy 7.7 MeV is bombarded on a thin gold foil, the closest distance from nucleus it can reach is ____\_\_\_\_ m . (Atomic number of gold =79= 79 and 14πϵ0=9×109\frac{1}{4\pi\epsilon_{0}}= 9 \times10^{9} in SI units)

The alpha carries charge 2e

The potential energy at closest approach is k(2e)(Ze)/r₀. Writing ke²Z/r₀ drops the factor 2 and halves the distance, and that halved value is usually an option.

Convert MeV to joules before using k = 9 × 10⁹

1 MeV = 1.6 × 10⁻¹³ J. Mixing MeV with SI constants gives an answer off by a power of ten. Working in MeV fm with e²/4πε₀ = 1.44 MeV fm avoids the conversion.

Closest approach is an upper limit on the radius

The alpha stops before touching the nucleus, so the nuclear radius is at most r₀. Read whether the question wants a radius or a diameter: the diameter is twice the radius.

Rare rebounds are not caused by faster alphas

All alphas in the beam have the same energy. A few bounce back because the nucleus is tiny, so only a few come in with an impact parameter near zero.
Drill 6 more on rutherford scattering and the distance of closest approach

Concept 2 of 2: Bohr's postulates and quantised angular momentum

Bohr kept Rutherford's nucleus and added rules. The electron may circle only in orbits where its angular momentum is a whole number of h/2π, and in those orbits it does not radiate. Light is given out or taken in only when the electron jumps between orbits, and the photon carries exactly the energy difference.

Definition

  • Postulate 1: the electron moves in a circle, held by the Coulomb pull of the nucleus.
  • Postulate 2: only orbits with L=mvr=nh2πL = mvr = \dfrac{nh}{2\pi}, n = 1, 2, 3, …, are allowed. L is a whole multiple of h/2πh/2\pi, not of h. L does not depend on Z.
  • Postulate 3 (frequency condition): in a jump, hν=Eupper−Elowerh\nu = E_{\text{upper}} - E_{\text{lower}}. Emission when the electron falls, absorption when it rises.
  • h and L have the same dimensions, [ML2T−1][ML^2T^{-1}].
  • The model works only for one-electron systems (H, He⁺, Li²⁺, …). It has no term for the repulsion between electrons.
  • Linear momentum in orbit n: p=mv=Lrp = mv = \dfrac{L}{r}.
  • With mv2r=kZe2r2\dfrac{mv^2}{r} = \dfrac{kZe^2}{r^2}: L2=mkZe2 rL^2 = mkZe^2\,r, so L∝rL \propto \sqrt{r}.
  • The rules give the energy of level n: En=−13.6 Z2n2E_n = -13.6\,\dfrac{Z^2}{n^2} eV (the next page works with it). So L fixes n, and n fixes the energy; or an energy fixes n, and n fixes L.
  • Another force: keep mvr=nh/2πmvr = nh/2\pi and replace the Coulomb balance with that force's own balance, then solve for r.

Bohr's quantisation and frequency condition

L=mvr=nh2π,hν=Eupper−ElowerL = mvr = \frac{nh}{2\pi}, \qquad h\nu = E_{\text{upper}} - E_{\text{lower}}

Worked example

An electron in a hydrogen atom has angular momentum 5h2π\dfrac{5h}{2\pi}. Find (a) its orbit, (b) its energy, and (c) how much its angular momentum changes when it drops to n = 3. Use h=6.6×10−34h = 6.6 \times 10^{-34} J s.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 2 Apr 2026 Shift 1 · Q19Moderate

Example 2 · Atoms · Rutherford Scattering and Bohr's Postulates

Angular momentum of an electron in a hydrogen atom is 3 hπ\frac{3\text{ }h}{\pi}, then the energy of the electron is ____\_\_\_\_ eV .

A multiple of h/2π, not of h

Bohr's rule is L = nh/2π. A statement that says angular momentum is an integral multiple of h is false, even though h and L have the same dimensions.

Read n off L before anything else

If L = 5h/π, rewrite it as 10h/2π: the orbit is n = 10. Then use the energy or radius formula. Plugging L straight into an energy formula has no meaning.

Higher orbit minus lower orbit

Bohr's frequency condition is hν = E_upper − E_lower for both emission and absorption. Writing E_lower − E_upper gives a negative frequency.

Bohr's model is for one electron only

It works for H, He⁺, Li²⁺ and other hydrogen-like ions. It leaves out the repulsion between electrons, so it fails for neutral helium and heavier atoms.
Drill 11 more on bohr's postulates and quantised angular momentum

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (2)

Watch out for (8)

Test yourself on Atoms

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Atoms chapter testDrill all 19 questions