On this page
Most textbook problems on Rutherford scattering stop at the head-on collision: the alpha particle stops, turns round, and you find the distance of closest approach. Real alpha particles mostly miss the nucleus by some distance, and that miss distance decides how far they are deflected. This post connects the impact parameter, the scattering angle and the closest approach, with numbers for gold.
What does the syllabus include?
The JEE (Main) 2026 syllabus lists the alpha-particle scattering experiment and Rutherford's model in Unit 18, and the CBSE Physics curriculum for 2026-27 has the same in Class XII Chapter 12. The NCERT chapter Atoms defines the impact parameter as "the perpendicular distance of the initial velocity vector of the a-particle from the centre of the nucleus", explains that small impact parameters give large deflections, and works the head-on case in Example 12.2. It also reports that about 1 in 8000 alpha particles were deflected by more than 90°. The formulas linking b and the angle go beyond NCERT, so treat them as JEE Advanced-style extensions. Check your exam year's syllabus.
How close does a head-on alpha particle get?
All of the kinetic energy K becomes electric potential energy at the turning point:
K = k(2e)(Ze)/d₀, so d₀ = 2kZe²/K.
A handy value is ke² ≈ 1.44 MeV fm, which follows from k = 8.99 × 10⁹ N m²/C² and e = 1.602 × 10⁻¹⁹ C. NCERT describes Geiger and Marsden using 5.5 MeV alpha particles on gold (Z = 79). Then:
d₀ = 1.44 × 2 × 79 / 5.5 ≈ 41.4 fm.
How does the impact parameter set the angle?
In his 1911 paper, Rutherford showed that the alpha particle follows a hyperbola, and that the deflection angle θ and the perpendicular distance p of the initial path from the nucleus satisfy cot(θ/2) = 2p/b, where his b is the head-on distance. In our notation, with impact parameter b and head-on distance d₀:
b = (d₀/2) cot(θ/2)
For 5.5 MeV alphas on gold:
θ = 60°: b = 20.7 × cot 30° ≈ 20.7 × 1.732 ≈ 35.8 fm.
θ = 90°: b = 20.7 × cot 45° ≈ 20.7 fm.
θ = 180°: cot 90° = 0, so b = 0, the head-on case.
Rutherford's own table gives the same pattern: p/b = 0.5 corresponds to 90°, and p/b = 1 to 53°.
How close does an off-centre particle get?
Two conservation laws give it. Angular momentum about the nucleus: mv₀b = mvr at the closest point, where the velocity is perpendicular to r. Energy: K = ½mv² + kqQ/r. Combining them gives
r² − d₀r − b² = 0, so r_min = [d₀ + √(d₀² + 4b²)] / 2.
Putting in b = (d₀/2) cot(θ/2) simplifies this to r_min = (d₀/2)(1 + cosec(θ/2)).
Worked example. For the 90° alpha above, b ≈ 20.7 fm and d₀ ≈ 41.4 fm:
r_min = [41.4 + √(41.4² + 4 × 20.7²)] / 2 ≈ (41.4 + 58.5)/2 ≈ 49.9 fm.
The shortcut gives 20.7 × (1 + 1/sin 45°) = 20.7 × 2.414 ≈ 49.9 fm. For the 60° alpha, r_min ≈ 20.7 × (1 + 2) ≈ 62.1 fm. Every off-centre particle stays further away than d₀.
Why does the angle distribution look the way it does?
Rutherford also showed that the fraction of particles deflected by more than an angle θ is proportional to cot²(θ/2), because it depends on the target area πb² around each nucleus. So the fraction beyond 90° is cot²45° / cot²30° = 1/3 of the fraction beyond 60°. That is why large-angle scattering is rare, consistent with NCERT's figure of about 1 in 8000 beyond 90°.
Which mistakes cost marks?
Charge e for the alpha particle. It is 2e.
Confusing b with r_min. The impact parameter is measured on the incoming straight line, far from the nucleus.
Using cot θ instead of cot(θ/2).
Leaving energy in MeV with k in SI units. Use joules throughout or ke² ≈ 1.44 MeV fm.
Calling d₀ the nuclear radius. NCERT gets about 30 fm for 7.7 MeV alphas, while gold's radius is about 6 fm. It is an upper limit.
A practice routine
Day 1: Derive d₀ and the quadratic for r_min without notes.
Day 2: Repeat the numbers for 7.7 MeV alphas and check d₀ ≈ 30 fm, as in NCERT.
Day 3: Find b for θ = 120° and 150°, and the matching r_min.
Day 4: Re-read the history in our gold foil experiment post, then do nuclear radius problems for comparison.
Day 5: Take a timed mixed test. Dunlosky and colleagues (2013) rated practice testing and spaced practice as the most useful study techniques.
To check your steps on problems like these, the Fermi Problem Library has worked problems with a step-by-step solver that explains the reasoning rather than printing an answer.
Tonight, plot b against θ for 5.5 MeV alphas on gold at 30°, 60°, 90°, 120° and 150°, and see how fast b shrinks.
