Estimating the Size of the Nucleus from Alpha Scattering
Physics · Atoms · NEET
In alpha scattering, we estimate the size of the nucleus by finding the distance of closest approach d. The alpha particle slows, stops, and turns back when all its kinetic energy K has turned into electric potential energy: d = 2Ze²/(4πε₀K). This d is an upper limit for the nucleus radius, because the alpha particle stops before it actually touches the nucleus. Memory hook: "Close, but no touch" — the alpha stops short, so the true nucleus is even smaller than d.
A head-on alpha particle slows as it climbs the electric hill of the +Ze nucleus, stops (v = 0) at distance d, then reverses. Equating kinetic energy K to potential energy gives d = 2Ze²/(4πε₀K). Since it stops before touching, d is an upper limit for the nucleus size (gold: d ≈ 30 fm, true radius ≈ 6 fm).
Your doubts, answered
Is the distance of closest approach the same as the size (radius) of the nucleus?
No. The distance of closest approach d is only an upper limit for the nucleus radius. At d the alpha particle stops and turns back, but it stops before it touches the nucleus surface. So the real nucleus radius is smaller than d. For gold, d comes out to about 30 fm, but the actual gold nucleus radius is only about 6 fm.
Why does the estimate give a value bigger than the true nucleus size?
Because the alpha particle is repelled by the positive nucleus and reverses direction while it is still some distance away. It never physically reaches the nuclear surface. So d = 2Ze²/(4πε₀K) measures where it turns around, not where the nucleus ends. That is why we say d is only an upper bound on nuclear size.
How exactly does closest approach set an upper limit on nuclear radius?
The alpha particle penetrates closest when it is aimed straight at the nucleus (head-on, impact parameter b = 0). Even in that best case it stops at distance d. Since it never gets closer than d, the nucleus must fit inside a sphere of radius d. Therefore radius of nucleus < d, giving an upper limit.
Which kinetic energy value do we use for the gold estimate?
NCERT uses the maximum kinetic energy of natural alpha particles, K = 7.7 MeV = 1.2 × 10⁻¹² J. Using a higher K makes d smaller, so a faster alpha particle gets closer and gives a tighter (smaller) upper limit on the nucleus size.
Does a heavier alpha particle come closer or stop farther away?
At the same speed v, use K = ½mv². Then d = 4Ze²/(4πε₀ m v²), so d is proportional to 1/m. A heavier particle at the same speed carries more energy, pushes closer, and gives a smaller d. This exact idea was asked in NEET 2016.
⚠️ The NEET trap ✗ Distance of closest approach = radius of the nucleus, so gold nucleus radius = 30 fm. ✓ Distance of closest approach is only an UPPER LIMIT. The alpha particle reverses before touching the nucleus, so the real gold nucleus radius (about 6 fm) is much smaller than d (about 30 fm). 🧠 NTA loves the word 'estimate'. The scattering method gives a maximum possible size, not the exact size. Read the question: 'closest approach' is not 'radius'.
Real NEET questions
NEET 2016 Phase 1
When an alpha-particle of mass m moving with velocity v bombards a heavy nucleus of charge Ze, its distance of closest approach from the nucleus depends on m as:
A · 1/m ✓
B · 1/sqrt(m)
C · 1/m²
D · m
Solution: At the closest approach the alpha particle stops, so all kinetic energy becomes electric potential energy. Step 1: ½mv² = (1/4πε₀)(2e)(Ze)/d. Step 2: Solve for d: d = (1/4πε₀)·4Ze²/(mv²). Step 3: Here Z, e, v are fixed, so d is proportional to 1/m. A heavier alpha particle at the same speed has more kinetic energy, gets closer, so d is smaller. Answer: 1/m (option A).
Solved Atoms NEET PYQs
Try the real previous-year questions from this chapter — each with the answer and a full solution.
What is the formula to estimate the size of the nucleus from scattering?
Use the distance of closest approach: d = 2Ze²/(4πε₀K), where K is the alpha particle's kinetic energy and Ze is the nuclear charge. This d is the upper limit of the nucleus size.
What is the estimated size of the gold nucleus?
For a 7.7 MeV alpha particle on gold (Z = 79), d = 3.0 × 10⁻¹⁴ m = 30 fm. This is an upper limit. The true gold nucleus radius is about 6 fm (1 fm = 10⁻¹⁵ m).
Why is the alpha scattering result an over-estimate of nuclear size?
The alpha particle is repelled and turns back before touching the nucleus. So d marks the turning point, not the nuclear surface, making d larger than the actual radius.
How can we get a smaller (better) estimate of the nucleus size?
Use alpha particles with higher kinetic energy K. Since d is proportional to 1/K, a faster alpha particle stops closer to the nucleus and gives a tighter upper limit.
What is the order of the size of a nucleus?
About 10⁻¹⁴ to 10⁻¹⁵ m, that is a few fermi (femtometre). The atom itself is about 10⁻¹⁰ m, so the nucleus is roughly 10⁵ times smaller than the atom.
