The answer in one line
The fraction of alphas scattered through less than ° is P = -. The rest, Q = -, bend by more than °.
The trick is P = 1 − Q. For the book values, Q = 0.40 and P = 0.60. This page recomputes both from every slider.
Problem Statement
A narrow beam of alpha particles, each with kinetic energy T = 600 keV, falls straight onto a gold foil. The foil has n = 1.1 × 10¹⁹ nuclei in every square centimetre. Find the fraction of alphas scattered through angles θ smaller than θ₀ = 20°.
Your values now: the foil is (Z = , A = ), T = keV, n = per cm², θ₀ = °.
We assume each alpha scatters only once. This means we can treat each nucleus on its own. The book data is a borderline case for this, and the page warns you when Q gets large.
We assume only the electric push between the alpha and one nucleus acts, and the nucleus stays still. We use this in the Rutherford formula.
We assume the alpha is slow enough for ordinary (non-relativistic) formulas. Its rest energy is 3727 MeV. Its speed is of the speed of light, safe below 0.1.
We assume the alpha never touches the nucleus. For a bend of exactly θ₀, its closest approach is fm. The nucleus has radius about fm.
The big idea in plain words
Look at the foil from the front, along the beam. Each gold nucleus owns a small square patch. Alphas land at random spots.
An alpha that lands close to the nucleus bends a lot. One that lands far away bends only a little. So there is a circle around each nucleus. Inside it, the alpha bends by more than θ₀. Outside it, the alpha bends by less than θ₀.
We want the share outside the circle. But there is a catch. The tiny bends near 0° are very common, and their count has no upper limit in this model. So we cannot add them up directly.
The way out is simple. We find the share inside the circle, which is easy. Then we subtract it from 1. The simulation draws the circle in purple and the rest of the patch in cyan.
Boxed derivation: how much one nucleus bends an alpha
The alpha has charge 2e and mass m. The nucleus has charge Ze. Here e = 1.602 × 10⁻¹⁹ C, and ε₀ = 8.85 × 10⁻¹² F/m sets the strength of electric force. The force is k/r², with k = 2Ze²/(4πε₀).
Step 1: the impulse. Far away the speed is v. After the bend it is v again, because energy is conserved. Only the direction changes. So the momentum change has size Δp = 2 m v sin(θ/2), along the line that bisects the two paths.
Step 2: the same impulse from the force. Call the angle of the alpha from that line ψ. Only the force component along the line adds up: Δp = ∫ (k/r²) cos ψ dt.
Step 3: angular momentum. It is conserved: m r² (dψ/dt) = m v b. So dt = r² dψ/(v b), and r² cancels: Δp = (k/(v b)) ∫ cos ψ dψ.
Step 4: the limits. The path turns through π − θ, so ψ runs from −(π − θ)/2 to +(π − θ)/2. The integral of cos ψ is 2 cos(θ/2). So Δp = 2k cos(θ/2)/(v b).
Step 5: set the two equal.
Here a = k/(m v²) = k/(2T) = Z × 1.44 MeV fm / T, because k = 2Z × 1.44 MeV fm. Numbers: a = × 1.44 / = fm. Result: b = a cot(θ/2).
Boxed derivation: the share inside the circle, then P
Step 1: the circle. An alpha bends by more than θ₀ when its impact parameter is smaller than b₀ = a cot(θ₀/2). Numbers: b₀ = × cot(°) = fm. The circle area is π b₀² = π a² cot²(θ₀/2).
Step 2: share of the foil. The foil has n nuclei per unit area, so each nucleus owns a patch of area 1/n. The share of the patch inside the circle is Q = n π a² cot²(θ₀/2). This needs the thin-foil assumption, so that circles do not overlap.
Step 3: units. n = per cm². The area π a² = fm². One fm² is 10⁻²⁶ cm². So π a² = × 10⁻²⁶ cm². Also cot²(°) = .
Step 4: the number. Q = × e-26 × = .
Step 5: why not add the small angles directly. The ring of impact parameters for angle θ has area dσ = π a² cos(θ/2) dθ / sin³(θ/2). With x = sin(θ/2), cos(θ/2) dθ = 2 dx, so the sum over angles from 0 to θ₀ is ∫ 2 dx / x³ = [−1/x²] from 0. At x = 0 this is infinite. So the direct sum has no finite value in this model.
Step 6: the same integral from the other side. From θ₀ to 180°, x runs from sin(θ₀/2) to 1, and the result is 1/sin²(θ₀/2) − 1 = cot²(θ₀/2). That matches Step 2. The fraction we want is the rest:
Result: P = 1 − = .
The solution viewed in each frame
Lab frame: the foil is at rest and the alpha comes in
This is the frame used above. The nucleus is treated as fixed. T = keV gives a = fm, then Q = and P = .
Centre-of-mass frame: the origin moves with the centre of mass
The nucleus recoils a little. In the frame of the centre of mass, the alpha and the nucleus have equal and opposite momentum, so the total momentum is zero. Each has momentum p₀/(1 + μ) = MeV/c, where μ = m_α/M = . This frame moves at constant velocity, so no pseudo-force is needed.
Two things change. The angle becomes θ* = θ + arcsin(μ sin θ), so θ₀ = ° becomes °. And the length becomes a* = a(1 + μ) = fm. Counting alphas does not depend on the frame. So Q = n π a*² cot²(θ₀*/2) = and P = 1 − Q = . The lab values are Q = and P = . They differ because gold is only 49 times heavier than the alpha. Try the Al preset to see a larger gap.
Alpha rest frame
In this frame the alpha stands still and the foil rushes at it. The alpha accelerates during the encounter, so this frame is not inertial. We would need a pseudo-force, and it adds nothing. We do not use it for the count.
What changes and what stays the same
In the lab frame the nucleus is fixed, the angle is θ and the length is a. In the CM frame both particles move, the angle is θ* and the length is a(1 + μ). The physical fraction is the same in every frame, apart from the small recoil term μ that the book ignores.
The solution by each method
Energy method
- We want to check that the alpha never reaches the nucleus. We use conservation of energy together with angular momentum: T = k/r_min + ½ m v_t², and m v b = m v_t r_min.
- This gives r_min = a (1 + 1/sin(θ/2)). For your θ₀ it is fm, far above the nuclear radius fm.
- So the Coulomb picture is safe, and P = 1 − Q = .
Momentum method (impulse)
- We want the cut radius b₀. The impulse 2 m v sin(θ/2) equals the force integral, which gives tan(θ/2) = a/b.
- At θ₀ = ° we get b₀ = fm.
- The circle share is Q = n π b₀² = , so P = .
Graph method
- Plot alphas per degree against θ (graph G1). The purple area from θ₀ to 180° is Q times the alphas sent.
- The curve rises without limit toward 0°, so the cyan area is not finite. We cannot read P from it directly.
- Graph G2 plots Q and 1 − Q against θ₀. Read both at your θ₀: Q = , P = .
Quantum-rule method
- We check whether wave effects matter. The de Broglie wavelength is λ = h/p = hc/(pc) = 1239.84 MeV fm / MeV = fm.
- This is much smaller than r_min = fm. So the alpha follows a path, and classical counting is safe.
Level-diagram method
No bound states or energy levels are involved. The alpha is free before and after, so this method does not apply.
Check method
- Units: n (per cm²) times π a² (cm²) has no units, as a fraction must.
- Limits: if n goes to 0, Q goes to 0 and P goes to 1, because a very thin foil scatters almost nothing. If θ₀ goes to 180°, cot² goes to 0, so P goes to 1, since every alpha is below 180°.
- If θ₀ goes to 0, the formula gives Q above 1. That is impossible, and the page warns you. The real Q stays below 1.
The impulse method is fastest. The check with limits is the most reliable in an exam.
What each slider changes
| Control | Changes | How |
|---|---|---|
| Alpha energy T | a, b₀, speed, Q, P | Q goes as 1/T² (inverse square) |
| Nuclei per area n | Q, P, foil thickness, cell size | Q proportional to n; P = 1 − Q |
| Beam I₀ | alphas sent, expected counts, noise | counts proportional; P none, as P is a chance for one alpha |
| Time τ | alphas sent, expected counts | counts proportional; P none |
| Cut angle θ₀ | b₀, the cyan and purple regions, Q, P | Q goes as cot²(θ₀/2); P rises with θ₀ |
| Foil material | Z, A, density, a, thickness, Q, P | Q goes as Z² |
| Frame | radii and angles drawn | answers differ only by the recoil term μ |
Parameter Table
| Quantity | Symbol | Now |
|---|---|---|
| Kinetic energy | T | keV |
| Length k/(2T) | a | fm |
| Cut radius | b₀ | fm |
| Nuclei per m² | n | |
| Foil thickness (from n and density) | d | µm |
| Cell side 1/√n | L | fm |
| Share beyond θ₀ | Q | |
| Fraction below θ₀ | P | |
| Cap solid angle 2π(1 − cos θ₀) | sr (% of the sphere) |
The cap covers only a small share of all directions, yet it holds most of the alphas. This is because Rutherford scattering is strongly peaked near the beam direction.
Common Exam Traps
- Trying to add all the small angles from 0 to θ₀. The sum is infinite. Count θ ≥ θ₀ and subtract from 1.
- Using cot instead of cot². Q has cot²(θ₀/2), and θ₀/2 = 10° here, not 20°.
- Using the charge e for the alpha. The alpha has charge 2e, so a = Z × 1.44 MeV fm / T (with the factor 2 already included).
- Mixing units: n is per cm², so convert π a² from fm² with 1 fm² = 10⁻²⁶ cm².
- Forgetting that T must be in MeV when using 1.44 MeV fm. 600 keV is 0.6 MeV.
- Quoting Q (0.40) as the answer. The question asks for the fraction below θ₀, which is P = 0.60.