The answer in one line
(a) About - alphas land in the window. (b) About - alphas bend by more than °.
For the book values the answers are 1.6 × 106 and 2.0 × 107. This page recomputes both from every slider.
Problem Statement
A narrow beam of alpha particles, each with kinetic energy T = 0.50 MeV, hits a gold foil of mass thickness ρd = 1.5 mg/cm². The beam carries I₀ = 5.0 × 10⁵ alphas per second and runs for τ = 30 minutes. Find (a) how many alphas are scattered into a window between 59° and 61° from the beam direction, and (b) how many alphas are scattered by more than θ₀ = 60°.
Your values now: the foil is (Z = , A = ), T = MeV, ρd = mg/cm², I₀ = per second, τ = min. The window runs from ° to ° and the cut angle is °.
We assume the foil is thin, so each alpha scatters at most once. This means we can add the effect of each nucleus separately.
We assume only the electric force between the alpha and one nucleus acts, and that the nucleus is heavy enough to stay still. We use this in the Rutherford formula.
We assume the alpha moves slowly, so ordinary (non-relativistic) formulas work. Its speed is of the speed of light, which is safe below 0.1.
We assume the alpha never touches the nucleus. Its closest approach is r_min = fm at the window angle, and the nucleus has radius about fm.
The big idea in plain words
Look down the beam at the foil. Each gold nucleus is a tiny target, and the alphas arrive at random spots. An alpha that passes close to a nucleus bends a lot. One that passes far away bends a little.
So the bend angle is set by one distance, the impact parameter b. It is how far the alpha's straight path would miss the nucleus. A small b gives a large angle.
That turns counting into geometry. All alphas inside a circle of radius b₀ around a nucleus bend by more than θ₀. The fraction of alphas that do this is the circle area divided by the area that one nucleus owns. For part (a), the alphas that land in the window are those that arrive in a thin ring, between radius b₂ and radius b₁.
Look at the target-plane panel in the simulation: the purple circle is part (b), the cyan ring is part (a).
Boxed derivation: how much one nucleus bends an alpha
Start with the alpha far away, moving at speed v along a line that misses the nucleus by b. Call the scattering angle θ.
Step 1: the impulse. Far away the speed is v. Far away after the bend the speed is also v, because energy is conserved and the potential energy returns to zero. The momentum changes only in direction, so the size of the momentum change is Δp = 2 m v sin(θ/2), pointing along the line that bisects the two paths.
Step 2: the same impulse from the force. Only the component of the force along that line adds up. Write the angle of the alpha from that line as ψ. The force is k/r² with k = 2Ze²/(4πε₀), and its component along the line is (k/r²) cos ψ. So Δp = ∫ (k/r²) cos ψ dt.
Step 3: angular momentum. Angular momentum is conserved: m r² dψ/dt = m v b. So dt = r² dψ/(v b), and the r² cancels:
Step 4: the limits. The path turns through an angle π − θ, so ψ runs from −(π − θ)/2 to +(π − θ)/2. The integral of cos ψ is 2 sin((π − θ)/2) = 2 cos(θ/2). So Δp = 2k cos(θ/2)/(v b).
Step 5: set the two impulses equal.
Here a = k/(m v²) = k/(2T). With k = 2Z × 1.44 MeV fm, this gives a = Z × 1.44 fm MeV / T. Right now a = fm.
Result: b = a cot(θ/2).
Boxed derivation: counting alphas
Step 1: area owned by one nucleus. The foil has n nuclei per square metre (n = ). Alphas hit the foil at random. One alpha meets about n × (area) nuclei, so the chance of passing within radius b₀ of some nucleus is n π b₀². This stays small for a thin foil.
Step 2: ring for a window. Alphas bent into the angle range θ to θ + dθ came from the ring between b and b − |db|. Its area is dσ = 2π b |db|. Taking b = a cot(θ/2), we get |db| = a dθ/(2 sin²(θ/2)). So
Write sin θ = 2 sin(θ/2) cos(θ/2) and dΩ = 2π sin θ dθ. Then dσ = π a² cos(θ/2) dθ / sin³(θ/2) becomes dσ = [a²/(4 sin⁴(θ/2))] dΩ. That is the Rutherford formula.
Step 3: part (a). The window is narrow, so treat dσ/dΩ as a constant. Count = (alphas sent) × n × (dσ/dΩ) × dΩ, with dΩ = 2π sin θ Δθ:
Numbers: I₀ τ = alphas, n = per m², a²/(4 sin⁴(θ/2)) = fm², dΩ = sr. The result is dN = . The exact integral over the window gives .
Step 4: part (b). Add up all rings from θ₀ to π. Write x = sin(θ/2). Then dx = ½ cos(θ/2) dθ, so cos(θ/2) dθ = 2 dx:
because 1/sin² − 1 = cot². (It is also just the circle: π b₀² = π a² cot²(θ₀/2).) With cot²(θ₀/2) = , the answer is N = .
Step 5: the unit traps. Δθ must be in radians (° = rad). τ must be in seconds ( s). ρd = mg/cm² = kg/m². And 1 fm² = 10⁻³⁰ m².
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, so a = Z × 1.44 fm MeV / T with T the alpha's energy in the lab. Results: (a) , (b) .
Centre-of-mass frame: the origin moves with the centre of mass
The nucleus is not truly fixed. It recoils. In the frame that moves with the centre of mass, the alpha and the nucleus have equal and opposite momentum p₀/(1 + μ) = , where μ = m_α/M = . No pseudo-force appears, because this frame moves at constant velocity.
Two things change. The angle in this frame is larger: θ* = θ + arcsin(μ sin θ), so the cut angle θ₀ = ° becomes θ₀* = °. And the length becomes a* = a(1 + μ) = fm. The counts are the same in any frame, so we use the same formula with starred values. Result: (a) , (b) . The difference from the lab result is % for (a) and % for (b). For gold it is tiny because gold is 49 times heavier than an alpha. Use the Al preset to see a real difference.
Alpha rest frame
In this frame the alpha stands still and the foil rushes at it. The nucleus now moves, so this frame is not inertial during the encounter (the alpha accelerates), and a pseudo-force would be needed. It adds nothing here, so we do not use it for counting.
The solution by each method
Energy method
Energy conservation gives the speed of approach and the closest distance: r_min = a (1 + 1/sin(θ/2)) = fm at the window angle. That is far bigger than the nucleus, so the alpha never touches it. This is the check that the Coulomb picture is valid. The target plane in the simulation shows b₀ = fm, b₁ = fm and b₂ = fm.
Momentum method (impulse)
This is Step 1 to 5 of the first boxed derivation: the impulse 2 m v sin(θ/2) equals the force integral. It gives tan(θ/2) = a/b, and everything else follows from it.
Graph method
Plot alphas per degree against θ (graph G1). The cyan strip is part (a): its area is height × width. The purple area from θ₀ to 180° is part (b). The curve falls steeply, so most of the purple area sits just beyond θ₀. Graph G2 plots part (b) against θ₀ directly.
Quantum-rule method
Wave effects matter only if the alpha's de Broglie wavelength is comparable to the distance of closest approach. Here λ = fm, much smaller than r_min = fm. So the alpha follows a path, and classical Rutherford 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
Check the size: the chance that one alpha scatters beyond θ₀ is n π a² cot²(θ₀/2) = . Multiply by the alphas sent () to get . Check the ratio: (b)/(a) is large because (b) covers every angle above θ₀, while (a) is a 2° ring. Check the units: 1 mg/cm² = 10⁻² kg/m², n = N_A (ρd)/A with N_A = 6.023 × 10²⁶ per kmol and A in kg per kmol.
What each slider changes
| Control | Changes | Does not change |
|---|---|---|
| Alpha energy T | a, all radii, both answers (as 1/T²) | nuclei per area |
| Mass thickness ρd | nuclei per area, both answers (proportional) | a, the angles |
| Beam I₀ and time τ | alphas sent, both answers (proportional), noise | the probability for one alpha |
| Window centre and width | answer (a), the cyan strip and ring | answer (b) |
| Cut angle θ₀ | answer (b), the purple region | answer (a) |
| Foil material | Z, A, a, n, and the nucleus size | the beam |
| Frame | radii and angles drawn (b* and θ*) | the counts |
Parameter Table
| Quantity | Symbol | Now |
|---|---|---|
| Length k/(2T) | a | fm |
| Nuclei per m² | n | |
| Alphas sent | I₀ τ | |
| Cross-section area per nucleus beyond θ₀ | π b₀² | × cot²(θ₀/2) fm² |
| Foil thickness | d | µm |
| Cell side (1/√n) | L | fm |
| (a) narrow window / exact | / | |
| (b) |
Common Exam Traps
- Using degrees for Δθ. It must be in radians: 2° = 0.0349 rad.
- Using minutes for τ. Use seconds: 30 min = 1800 s.
- Forgetting the ring. dΩ = 2π sin θ Δθ covers the whole circle around the beam, not just one direction.
- Forgetting to convert ρd from mg/cm² to kg/m² (a factor 10⁻²), or fm² to m² (10⁻³⁰).
- Using the narrow-window formula for part (b). Part (b) needs the integral over all angles above θ₀, which gives cot²(θ₀/2).
- Mixing up what the 1/T² means: doubling the energy cuts both answers to a quarter.
- Treating a as the closest approach. The closest approach in a head-on hit is 2a.