The answer in one line
The effective cross-section of a nucleus for alphas scattered through more than ° is σ = - barn, which is - kilobarn.
For the book values (uranium, T = 1.5 MeV, θ₀ = 60°) the book gives 737 b = 0.737 kb. With the constants used here the page gives 735 b. The small gap is rounding in a = Z × 1.44 / T. This page recomputes σ from every slider.
Problem Statement
Find the effective cross-section of a uranium nucleus for scattering of alpha particles with kinetic energy T = 1.5 MeV through angles larger than θ₀ = 60°.
Your values now: nucleus (Z = , A = ), T = keV, θ₀ = °.
We assume only the electric push between the alpha and the nucleus acts, and the nucleus stays at rest. We use this in the Rutherford formula. The CM tab shows what recoil changes.
We assume the alpha never touches the nucleus. For a bend of exactly θ₀ the closest approach is fm, and the nucleus radius is about fm (alpha plus nucleus).
We assume the nucleus is bare. In a real atom the electrons shield it for very small angles. The page warns you when that matters.
We assume the alpha is slow enough for ordinary formulas. Its speed is of the speed of light, safe below 0.1.
The big idea in plain words
Look at the nucleus from the front, along the beam. Draw an invisible circle around it. An alpha that arrives inside the circle bends by more than θ₀. An alpha that arrives outside bends less. The cross-section σ is simply the area of that circle.
So we need only one thing: the radius b₀ of the circle. The target plane in the simulation draws this circle in purple. Every alpha is a dot, purple if it falls inside and cyan if outside.
Boxed derivation: the radius of the circle
The alpha has charge 2e and mass m. The nucleus has charge Ze. The force is k/r², with k = 2Ze²/(4πε₀) and e²/(4πε₀) = 1.44 MeV fm.
Step 1: the impulse. Far away the speed is v. After the bend it is v again, because energy is conserved. So the momentum change has size Δp = 2 m v sin(θ/2).
Step 2: the same impulse from the force. Call ψ the angle from the line that bisects the two paths. Then Δp = ∫ (k/r²) cos ψ dt.
Step 3: angular momentum. m r² (dψ/dt) = m v b, where b is the impact parameter. So dt = r² dψ/(v b), and r² cancels: Δp = (k/(v b)) ∫ cos ψ dψ.
Step 4: the limits. ψ runs from −(π − θ)/2 to +(π − θ)/2, and 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 the 2 in the alpha charge cancels the 2 in 2T. Numbers: a = × 1.44 / = fm. Result: b = a cot(θ/2).
Boxed derivation: the cross-section
Step 1: the circle. An alpha bends by more than θ₀ when b is smaller than b₀ = a cot(θ₀/2). With θ₀ = °: cot(°) gives b₀ = fm.
Step 2: the area.
Step 3: units. 1 fm = 10⁻¹⁵ m, so 1 fm² = 10⁻³⁰ m². Then σ = m². One barn is 10⁻²⁸ m², so 1 fm² = 0.01 b. Therefore σ = b = kb.
Step 4: the same result from rings. The ring between θ and θ + dθ has area dσ = π a² cos(θ/2) dθ / sin³(θ/2). With x = sin(θ/2) we get cos(θ/2) dθ = 2 dx. The integral of 2 dx/x³ from sin(θ₀/2) to 1 equals cot²(θ₀/2). So the area of all rings beyond θ₀ is again π a² cot²(θ₀/2). Graph G1 shows this as the purple area.
The solution viewed in each frame
Lab frame: the nucleus is at rest and the alpha comes in
This is the frame used above. T = keV gives a = fm and b₀ = fm, so σ = b.
Centre-of-mass frame: the origin moves with the centre of mass
The nucleus recoils a little. In the CM frame the alpha and the nucleus carry equal and opposite momentum, each 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 °. The length becomes a* = a(1 + μ) = fm. Then σ* = π a*² cot²(θ₀*/2) = b. The lab value is b, a difference of %.
The two frames look almost the same for uranium because the nucleus is about 60 times heavier than the alpha. The centre of mass sits very near the nucleus, and the nucleus moves only % as far as the alpha. The page says this in a banner. Use the preset "Light target Al (CM)" to see a real difference.
Alpha rest frame
The alpha accelerates during the encounter, so this frame is not inertial. It needs a pseudo-force and adds nothing, so we do not use it.
The solution by each method
Energy method (closest approach)
- We check that the alpha never reaches the nucleus. Use T = k/r_min + ½ m v_t² together with m v b = m v_t r_min.
- This gives r_min = a (1 + 1/sin(θ/2)). At θ₀ it is fm, compared with the nucleus radius fm.
- If r_min is larger, the Coulomb picture is safe and σ = b.
Momentum method (impulse)
- We want b₀. The impulse 2 m v sin(θ/2) equals the force integral, which gives tan(θ/2) = a/b.
- At θ₀ = °, b₀ = fm.
- σ = π b₀² = b.
Graph method
- Graph G1 plots barns per degree against θ. The purple area from θ₀ to 180° is σ.
- Graph G2 plots σ against θ₀ on a log axis. Read it at your θ₀: σ = b. The grey dashed curve is the book case, so you see how your change moved it.
Quantum-rule method
- We check that waves do not matter. The de Broglie wavelength is λ = 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: π a² has the unit of area, as a cross-section must.
- Limits: if θ₀ goes to 180°, cot² goes to 0, so σ goes to 0. If θ₀ goes to 0, σ grows without bound, because the nucleus bends distant alphas by tiny angles. If T doubles, σ becomes a quarter.
- Numerical check: the simulation integrates the real path for b = b₀ step by step. It exits at °, and θ₀ is °. This confirms the formula for b₀.
The impulse method is fastest. The limits check is the most reliable in an exam.
The statistics demonstration
A cross-section can be measured. Fire N alphas at random points in a disc of radius R, and count how many land inside the purple circle. Then σ_measured = π R² × (hits / fired). Each alpha is a dot in the target plane. Graph G3 shows σ_measured against the number fired, with the cyan band marking the expected scatter of one standard deviation. The band shrinks about as 1/√hits. With alphas fired at R = fm, about hits are expected. The beam disc has area fm².
What each slider changes
| Control | Changes | How |
|---|---|---|
| Alpha energy T | a, b₀, σ, r_min, speed, paths, G1, G2 | σ goes as 1/T² |
| Cut angle θ₀ | b₀, σ, purple regions everywhere | σ goes as cot²(θ₀/2); falls as θ₀ rises |
| Alphas fired N | dots, measured σ, G3 band | scatter about 1/√hits; theory σ unchanged |
| Beam disc radius R | disc, hit share (b₀/R)², measured scatter | theory σ unchanged: σ belongs to the nucleus |
| Nucleus | Z, A, a, b₀, σ, μ, nucleus size | σ goes as Z² |
| Target σ (given-the-answer mode) | solves T | T = T_now × √(σ_now/σ_target) |
| Zoom, frame | picture scale; circle drawn as b or b* | zoom none on σ; frame changes σ only by μ |
Parameter Table
| Quantity | Symbol | Now |
|---|---|---|
| Kinetic energy | T | keV |
| Length Z × 1.44 / T | a | fm |
| Cut radius | b₀ | fm |
| Cross-section | σ | fm² = b |
| dσ/dΩ at θ₀ | b/sr | |
| Speed | v | × 10⁷ m/s |
| Beam disc area | π R² | fm² |
Common Exam Traps
- Using the wrong charge. The alpha has charge 2e, so a = Z × 1.44 MeV fm / T. Do not also put a 2 in the denominator.
- Using cot instead of cot². σ has cot²(θ₀/2), and θ₀/2 = 30° here, not 60°.
- Forgetting that T must be in MeV when you use 1.44 MeV fm.
- Unit slips: 1 fm² = 10⁻³⁰ m² = 0.01 b, and 1 b = 10⁻²⁸ m².
- Thinking a bigger beam gives a bigger σ. The cross-section belongs to the nucleus. A bigger beam only changes how many alphas hit.