The answer in one line
The fraction of protons scattered through more than ° is Q = -. Out of protons sent, about - do this.
For the book values, Q = 1.4 × 10⁻³. That is about 1.4 protons in every 1000. This page recomputes it from every slider.
Problem Statement
A narrow beam of protons, each with kinetic energy T = 1.4 MeV, falls straight onto a brass foil. The foil has mass thickness ρd = 1.5 mg/cm². This means each square centimetre of foil has 1.5 mg of metal. Copper and zinc are in the weight ratio 7 : 3. Find the fraction of protons scattered through angles larger than θ₀ = 30°.
Your values now: element 1 is (Z = , A = ), with weight share . Element 2 is (Z = , A = ), with weight share . T = keV, ρd = mg/cm², θ₀ = °.
We assume each proton scatters only once. This means the two metals add their effects separately. The page warns you when Q is large.
We assume only the electric push between the proton and one nucleus acts, and that nucleus stays still. We use this in the Rutherford formula.
We assume the metals are mixed so evenly that each kind of nucleus is spread across the foil. We use this to count nuclei per area for each metal.
We assume the proton is slow enough for ordinary (non-relativistic) formulas. Its rest energy is 938 MeV. Its speed is of the speed of light, safe below 0.1.
We assume the proton never touches a nucleus. For a bend of exactly θ₀, the closest approach is fm for element 1 and fm for element 2. The nuclei have radii about fm and fm.
The big idea in plain words
Look at the foil from the front, along the beam. Every nucleus has an invisible circle around it. A proton that lands inside the circle bends by more than θ₀. A proton that lands outside bends less.
The fraction we want is the total area of all these circles, divided by the area of the foil. Copper and zinc have different circles, because their nuclei have different charge. Zinc has charge 30e and copper 29e. So zinc's circle is slightly larger.
So we do three jobs. We count the nuclei of each metal per square metre. We find the circle size for each. Then we add the circle areas. The simulation draws both circles to scale, magnified so you can see them.
Boxed derivation: how much one nucleus bends a proton
The proton has charge e and mass m. A 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 = Ze²/(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 proton 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, where b is the impact parameter (how far the straight path would miss the nucleus). 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 / (2T), because e²/(4πε₀) = 1.44 MeV fm. For element 1: a₁ = × 1.44 / (2 × ) = fm. For element 2: a₂ = × 1.44 / (2 × ) = fm. Result: b = a cot(θ/2).
Boxed derivation: adding up the circles
Step 1: the circle for each metal. A proton bends by more than θ₀ when b is smaller than b₀ = a cot(θ₀/2). So the circle area is π a² cot²(θ₀/2). Numbers: b₀ = fm for element 1, and fm for element 2. Also cot²(°) = .
Step 2: nuclei per area. Take a patch of foil of area 1 m². Its mass is ρd = kg. Element 1 has weight share w₁ = , so its mass is w₁ ρd. One kmol of element 1 has mass A₁ kg and holds N_A = 6.023 × 10²⁶ atoms. So
Step 3: the share of the foil covered by circles. One m² holds n₁ circles of area π b₀₁² and n₂ circles of area π b₀₂². The total circle area per m² is n₁ π b₀₁² + n₂ π b₀₂². This is also the chance Q that one proton lands inside a circle, because the foil is 1 m² and the proton lands at random. This needs a thin foil, so the circles do not overlap.
Step 4: the numbers. n₁ a₁² = × ( fm)² and n₂ a₂² = × ( fm)². Then Σ n a² = fm²/m², and 1 fm² = 10⁻³⁰ m². So Q = π × × × 10⁻³⁰ = .
Step 5: where each part comes from. Element 1 gives Q₁ = . Element 2 gives Q₂ = . Their sum is Q. The ring-sum version of this result starts from dσ = π a² cos(θ/2) dθ / sin³(θ/2). With x = sin(θ/2), cos(θ/2) dθ = 2 dx, and the integral of 2 dx/x³ from sin(θ₀/2) to 1 equals cot²(θ₀/2). This gives the same circle area, so the two views agree.
A note on the book. The book labels n₁ with 0.7 for zinc (Z = 30) and n₂ with 0.3 for copper. But the problem says copper is 7 and zinc is 3. Using the weights as the problem states them, Q = 1.40 × 10⁻³. The book's swapped weights give 1.42 × 10⁻³. Both are close to the book's 1.43 × 10⁻³, so the idea is the same.
The solution viewed in each frame
Lab frame: the foil is at rest and the proton comes in
This is the frame used above. The nuclei are treated as fixed. T = keV gives a₁ = fm and a₂ = fm, so Q = .
Centre-of-mass frame: the origin moves with the centre of mass
A nucleus recoils a little. In the frame of the centre of mass, the proton and the nucleus have equal and opposite momentum, so the total momentum is zero. For element 1, each has momentum p₀/(1 + μ₁) = MeV/c, where μ₁ = m_p/M₁ = . This frame moves at constant velocity, so no pseudo-force is needed.
Two things change for each metal. The angle becomes θ* = θ + arcsin(μ sin θ). For element 1, θ₀ = ° becomes °. The length becomes a* = a(1 + μ), so a₁* = fm. Counting protons does not depend on the frame. So Q = π Σ n a*² cot²(θ₀*/2) = . The lab value is . The two differ by %, because copper and zinc nuclei are about 64 times heavier than a proton. Try the Al + Be preset for a bigger gap.
Proton rest frame
In this frame the proton stands still and the foil rushes at it. The proton 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 nuclei are 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 proton never reaches a 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)). At θ₀ it is fm for element 1, and fm for element 2. Both are above the nuclear radius.
- So the Coulomb picture is safe, and Q = .
Momentum method (impulse)
- We want the cut radius b₀ for each metal. The impulse 2 m v sin(θ/2) equals the force integral, which gives tan(θ/2) = a/b.
- At θ₀ = °: b₀₁ = fm and b₀₂ = fm.
- Q = n₁ π b₀₁² + n₂ π b₀₂² = + = .
Graph method
- Plot protons per degree against θ (graph G1). The purple area from θ₀ to 180° is Q times the protons sent.
- Graph G2 plots Q against θ₀ on a log axis. Read it at your θ₀: Q = .
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, so the proton follows a path, and classical counting is safe.
Level-diagram method
No bound states or energy levels are involved. The proton is free before and after, so this method does not apply.
Check method
- Units: n (per m²) times π a² (m²) has no units, as a fraction must.
- Limits: if ρd goes to 0, Q goes to 0, because an empty foil scatters nothing. If θ₀ goes to 180°, cot² goes to 0, so Q goes to 0. If T gets large, Q falls as 1/T².
- If w₁ = 1, the foil is pure element 1, and Q = Q₁ alone. If w₁ = 0, it is pure element 2.
The impulse method is fastest. The check with limits is the most reliable in an exam.
What each slider changes
| Control | Changes | How |
|---|---|---|
| Proton energy T | a₁, a₂, circles, speed, Q | Q goes as 1/T² (inverse square) |
| Mass thickness ρd | n₁, n₂, Q, foil thickness, nucleus spacing | Q proportional to ρd |
| Weight share w₁ | n₁, n₂, Q₁, Q₂, density, the mosaic mix | Q is a weighted sum of w₁ Z₁²/A₁ and w₂ Z₂²/A₂ |
| Beam I₀ and time τ | protons sent, expected counts, noise | counts proportional; Q none, as Q is a chance for one proton |
| Cut angle θ₀ | circles, regions, Q | Q goes as cot²(θ₀/2); falls as θ₀ rises |
| Elements 1 and 2 | Z, A, density, a, n, circles, Q | Q goes as Z²/A |
| Zoom, frame | mosaic scale; circles drawn as b or b* | zoom none on Q; frame changes Q only by μ |
Parameter Table
| Quantity | Symbol | Now |
|---|---|---|
| Kinetic energy | T | keV |
| Lengths k/(2T) | a₁, a₂ | fm, fm |
| Cut radii | b₀₁, b₀₂ | fm, fm |
| Nuclei per m² | n₁, n₂ | , |
| Mixed density, thickness | ρ, d | g/cm³, µm |
| Spacing between nuclei | 1/√n | fm |
| Parts of Q | Q₁, Q₂ | , |
| Answer | Q |
The purple region covers % of all directions only for the small cap θ < θ₀. The protons that go beyond θ₀ are few because scattering is strongly peaked near the beam.
Common Exam Traps
- Using the wrong charge. A proton has charge e, not 2e. So a = Z × 1.44 MeV fm / (2T).
- Treating the foil as one metal. Each metal has its own n and its own Z², so add n₁ Z₁² + n₂ Z₂².
- Using the weight ratio 7 : 3 as a number ratio. It is a mass ratio, so n₁ : n₂ = (0.7/A₁) : (0.3/A₂).
- Mixing up which metal has which share. Copper is 7, zinc is 3, in this problem.
- Using cot instead of cot². Q has cot²(θ₀/2), and θ₀/2 = 15° here, not 30°.
- Unit slips: ρd = 1.5 mg/cm² = 1.5 × 10⁻² kg/m², T must be in MeV with 1.44 MeV fm, and 1 fm² = 10⁻³⁰ m².