IRODOV 6.2 · QUESTION 02 · RUTHERFORD SCATTERING

How far off-centre must an alpha particle aim to bend by 60° around a gold nucleus?

The answer in one line: b = (q₁q₂/4πε₀)·cot(θ/2) / (2T) ≈ 0.730 pm (730 fm, the book gives 0.73 pm). With the sliders now:

Problem statement

An alpha particle with kinetic energy T = 0.27 MeV is deflected through θ = 60° by a gold foil. Find the matching aiming parameter b. The aiming parameter is the sideways gap between the nucleus and the straight line the alpha first travels along.

The big idea in plain words

An alpha particle is a helium nucleus with two positive charges. A gold nucleus has 79. Like charges push apart. If the alpha aims right at the nucleus, it bounces straight back. If it aims off to the side, it only bends. A bigger sideways gap gives a smaller bend. So the bend angle tells us the gap.

Symbols and constants: q₁ = 2e and q₂ = Ze = 79e are the two charges. e = 1.602×10⁻¹⁹ C is the charge of one proton. ε₀ = 8.854×10⁻¹² F/m sets how strong electric force is. m = 3727.4 MeV/c² is the alpha mass (6.645×10⁻²⁷ kg). M = 183 476 MeV/c² is the gold mass (196.97 u), about 49 times bigger. 1 MeV = 1.602×10⁻¹³ J. 1 fm = 10⁻¹⁵ m, 1 pm = 10⁻¹² m = 1000 fm. hc = 1239.84 MeV·fm (the same number as eV·nm). The electron rest energy 0.511 MeV is not needed, since no electron is involved.

Solution in each frame

1. Which formulas are safe. We assume the gold nucleus stays still. This means gold, 49 times heavier, barely recoils, and we use it in every step. We assume only the electric push between two point charges acts, so the gold's own electrons and the other atoms do not matter. We also assume the alpha meets one nucleus at a time. Next we check the speed. The ratio is T/(mc²) = 0.27 MeV / 3727.4 MeV = 7.2×10⁻⁵, far below 1. So we assume the slow-speed formulas T = ½mv² and p = mv are good.

\(v=c\sqrt{2T/(mc^2)}=2.998\times10^{8}\sqrt{2\times0.27/3727.4}=3.61\times10^{6}\ \text{m/s}\ (0.0120\,c)\)

2. The force. We want the push between the charges. The named law is Coulomb's law: the force is q₁q₂/(4πε₀r²), where r is the gap between them. We write it as k/r² with k = q₁q₂/(4πε₀). First we find e²/(4πε₀) in handy units:

\(\dfrac{e^2}{4\pi\varepsilon_0}=\dfrac{(1.602\times10^{-19}\ \text{C})^2}{4\pi\times8.854\times10^{-12}\ \text{F/m}}=2.307\times10^{-28}\ \text{J·m}=\dfrac{2.307\times10^{-28}}{1.602\times10^{-13}}\ \text{MeV·m}=1.440\ \text{MeV·fm}\)
\(k=2\times79\times1.440\ \text{MeV·fm}=227.5\ \text{MeV·fm}\)

The work to bring the alpha in from far away is \(U(r)=\int_r^\infty \dfrac{k}{r'^2}dr'=\dfrac{k}{r}\). This is its electric energy.

3. What stays fixed. We want a link between the sideways gap and the path. The force points along the line to the nucleus, so it cannot twist the alpha sideways. This is conservation of angular momentum. At the start the alpha has speed v and gap b, so L = mvb. Later, at distance r and polar angle φ, L = mr²φ̇ (φ̇ is how fast φ changes). Equating them and cancelling m gives r²φ̇ = vb.

4. The picture. We want the angle between the first and last directions. We assume the path is a mirror image about the line from the nucleus to the closest point. This means running the motion backwards gives the same path, because Coulomb force does not care about the direction of time. Measure φ from that line. The alpha starts at φ = −φ₀ and ends at +φ₀. At the start it heads inward and at the end it heads outward along the final angle. So the turn is θ = π − 2φ₀, giving φ₀ = π/2 − θ/2.

5. The push along the mirror line. We use the impulse-momentum law: the change in momentum equals the force added up over time. Along the mirror line, momentum goes from −mv cosφ₀ to +mv cosφ₀, a change of 2mv cosφ₀. The force along that line is (k/r²)cosφ. From step 3, dt = r²dφ/(vb).

\(2mv\cos\varphi_0=\displaystyle\int F_{axis}\,dt=\int_{-\varphi_0}^{\varphi_0}\frac{k\cos\varphi}{r^2}\cdot\frac{r^2\,d\varphi}{vb}=\frac{k}{vb}\Big[\sin\varphi\Big]_{-\varphi_0}^{\varphi_0}=\frac{2k\sin\varphi_0}{vb}\)
\(\tan\varphi_0=\dfrac{mv^2b}{k}=\dfrac{2Tb}{k}\). Since \(\tan\varphi_0=\cot(\theta/2)\): \(\;b=\dfrac{k}{2T}\cot\dfrac{\theta}{2}\)

6. The number. We put in the values. cot 30° = 1.732.

\(b=\dfrac{k\cot(\theta/2)}{2T}=\dfrac{227.5\ \text{MeV·fm}\times1.732}{2\times0.27\ \text{MeV}}=\dfrac{394.1}{0.54}\ \text{fm}=729.8\ \text{fm}=0.730\ \text{pm}\)

This is about 100 times the gold nucleus radius (about 7 fm, from the rule 1.2 fm × 197^{1/3}). So the alpha stays well outside the nucleus, and the point-charge model holds.

b ≈ 0.730 pm

What changes and what stays the same: who moves (alpha only, both bodies, or gold only), the momentum arrows (the gold carries 49 times more momentum in the alpha frame), and the 2% mass factor μ/m change. The gap b, the angle θ, the closest approach and the force law stay the same.

Solution by each method

Energy method.

  1. Energy is conserved: T = ½m(ṙ² + r²φ̇²) + k/r. With r²φ̇ = vb, the sideways part is ½mv²b²/r² = Tb²/r². So ½mṙ² = T(1 − k/(Tr) − b²/r²). Write a = k/(2T) = 227.5/0.54 = 421.3 fm.
  2. At the closest point ṙ = 0, so r² − 2a r − b² = 0, giving r_min = a + √(a²+b²).
  3. The sweep angle is dφ = (vb/r²)dr/ṙ with ṙ = v√(1 − 2a/r − b²/r²). Put u = 1/r, so dr/r² = −du: \(\varphi_0=\int_0^{u_m}\dfrac{b\,du}{\sqrt{1-2au-b^2u^2}}\).
  4. Complete the square: \(1-2au-b^2u^2=C^2-s^2\) with s = b(u + a/b²) and C = √(a²+b²)/b. Then du = ds/b and \(\varphi_0=\int_{a/b}^{C}\dfrac{ds}{\sqrt{C^2-s^2}}=\dfrac\pi2-\arcsin\dfrac{a}{\sqrt{a^2+b^2}}\), so tanφ₀ = b/a.
  5. With φ₀ = π/2 − θ/2: b = a cot(θ/2) = 421.3 fm × 1.732 = 729.8 fm. Also r_min = 421.3 + 842.6 = 1264 fm.
b ≈ 0.730 pm

Momentum method.

  1. The force is central, so L = mvb stays fixed, and r²φ̇ = vb.
  2. Along the mirror line, the momentum change is 2mv cosφ₀ (from −mv cosφ₀ to +mv cosφ₀).
  3. The impulse is \(\int\frac{k\cos\varphi}{r^2}\frac{r^2d\varphi}{vb}=\frac{2k\sin\varphi_0}{vb}\).
  4. Equating: tanφ₀ = 2Tb/k = cot(θ/2), so b = (227.5 × 1.732)/(2 × 0.27) fm = 729.8 fm.
b ≈ 0.730 pm

Graph method.

  1. Plot U(r) = k/r (derived in lab step 2) and draw a flat line at T = 0.27 MeV (see the lower graph).
  2. They cross where k/r₀ = T. This is the head-on stopping point: r₀ = 227.5/0.27 = 842.6 fm.
  3. Half of this is a = r₀/2 = 421.3 fm. The two straight-line ends of the path meet at x = −a, with tanφ₀ = b/a.
  4. On the upper graph, b = a·cot(θ/2). At θ = 60° the curve stands 1.732 times a: b = 1.732 × 421.3 = 729.8 fm.
b ≈ 0.730 pm

Quantum-rule method. This checks the classical picture. It does not find b alone.

  1. Momentum: pc = √(2mc²T) = √(2 × 3727.4 × 0.27) = 44.86 MeV.
  2. de Broglie wavelength λ = h/p = hc/(pc) = 1239.84 MeV·fm / 44.86 MeV = 27.6 fm.
  3. We assume a classical path is fine when λ is much smaller than b. Here λ/b = 27.6/729.8 = 0.038. That is small, so the aiming-parameter idea is sound.
  4. The angular momentum in quanta is l = pb/ħ = 44.86 × 729.8 / 197.3 = 166. This is a large number, which also means classical.
b ≈ 0.730 pm (classical path justified)

Level-diagram method. Does not suit this problem. The alpha is not in a bound level and no photon is made.

Check method.

  1. Units: MeV·fm / MeV = fm, and cot is a pure number ✓.
  2. θ → 180° (head-on): cot 90° = 0, so b → 0 ✓.
  3. θ → 0: cot grows without limit, so b → ∞. A far-away alpha barely bends ✓.
  4. T → ∞ or Z → 0: b → 0. A fast alpha, or a weak push, needs a close aim to bend ✓.
  5. At θ = 90°, b = a = r₀/2 = 421 fm, matching the graph ✓.
b ≈ 0.730 pm

Fastest: the momentum method (four lines). Most reliable in an exam: the check method, with the energy method as an independent derivation.

Parameter table

SymbolMeaningValueLive (slider)
Talpha kinetic energy0.27 MeV
θdeflection angle60°
Ztarget charge number79
kq₁q₂/4πε₀227.5 MeV·fm
ak/2T (b at 90°)421.3 fm
r₀head-on stopping distance k/T842.6 fm
r_minclosest approach1264 fm
valpha speed0.0120 c
λde Broglie wavelength27.6 fm
A, Rtarget mass number and nuclear radius (1.2 fm × A^(1/3))197, 7.0 fm
screeningdistance beyond which the electron cloud hides the nucleusabout 10 900 fm
baiming parameter (answer)0.730 pm

What the sliders show

Z (charge of the target). The push constant k = 2Z × 1.44 MeV·fm grows with Z. So a = k/2T and b = a·cot(θ/2) grow in proportion to Z. Doubling Z doubles b for the same angle. The nucleus itself grows slowly, as A^(1/3), and the screening length of the electron cloud shrinks as Z^(−1/3). Try Z = 6, 29 and 79 at fixed scale.

T (energy of the alpha). a and b are proportional to 1/T. A faster alpha needs a closer aim to bend by the same angle. Switch to "Given b, find θ" and keep b fixed: tan(θ/2) = a/b = k/(2Tb), so a faster alpha is bent less. At b = 730 fm on gold, 0.27 MeV gives 60° but 5 MeV gives only about 3.6°. It goes almost straight.

Can any angle happen for a given element? For a point charge with the full Coulomb force, yes: every angle between 0° and 180° has exactly one b. Three real limits cut this down.

  1. Very small angles need a very large b. Beyond the screening length the electron cloud cancels the nuclear charge, so the smallest real angle is about 2·atan(a / screening length). For gold at 0.27 MeV this is about 4°.
  2. Very large angles need a close approach, r_min = a(1 + 1/sin(θ/2)). The closest possible is the head-on value 2a = k/T. If this is smaller than the contact distance R_c = R_nucleus + R_alpha, the alpha touches the nucleus, the strong nuclear force acts, and the Rutherford formula fails. For gold this needs T above about 25 MeV. Light targets reach it sooner: carbon at a few MeV.
  3. A target lighter than the alpha cannot throw it far sideways. For hydrogen the largest lab angle is about 15° (tick "Include nucleus recoil" to see it).

Why most alphas at high energy go straight. For a given bend, b is proportional to 1/T, and the target area that gives a bend of θ or more is π·b², which falls as 1/T². So in a foil, a fast beam is bent far less often. The de Broglie wavelength falls only as T^(−1/2), so λ/b grows with T. At very high T the sharp-path picture also starts to weaken (see the green and blue marks on the ruler).

Common exam traps