IRODOV 6.3 · QUESTION 03 · HEAD-ON COLLISION

How close can an alpha particle get to a lead nucleus, and to a free lithium nucleus?

The answer in one line: (a) r_min = k/T ≈ 0.590 pm (590 fm, the book gives 0.591 pm). (b) r_min = (k/T)(1 + m₁/m₂) ≈ 0.034 pm (33.9 fm). With the sliders now:

Problem statement

An alpha particle with kinetic energy T = 0.40 MeV hits a nucleus head-on. Find the smallest distance between them for (a) a stationary Pb nucleus and (b) a stationary free Li nucleus. Head-on means the alpha aims straight at the centre. Free means the nucleus is not held and can recoil (be pushed away).

The big idea in plain words

Both nuclei carry positive charge, so they push each other apart. The alpha slows down as it climbs this hill of electric energy. At the closest point it stops closing in, and then it bounces back. A heavy lead nucleus hardly moves, so the alpha's whole energy goes into the hill. A light lithium nucleus gets shoved away. Some energy goes into its motion, so the alpha cannot get as deep as the energy alone would allow, and the answer shifts.

Symbols and constants: q₁ = 2e for the alpha, q₂ = Z₂e for the target (Z₂ = 82 for Pb, 3 for Li). e = 1.602×10⁻¹⁹ C is the proton charge. ε₀ = 8.854×10⁻¹² F/m sets the strength of electric force. m₁ and m₂ are the masses. We use mass number × 931.5 MeV/c², so m₁ = 4 (3726 MeV/c²), Pb is 208 and Li is 7. The ratio m₁/m₂ is 4/208 and 4/7. 1 MeV = 1.602×10⁻¹³ J, 1 fm = 10⁻¹⁵ m, 1 pm = 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 want to know if simple formulas work. We assume the nuclei are point charges that only push each other with the electric force. This means we ignore nuclear contact, which is fine if the answer is far bigger than a nucleus (about 1.2 fm × A^{1/3}, so 7 fm for Pb and 2.3 fm for Li). We assume the path is a straight line, which is what head-on means. We assume classical physics, which the quantum method below tests. Last, the speed check: T/(m₁c²) = 0.40 MeV / 3726 MeV = 1.1×10⁻⁴, far below 1. So we assume the slow-speed forms T = ½m₁v² and p = mv are good.

2. The electric energy. We want the energy stored when the nuclei are a gap r apart. The named law is Coulomb's law: force = k/r² with k = q₁q₂/(4πε₀). First we find e²/(4πε₀) in handy units:

\(\dfrac{e^2}{4\pi\varepsilon_0}=\dfrac{(1.602\times10^{-19}\ \text{C})^2}{4\pi\times8.854\times10^{-12}\ \text{F/m}}=2.307\times10^{-28}\ \text{J·m}=\dfrac{2.307\times10^{-28}}{1.602\times10^{-13}}\ \text{MeV·m}=1.440\ \text{MeV·fm}\)
\(k_{Pb}=2\times82\times1.440=236.2\ \text{MeV·fm},\qquad k_{Li}=2\times3\times1.440=8.64\ \text{MeV·fm}\)

The energy is the work needed to bring the alpha in from far away: \(U(r)=\int_r^\infty\dfrac{k}{x^2}dx=\Big[-\dfrac kx\Big]_r^\infty=\dfrac kr\).

3. Case (a), lead. We assume Pb is so heavy (208 against 4) that it stays still. This means all of T goes into U. Conservation of energy at the closest point, where the alpha is at rest, gives T = k/r_min:

\(r_{min}=\dfrac kT=\dfrac{236.2\ \text{MeV·fm}}{0.40\ \text{MeV}}=590.4\ \text{fm}=0.590\ \text{pm}\)

4. Case (b), free lithium. Now the target also moves, so we need two laws. Gap d shrinks until it stops shrinking. At that moment the two speeds are equal, call it V. Momentum is conserved (no outside push), so m₁v = (m₁+m₂)V, giving V = m₁v/(m₁+m₂). Then energy is conserved:

\(T=\tfrac12(m_1+m_2)V^2+\dfrac{k}{r_{min}},\quad \tfrac12(m_1+m_2)V^2=\dfrac{m_1^2v^2}{2(m_1+m_2)}=T\dfrac{m_1}{m_1+m_2}\)
\(\dfrac{k}{r_{min}}=T\dfrac{m_2}{m_1+m_2}\;\Rightarrow\;r_{min}=\dfrac kT\left(1+\dfrac{m_1}{m_2}\right)=\dfrac{8.64\ \text{MeV·fm}}{0.40\ \text{MeV}}\left(1+\dfrac47\right)=21.6\times1.571=33.9\ \text{fm}=0.034\ \text{pm}\)

This is about 15 times the Li radius, so the point-charge model holds.

(a) 0.590 pm (b) 0.034 pm

What changes and what stays the same: who moves (target still, both bodies, or only the target), the momentum arrows (zero total in the CM; the target carries m₂/m₁ times more in the alpha frame) and the energy shares change. The closest distance and the force law stay the same.

Solution by each method

Energy method.

  1. Split the alpha's energy. With V = m₁v/(m₁+m₂): ½m₁v² = ½(m₁+m₂)V² + ½μv². The CM part cannot become electric energy.
  2. Check: ½(m₁+m₂)V² = T·m₁/(m₁+m₂), so the rest is T·m₂/(m₁+m₂) = 0.40 × 7/11 = 0.2545 MeV.
  3. At closest approach, k/r_min = 0.2545 MeV, so r_min = 8.64/0.2545 = 33.9 fm (b).
  4. If the target is fixed, V = 0 and all of T counts: r_min = 236.2/0.40 = 590.4 fm (a).
(a) 0.590 pm (b) 0.034 pm

Momentum method.

  1. Total momentum is conserved: m₁v = m₁v₁ + m₂v₂ at all times.
  2. The gap d = x₂ − x₁ is smallest when ḋ = 0, so v₁ = v₂ = V and V = m₁v/(m₁+m₂) = (4/11)v.
  3. Energy then reads T = ½(m₁+m₂)V² + k/r_min = (4/11)T + k/r_min.
  4. So k/r_min = (7/11)T = 0.2545 MeV and r_min = 33.9 fm (b). For fixed Pb, the target absorbs the momentum but no energy, so r_min = k/T = 590.4 fm (a).
(a) 0.590 pm (b) 0.034 pm

Graph method.

  1. Plot U(r) = k/r (derived in lab step 2), the curve in the upper right graph.
  2. Draw a flat line at T = 0.40 MeV. For Pb the curve meets it at r = 236.2/0.40 = 590.4 fm (a).
  3. For Li a part of T is locked in the CM motion. The usable line sits lower, at T′ = 0.2545 MeV (green line).
  4. It meets the curve at r = 8.64/0.2545 = 33.9 fm (b). The gap between the two lines is 0.1455 MeV, the locked energy.
(a) 0.590 pm (b) 0.034 pm

Quantum-rule method. This tests the classical picture. It does not find r_min alone.

  1. Momentum: pc = √(2m₁c²T) = √(2 × 3726 × 0.40) = 54.6 MeV.
  2. For relative motion the momentum is p′ = (m₂/(m₁+m₂))p, about 54.6 MeV/c for Pb and 34.7 MeV/c for Li.
  3. de Broglie wavelength λ = h/p = hc/(pc): Pb 22.7 fm, Li 1239.84/34.7 = 35.7 fm.
  4. For Pb, λ/r_min = 0.04, so the classical path is sound. For Li, λ is about the same size as r_min = 33.9 fm. We assume classical physics anyway, as the book does, but the Li answer is only a rough guide.
(a) 0.590 pm (b) 0.034 pm (classical)

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 ✓.
  2. m₂ → ∞: 1 + m₁/m₂ → 1 and (b) becomes (a) ✓.
  3. Li case: the factor 1.571 makes r_min bigger than k/T = 21.6 fm. A moving target means a less deep approach ✓.
  4. T → ∞: r_min → 0. A faster alpha gets deeper ✓. Z₂ → 0: r_min → 0 ✓.
  5. m₂ → 0: r_min → ∞. A feather-light target is simply brushed aside ✓.
(a) 0.590 pm (b) 0.034 pm

Fastest: the CM-frame method (T′ = k/r). Most reliable in an exam: the momentum method plus the limit checks.

Parameter table

SymbolMeaningValueLive (slider)
Talpha kinetic energy0.40 MeV
Z₂, A₂target charge, mass number82, 208 / 3, 7
m₁/m₂mass ratio4/208 / 4/7
kq₁q₂/4πε₀236.2 / 8.64 MeV·fm
T′usable energy, T·m₂/(m₁+m₂)0.2545 MeV (Li)
r_minclosest distance (answer)0.590 pm / 0.034 pm
λde Broglie wavelength (relative)22.7 / 35.7 fm

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