3.8.1.5
Closest approach gives an upper limit for nuclear radius
Nuclear Structure & Radiation — AQA A-Level Physics
- In Rutherford scattering, alpha particles that rebound at 180° come closest to the nucleus.
- At closest approach r, all kinetic energyThe capacity to do work. Measured in joules (J).The energyThe capacity to do work. Measured in joules (J). an object possesses due to its motion. has been converted to electrostatic potential energyThe capacity to do work. Measured in joules (J)..
- This gives an upper limit for the nuclear radius — the actual radius is smaller.
$$E_k = \frac{Qq}{4\pi\varepsilon_0 r}$$
- $Eₖ$: initial kinetic energyThe energy an object possesses due to its motion. of the alpha particle (J)
- $Q$: chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). of the alpha particle = 2e (C)
- $q$: chargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). of the target nucleus = Ze (C)
- $ε₀$: permittivity of free space (F m⁻¹)
- $r$: distance of closest approach (m)
- Rearranging: $r = Qq / (4\pi\varepsilon_{0} Eₖ). For an \alpha particle hitting gold (Z = 79)$ with 5 MeV, r ≈ 45 fm. The actual gold nucleus radius is about 6.6 fm.
- Advantages: simple mathematics, gives a good upper limit estimate.
- Disadvantages: always overestimates radius, cannot account for the strong nuclear force at very close range, alpha particles have finite size, the target nucleus recoils.
Worked Example
An alpha particle is accelerated to 2.55 × 10⁷ m s⁻¹ and aimed at an aluminium-27 nucleus (Z = 13). Calculate the distance of closest approach.
Show Solution
1
List known values
- Mass of alpha: $m = 4 \times 1.66 \times 10^{-27} = 6.64 \times 10^{-27}$ kg
- Speed: $v = 2.55 \times 10^{7}$ m s⁻¹
- ChargeA property of matter that causes it to experience a force in an electromagnetic field. Measured in coulombs (C). of alpha: $Q = 2 \times 1.6 \times 10^{-19}$ C
- Charge of Al nucleus: $q = 13 \times 1.6 \times 10^{-19}$ C
- $\varepsilon_0 = 8.85 \times 10^{-12}$ F m⁻¹
2
Set kinetic energyThe energy an object possesses due to its motion. equal to potential energy
$$\frac{1}{2}mv^2 = \frac{Qq}{4\pi\varepsilon_0 r}$$
3
Rearrange for r and substitute
$$r = \frac{Ze^2}{\pi\varepsilon_0 mv^2} = \frac{13 \times (1.6 \times 10^{-19})^2}{\pi \times 8.85 \times 10^{-12} \times 6.64 \times 10^{-27} \times (2.55 \times 10^{7})^2}$$
4
Calculate
$$r = 2.77 \times 10^{-15} \text{ m}$$
Answer
$r = 2.77 \times 10^{-15}$ m