Nuclear & Particle Physics - Alpha/Rutherford Scattering

When alpha particles are directed onto atoms in a thin metal foil, some make very close collisions with the nuclei of the atoms and are scattered at large angles. If an alpha particles with an initial kinetic energy of 5 MeV happens to be scattered through an angle of 180^o, which of the following must have been its distance of the closest approach to the scattering nucleus? (Assume that the metal foil is made of silver, with Z = 50.)

A. 1.22 × 50^1/3 fm
B. 2.9 × 10⁻¹⁴ m
C. 1.0 × 10⁻¹² m
D. 3.0 × 10⁻⁸ m
E. 1.7 × 10⁻⁷ m

(GR9677 #19)

Solution:

"When alpha particles are directed onto atoms in a thin metal foil, some make very close collisions with the nuclei of the atoms and are scattered at large angles."→ Rutherford Scattering: the discovery of nucleus.

Rutherford estimated the radius of a silver nucleus to be 2 × 10⁻¹⁴ m, by observing the angular dependence of alpha-particle scattering (source).

Answer: B

Calculation:

Conservation of Energy: U = T

U = kq_α q_s / r = T
r =  kq_α q_s / T

Given:
T = 5 MeV = 5 × 10⁶ eV
Z_s = 50 → q_s = Z_αe = 50e
Z_α = 2 → q_α = Z_αe = 2e 

k = 1/4πɛ₀ = 1/(4 × 3.14 × 8.85 × 10⁻¹²)  ≈ 10¹⁰  Nm²/C²

r = (10¹⁰ Nm²/C²× 50e × 2e) / (5 × 10⁶ eV)
= 2 × 10⁵ e Nm²/VC² 

With e = 1.60 × 10⁻¹⁹ C

1 Volt = 1 Nm/C

r = 2 × 10⁵ × 1.60 × 10⁻¹⁹ m ≈ 3 × 10⁻¹⁴ m