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 180o, 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 × 501/3 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)
"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 = kqα qs / r = T
r = kqα qs / T
Given:
T = 5 MeV = 5 × 106 eV
Zs = 50 → qs = Zαe = 50e
Zα = 2 → qα = Zαe = 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
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