Question

An α-particle of energy 10 MeV is scattered through 180^o by a fixed uranium nucleus. Calculate the order of distance of the closest approach?

(a) 10^-20cm

(b) 10^-12cm

(c) 10^-11cm

(d) 10^12cm

This question was posed to me in class test.

This intriguing question comes from Alpha-Particle Scattering and Rutherford’s Nuclear Model of Atom in section Atoms of Physics – Class 12

Answer 1

Correct answer is (b) 10^-12cm

To elaborate: r0 = \(\frac {(2Ze^2)}{(4\pi \varepsilon_0 (\frac {1}{2}m v^2))}\)

r0 = \(\frac {9 \times 10^9 \times 2 \times 92 \times (1.6 \times 10^{-19})^2}{10 \times 1 \times 10^{-13}}\) J

r0 = 4.239 × 10^-14 m

r0 = 4.2 × 10^-12 cm.