Question

An alpha particle of energy 5 MeV is scattered through 180° by a fixed uranium nucleus. The distance of the closest approach is of the order of-

• A. 10^–15 cm
• B. 10^–13 cm
• C. 10^–12 cm
• D. 10^–19 cm

Answer 1

Answer: (C)

10^–12 cm

Answer 2

r₀ = $\frac{2kZe^{2}}{\frac{1}{2}mv^{2}} = \frac{2 \times 9 \times 10^{9} \times 92 \times (1.6 \times 10^{–19})^{2}}{5 \times 10^{6} \times 1.6 \times 10^{–19}}m$s