Question

The maximum KE found in a particles of natural origin is 7.7 MeV. Then, if these are used in alpha scattering experiment, the distance of closest approach from gold nucleus is about –

• A. 1.5 × 10^–15 m
• B. 3 × 10^–14m
• C. 41 × 10^–15m
• D. 3 × 10^–16m

Answer 1

Answer: (B)

3 × 10^–14m

Answer 2

K.E_a = 7.7 MeV

d_c=$\frac{K(79e)(2e)}{K.E_{\alpha}}$=$\frac{9 \times 10^{9} \times 79 \times 2 \times e^{2}}{7.7 \times 10^{6} \times e}$

=$\frac{9 \times 79 \times 2 \times 10^{9} \times 1.6 \times 10^{- 19}}{7.7 \times 10^{6}}$

=$\frac{9 \times 79 \times 2 \times 1.6}{7.7}$× 10^–16 m = 3 × 10^–14 m