Q. $\alpha -$particles of energy $400\ KeV$ are bombarded on nucleus of $\ _{82}Pb$. In scattering of $\alpha -$particles, its minimum distance from nucleus will be
A) $0.59\ nm$
B) $0.59\ \mathring{\mathrm{A}}$
C) $5.9\ pm$
D) $0.59\ pm$
Explanation
(d) Suppose closest distance is $r$, according to conservation of energy
$400 \times 10^{3} \times 1.6 \times 10^{- 19} = 9 \times 10^{9}\frac{(ze)(2e)}{r}$
$\Rightarrow 6.4 \times 10^{- 14} = \frac{9 \times 10^{9} \times (82 \times 1.6 \times 10^{- 19}) \times (2 \times 1.6 \times 10^{- 19})}{r}$
$\Rightarrow r = 5.9 \times 10^{- 13}m = 0.59\ pm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.