Q. An $\alpha$-particle of $5\ MeV$ energy strikes with a nucleus of uranium at stationary at an scattering angle of $180{^\circ}$. The nearest distance upto which $\alpha$-particle reaches the nucleus will be of the order of
A) $1\ \mathring{\mathrm{A}}$
B) $10^{- 10}cm$
C) $10^{- 12}cm$
D) $10^{- 15}cm$
Explanation
(c) At closest distance of approach
Kinetic energy = Potential energy
$\Rightarrow 5 \times 10^{6} \times 1.6 \times 10^{- 19} = \frac{1}{4\pi\varepsilon_{0}} \times \frac{(ze)(2e)}{r}$
For uranium $z = 92$, so $r = 5.3 \times 10^{- 12}cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.