An alpha particle of energy \( \displaystyle\frac{1}{2} \) mv2 bombards a heavy target nucleus of charge Ze. Then, the distance of closest approach for the alpha particle will be proportional to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → v−2
At the distance of closest approach the alpha particle momentarily stops, so all its kinetic energy has become electrostatic potential energy. The alpha particle has charge 2e:
\( \displaystyle \frac{1}{2}mv^{2} = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r_0} \)
\( \displaystyle r_0 = \frac{1}{4\pi\varepsilon_0}\,\frac{4Ze^{2}}{mv^{2}} \)
So \( r_0 \propto v^{-2} \) (and \( r_0 \propto Ze \)).