When an alpha-particle of mass \( M \) moving with velocity \( v \) bombards a heavy nucleus of charge \( Ze \), its distance of closest approach from the nucleus depends on \( M \) as
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \propto \displaystyle\frac{1}{M} \)
At the distance of closest approach the alpha particle stops for an instant, so all its kinetic energy has become electrostatic potential energy (conservation of energy). The alpha particle has charge \( 2e \).
\( \displaystyle \frac{1}{2}Mv^{2} = \frac{1}{4\pi\varepsilon_{0}}\,\frac{(2e)(Ze)}{r_{0}} \)
\( \displaystyle r_{0} = \frac{4Ze^{2}}{4\pi\varepsilon_{0}\,M v^{2}} \;\Rightarrow\; r_{0} \propto \frac{1}{M} \)
The velocity \( v \) is given as fixed, so \( r_{0} \) varies inversely with the mass.