Determine the distance of closest approach when an alpha particle of kinetic energy \( 4.5\,\mathrm{MeV} \) strikes a nucleus of \( Z=80 \), stops and reverses its direction
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 5.12\times10^{-14}\,\mathrm{m} \)
At the distance of closest approach the alpha particle momentarily stops, so all its kinetic energy has become electrostatic potential energy of the alpha particle (charge 2e) and the nucleus (charge Ze):
\( \displaystyle K = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r_0} \Rightarrow r_0 = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^{2}}{K} \)
\( K = 4.5\,\mathrm{MeV} = 4.5\times10^{6}\times1.6\times10^{-19}\,\mathrm{J} \)
\( Z = 80 \)
\( 1/4\pi\varepsilon_0 = 9\times10^{9}\,\mathrm{N\,m^{2}C^{-2}} \)
One factor of e cancels between the numerator and K:
\( \displaystyle r_0 = \frac{9\times10^{9}\times2\times80\times1.6\times10^{-19}}{4.5\times10^{6}} \)
\( \displaystyle r_0 = \frac{2.304\times10^{-7}}{4.5\times10^{6}} = 5.12\times10^{-14}\,\mathrm{m} \)
Note: NTA picked this question from an NCERT (2026-27) example and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Alpha-Particle Trajectory – …
