In a Geiger Marsden experiment, an alpha particle of energy 6 MeV hits a nucleus of \( z = 72 \). The distance of closest approach nearly is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 35\,\mathrm{fm} \)
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 \( Ze \)).
\( \displaystyle K = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{r_0} \;\Rightarrow\; r_0 = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^{2}}{K} \)
\( K = 6\,\mathrm{MeV} = 6\times10^{6} \times 1.6\times10^{-19}\,\mathrm{J} = 9.6\times10^{-13}\,\mathrm{J} \)
\( Z = 72 \)
\( \displaystyle r_0 = \frac{9\times10^{9} \times 2 \times 72 \times (1.6\times10^{-19})^{2}}{9.6\times10^{-13}}\,\mathrm{m} \)
\( \displaystyle r_0 = \frac{3.32\times10^{-26}}{9.6\times10^{-13}}\,\mathrm{m} = 3.46\times10^{-14}\,\mathrm{m} \approx 35\,\mathrm{fm} \)
Shortcut: Use \( \displaystyle\frac{e^{2}}{4\pi\varepsilon_0} = 1.44\,\mathrm{MeV\,fm} \): \( r_0 = \displaystyle\frac{2 \times 72 \times 1.44\,\mathrm{MeV\,fm}}{6\,\mathrm{MeV}} = 34.6\,\mathrm{fm} \approx 35\,\mathrm{fm} \).
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 – …
