In Geiger Marsden alpha scattering experiments, a beam of alpha particles emitted from \( {}^{214}_{83}\mathrm{Bi} \) were directed to strike a gold foil and various observations were made. Now answer the following questions.
An alpha particle of energy 5 MeV is scattered through \( 180^\circ \) by a fixed Uranium Nucleus. The distance of closest approach is of the order of :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10^{-12}\,\mathrm{cm} \)
An alpha particle scattered through \( 180^\circ \) has made a head-on collision. It stops momentarily at the distance of closest approach \( d \), where all its kinetic energy has become electrostatic potential energy (energy conservation).
\( \displaystyle K = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{d} \;\Rightarrow\; d = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^{2}}{K} \)
\( Z = 92 \text{ (uranium)} \)
\( e = 1.6\times10^{-19}\,\mathrm{C} \)
\( 1/4\pi\varepsilon_0 = 9\times10^{9}\,\mathrm{N\,m^2\,C^{-2}} \)
\( \displaystyle K = 5\,\mathrm{MeV} = 5\times10^{6} \times 1.6\times10^{-19}\,\mathrm{J} = 8\times10^{-13}\,\mathrm{J} \)
\( \displaystyle d = \frac{9\times10^{9} \times 2 \times 92 \times (1.6\times10^{-19})^{2}}{8\times10^{-13}} = 5.3\times10^{-14}\,\mathrm{m} \)
\( \displaystyle d = 5.3\times10^{-14}\,\mathrm{m} = 5.3\times10^{-12}\,\mathrm{cm} \)
So the distance of closest approach is of the order of \( 10^{-12}\,\mathrm{cm} \).
Shortcut: Use \( e^{2}/4\pi\varepsilon_0 = 1.44\,\mathrm{MeV\,fm} \): \( d = 2 \times 92 \times 1.44/5 \approx 53\,\mathrm{fm} = 5.3\times10^{-14}\,\mathrm{m} = 5.3\times10^{-12}\,\mathrm{cm} \).
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 – …
