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.
In Geiger Marsden experiments, If an alpha particle of energy 5 MeV hits a nucleus of \( z = 80 \), the distance of closest approach is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 46 fm
At the distance of closest approach the alpha particle (charge \( 2e \)) stops for an instant, so its whole kinetic energy has turned into electrostatic potential energy with the nucleus (charge \( Ze \)).
\( \displaystyle K = \frac{1}{4\pi\varepsilon_0}\,\frac{(2e)(Ze)}{d} \;\Rightarrow\; d = \frac{1}{4\pi\varepsilon_0}\,\frac{2Ze^{2}}{K} \)
\( Z = 80 \)
\( 1/4\pi\varepsilon_0 = 9\times10^{9}\,\mathrm{N\,m^2\,C^{-2}} \)
\( e = 1.6\times10^{-19}\,\mathrm{C} \)
\( \displaystyle K = 5\,\mathrm{MeV} = 5\times10^{6} \times 1.6\times10^{-19}\,\mathrm{J} = 8\times10^{-13}\,\mathrm{J} \)
\( \displaystyle d = \frac{9\times10^{9} \times 2 \times 80 \times (1.6\times10^{-19})^{2}}{8\times10^{-13}} = 4.6\times10^{-14}\,\mathrm{m} \)
\( \displaystyle d = 46\times10^{-15}\,\mathrm{m} = 46\,\mathrm{fm} \)
Shortcut: With \( e^{2}/4\pi\varepsilon_0 = 1.44\,\mathrm{MeV\,fm} \) and \( K \) in MeV, \( d = 2Z \times 1.44/K = 2 \times 80 \times 1.44/5 = 46\,\mathrm{fm} \).
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 – …
