An α-particle moves head-on towards a gold nucleus (\(Z=79\)). It comes to rest for a moment at a distance of \(3.0\times10^{-14}\ \mathrm{m}\) from the centre of the nucleus. The electrostatic force on the α-particle at this point is about (Take \(1/4\pi\varepsilon_0=9\times10^{9}\ \mathrm{N\,m^{2}\,C^{-2}}\) and \(e=1.6\times10^{-19}\ \mathrm{C}\).)
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(40\ \mathrm{N}\)
The nucleus and the α-particle repel each other by Coulomb’s law. The α-particle has a charge \( 2e \), and the gold nucleus has a charge \( Ze \).
\( \begin{aligned} F &= \frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r^{2}} \\ &= \frac{9\times10^{9}\times2\times79\times(1.6\times10^{-19})^{2}}{(3.0\times10^{-14})^{2}} \\ &= \frac{3.64\times10^{-26}}{9.0\times10^{-28}} \\ &= 40\ \mathrm{N} \end{aligned} \)
This is a very large force for a particle of such small mass. It is strong enough to turn the α-particle back along its path.
NCERT (2026-27):
Alpha-particle Scattering and Rutherford’s Nuclear Model of Atom – “The magnitude of this force is”
\( \displaystyle F=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{r^{2}} \)\( \qquad (12.1) \)
Alpha-particle Scattering and Rutherford’s Nuclear Model of Atom – “where r is the distance between the α-particle and the nucleus. …”