An α-particle of kinetic energy 7.7 MeV moves head-on towards a gold nucleus. Its distance of closest approach is 30 fm. A proton of the same kinetic energy now moves head-on towards a gold nucleus. The distance of closest approach of the proton is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 15 fm
At the distance of closest approach, the particle stops for a moment. All its kinetic energy is then stored as electric potential energy of the particle and the nucleus. For a particle of charge \( q \) and a nucleus of charge \( Ze \),
\( \begin{aligned} K &= \frac{1}{4\pi\varepsilon_0}\frac{qZe}{d} \\ d &= \frac{qZe}{4\pi\varepsilon_0K} \end{aligned} \)
Here \( K \) and \( Z \) are the same for both particles, so \( d\propto q \). The proton has charge \( e \). The α-particle has charge \( 2e \).
\( \begin{aligned} d_p &= \frac{e}{2e}\times d_\alpha \\ &= \frac{1}{2}\times30\ \mathrm{fm} \\ &= 15\ \mathrm{fm} \end{aligned} \)
Hence, the proton stops at 15 fm from the gold nucleus. This is half the distance for the α-particle.
NCERT (2026-27):
Example 12.2 – “Let d be the centre-to-centre distance between the α-particle and the gold nucleus when the α-particle is at its stopping point. Then we can write the conservation of energy Ei = Ef as”
\( \displaystyle K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d} \)\( \displaystyle =\frac{2Ze^{2}}{4\pi\varepsilon_0d} \)