An α-particle of kinetic energy 6 MeV moves head-on towards a heavy nucleus. It comes to rest for a moment at a distance \(d\) from the centre of the nucleus. Assume that the nucleus stays at rest. When the α-particle was at a distance \(3d\) from the centre of the nucleus, its kinetic energy was
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4 MeV
The total mechanical energy of the α-particle and the nucleus stays constant during the motion. Their electric potential energy at a distance \( r \) is \( U=2Ze^{2}/4\pi\varepsilon_0r \), so \( U\propto1/r \). At the distance \( d \), all the kinetic energy has turned into potential energy. So \( U(d)=6\ \mathrm{MeV} \).
\( \begin{aligned} U(3d) &= \frac{d}{3d}\times U(d) \\ &= \frac{1}{3}\times6\ \mathrm{MeV} \\ &= 2\ \mathrm{MeV} \end{aligned} \)
\( \begin{aligned} K &= 6\ \mathrm{MeV}-U(3d) \\ &= 6\ \mathrm{MeV}-2\ \mathrm{MeV} \\ &= 4\ \mathrm{MeV} \end{aligned} \)
Therefore, the α-particle had a kinetic energy of 4 MeV at the distance \( 3d \).
NCERT (2026-27):
Example 12.2 – “… The key idea here is that throughout the scattering process, the total mechanical energy of the system consisting of an α-particle and a gold nucleus is conserved. …”
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”
\( \displaystyle K=\frac{1}{4\pi\varepsilon_0}\frac{(2e)(Ze)}{d}= \)\( \displaystyle \frac{2Ze^{2}}{4\pi\varepsilon_0d} \)