The impact parameter for an alpha-particle approaching a target nucleus is minimum when the scattering angle \( \theta \) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( \pi\,\mathrm{rad} \)
In Rutherford scattering the impact parameter \( b \) is the perpendicular distance of the initial line of motion of the alpha particle from the nucleus. It is related to the scattering angle by
\( \displaystyle b = \frac{Z e^{2}\cot(\theta/2)}{4\pi\varepsilon_0\,K} \)
where \( K \) is the kinetic energy of the alpha particle. The smallest possible value is \( b = 0 \), which needs \( \cot(\theta/2) = 0 \), that is \( \theta/2 = \pi/2 \).
\( \displaystyle \theta = \pi\,\mathrm{rad} \)
A head-on collision (zero impact parameter) sends the alpha particle straight back, so the scattering angle is \( \pi \).
NCERT (2026-27):
Alpha-Particle Trajectory – “The trajectory traced by an α-particle depends on the impact parameter, b of collision. The impact parameter is the perpendicular distance of the initial velocity vector of the α-particle from the centre of the nucleus (Fig. 12.4).”
Alpha-Particle Trajectory – “… It is seen that an α-particle close to the nucleus (small impact parameter) suffers large scattering. In case of head-on collision, the impact parameter is minimum and the α-particle rebounds back (θ ≅ π).”