Correct Option: (a) $0^{\circ}$
Step 1 (Formula / Governing Law): [0.5 Mark]
The relationship between the impact parameter ($b$) and the scattering angle ($\theta$) for an alpha particle scattering by a nucleus is given by the Rutherford scattering formula:
$$b = \frac{1}{4\pi\epsilon_0} \cdot \frac{Ze^2 \cot(\theta/2)}{K.E.}$$Where $Ze$ is the charge of the nucleus, $K.E.$ is the kinetic energy of the alpha particle, and $\theta$ is the scattering angle.
Step 2 (Step-by-Step Mathematical Working): [0.25 Mark]
- From the formula, we observe that $b \propto \cot(\theta/2)$.
- To maximize $b$, we need to maximize the value of $\cot(\theta/2)$.
- As $\theta \to 0^{\circ}$, the value of $\theta/2 \to 0^{\circ}$.
- Since $\cot(0^{\circ})$ is undefined (approaches infinity), the impact parameter $b$ is maximum when the scattering angle $\theta$ is minimum.
- Among the given options, $0^{\circ}$ is the smallest possible scattering angle.
Step 3 (Final Concluded Answer): [0.25 Mark]
The impact parameter is maximum when the alpha particle is aimed far away from the nucleus, resulting in the least amount of deflection. Therefore, the scattering angle is $0^{\circ}$.
💡 EL10 Concept Insight:
Think of the impact parameter as the "aiming distance" from the center of the nucleus; if you aim the alpha particle very far away (large $b$), it barely feels the nucleus's push and continues almost straight, meaning it barely turns at all (scattering angle near $0^{\circ}$).