Alpha Particle Rebound in Rutherford Scattering Experiment

To explain why only a few $\alpha$-particles rebound back in Rutherford's $\alpha$-particle scattering experiment, we analyze the physics behind the trajectory and scattering of $\alpha$-particles:

1. Theoretical Background

The relationship between the impact parameter $b$ (the perpendicular distance of the initial velocity vector of the $\alpha$-particle from the center of the nucleus) and the scattering angle $\theta$ is given by: $b = \frac{1}{4\pi\varepsilon_0} \frac{Z e \cdot q_\alpha}{m v^2} \cot\left(\frac{\theta}{2}\right)$ where:

• $Z = 79$ is the atomic number of Gold ($\text{Au}$),
• $q_\alpha = +2e$ is the charge of the $\alpha$-particle,
• $K = \frac{1}{2}m v^2$ is the kinetic energy of the incident $\alpha$-particle.

A rebound corresponds to large-angle scattering ($\theta \approx 180^\circ$).

---

2. Evaluation of Statements

• Statement A: The radius of a gold atom is of the order of $10^{-10}\text{ m}$, whereas the radius of its nucleus is of the order of $10^{-14}\text{ m}$. Because the nucleus occupies an extremely small fraction ($\sim 10^{-12}$ to $10^{-15}$) of the atom's volume, most of the space inside the atom is empty. Consequently, only a tiny fraction of $\alpha$-particles pass close enough to the nucleus to experience significant electrostatic repulsion. (Correct)

• Statement B: An $\alpha$-particle carries a charge of $+2e$, whereas a gold nucleus carries a charge of $+79e$. Their charges are not equal. (Incorrect)

• Statement C: For an $\alpha$-particle to rebound ($\theta \to 180^\circ$), we evaluate the impact parameter: $\lim_{\theta \to 180^\circ} \cot\left(\frac{\theta}{2}\right) = \cot(90^\circ) = 0 \implies b \to 0$ Therefore, the impact parameter $b$ must be minimum (nearly zero) for rebounding to occur. Since $b \approx 0$ represents a very narrow cross-sectional area, only a few $\alpha$-particles have a minimum impact parameter. (Correct)

• Statement D: The incident $\alpha$-particles emitted from the radioactive source have uniform kinetic energy ($\approx 5.5\text{ MeV}$). The phenomenon of rebounding is governed by trajectory geometry, not by a fraction of particles having higher kinetic energy. (Incorrect)

• Statement E: Rebounding ($\theta \approx 180^\circ$) occurs when the trajectory of the $\alpha$-particle is aligned almost directly with the center of the target nucleus, which corresponds to a head-on collision ($b \approx 0$). Since the target nucleus is extremely small, only a few $\alpha$-particles undergo head-on collisions. (Correct)

---

Conclusion

Statements A, C, and E correctly explain the observation.

Correct Option: D (A, C, E Only)