Q12.2:

Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?

1. Theoretical Foundation: Rutherford Alpha-Scattering Model

The Rutherford scattering formula for the differential cross-section, which describes the probability of an $\alpha$-particle ($Z_2 = 2$) being scattered at an angle $\theta$ by a target nucleus with atomic number $Z_1$, is given by: $$\frac{d\sigma}{d\Omega} = \left( \frac{1}{4\pi\epsilon_0} \right)^2 \left( \frac{Z_1 Z_2 e^2}{2mv^2 \sin^2(\theta/2)} \right)^2$$ [Per Coulomb's Law and the conservation of energy and momentum in a central potential field].

2. Comparative Parameters: Gold vs. Hydrogen

To analyze the expected results, we compare the scattering center properties of a Gold nucleus ($Au$) and a Hydrogen nucleus ($H$):

3. Step-by-Step Analysis of Expected Phenomena

Step 1: Scattering Probability (Cross-Section)

Since the scattering cross-section is proportional to $Z_1^2$, the probability of scattering for hydrogen ($Z_1 = 1$) compared to gold ($Z_1 = 79$) is reduced by a factor of: $$\frac{\sigma_H}{\sigma_{Au}} = \left( \frac{1}{79} \right)^2 \approx \frac{1}{6241}$$ [Per the Rutherford Scattering Cross-Section proportionality]. Consequently, most $\alpha$-particles will pass through the hydrogen sheet with negligible deviation.

Step 2: Recoil Effects and Conservation Laws

In the original experiment, the Gold nucleus ($M \approx 197m_\alpha$) is nearly stationary due to its large mass. In the case of Hydrogen ($M \approx 0.25m_\alpha$), the conservation of momentum and energy dictates: $$m_\alpha v_i = m_\alpha v_f \cos\theta + M V \cos\phi$$ $$0 = m_\alpha v_f \sin\theta - M V \sin\phi$$ Because the target mass $M$ is smaller than the projectile mass $m_\alpha$, the hydrogen nucleus will undergo significant recoil. This is a head-on collision scenario where energy transfer to the nucleus is maximized, significantly altering the kinematics of the $\alpha$-particle.

Step 3: Nuclear Size and Multiple Scattering

Because the hydrogen nucleus is merely a single proton, the impact parameter $b$ required for large-angle deflection is significantly lower. Furthermore, the electron cloud of the hydrogen atom will play a more significant role in energy loss (ionization) relative to the nucleus itself compared to the high-$Z$ gold atom.

4. Visualization of Scattering Dynamics

Summary of Expected Results

• Reduced Scattering Frequency: The number of $\alpha$-particles scattered at large angles will decrease by a factor of roughly $6241$ due to the lower nuclear charge of the hydrogen target.
• Kinematic Recoil: Unlike the gold foil where the target remains stationary, the light hydrogen nucleus will recoil significantly upon collision, leading to different energy spectra for the outgoing $\alpha$-particles.
• Ionization Dominance: Due to the low mass density and atomic structure, the energy loss per unit path length will be dominated by interactions with the electron cloud rather than purely nuclear scattering.

Final Solution: The experiment would show a drastic reduction in wide-angle scattering events (by a factor of $\approx 6200$) and significant momentum transfer to the hydrogen nuclei, causing them to recoil, unlike the stationary behavior observed with gold.