E1-3. Rutherford Scattering, Closest Approach, and DeviationsHL ONLY
1. Electrostatic Energetics of Scattering
In a head-on collision during Rutherford scattering, an incoming alpha particle is fired directly along a line connecting its center to the target gold nucleus. As the positive $\alpha$-particle approaches the massive positive target nucleus, it experiences a powerful repulsive Coulomb force that slows it down.
- At an infinite distance away, the alpha particle possesses only kinetic energy: $$E_k = \frac{1}{2}m_{\alpha}v^2$$.
- As it flies closer, its kinetic energy converts into electric potential energy ($E_p$).
- At its point of closest approach ($d$), the alpha particle slows to a stop momentarily, reducing its speed and kinetic energy to zero before reversing direction.
By applying the law of conservation of mechanical energy, we can equate the initial kinetic energy to the peak electric potential energy at distance $d$:
Where the specific components are defined as:
- $Q$ (Charge of incoming $\alpha$-particle): Equal to $+2e$ (since it contains 2 protons).
- $q$ (Charge of target nucleus): Equal to $+Ze$ (where $Z$ is the atomic number of the foil target).
- $k$ (Coulomb Constant): $\frac{1}{4\pi\varepsilon_0} \approx 8.99 \times 10^9\text{ N m}^2\text{ C}^{-2}$.
- $e$ (Elementary Charge): $1.60 \times 10^{-19}\text{ C}$.
Substituting these charge values into our balance gives us a tool to estimate the maximum upper bound of a nuclear radius:
2. Deviations from Pure Coulomb Scattering
Rutherford's classical mathematical model assumed that alpha particles and target nuclei interact only through electrostatic repulsion. This assumption works perfectly for low to moderately high particle energies. However, striking deviations occur when the alpha particle is given an extremely high initial kinetic energy:
- Low to Moderate Energies ($E_k < 25\text{ MeV}$): The alpha particle lacks the energy to overcome the Coulomb barrier. It stops far outside the nucleus, feeling only the electrostatic repulsion. The relationship between scattering angle and particle count matches Rutherford's formula perfectly.
- Extremely High Energies ($E_k > 27.5\text{ MeV}$): If the alpha particle has enough kinetic energy, its distance of closest approach $d$ becomes smaller than the actual nuclear radius ($d < R$). When this happens, the alpha particle penetrates the core boundary and enters the range of the strong nuclear force.
- The strong nuclear force is incredibly powerful but acts only over short distances ($\approx 1\text{--}2\text{ fm}$). It is a purely attractive force between nucleons.
- Once the attractive strong force takes over, it counteracts the electrostatic repulsion. As a result, the number of alpha particles scattered at large backward angles drops sharply to zero compared to the purely classical prediction.
Scattering Profile Deviation at High Energy
Scattering Profile Deviation at High Energy
This threshold energy ($27.5\text{ MeV}$) provides an elegant method to calculate the precise boundary where the strong force takes over, giving physicists an accurate measurement of the actual size of the target nucleus.
3. Examples
Example 1: Classic Head-on Distance of Closest Approach
Problem: An alpha particle ($m_{\alpha} = 6.64 \times 10^{-27}\text{ kg}$) is accelerated to a kinetic energy of $5.00\text{ MeV}$. It travels along a straight line heading directly into a stationary gold nucleus (${}_{79}^{197}\text{Au}$). Assuming the gold nucleus remains stationary and experiences only classical Coulomb interaction, calculate the exact distance of closest approach ($d$).
Solution:
Step 1: Convert the initial kinetic energy from MeV to standard Joules: $$ \begin{aligned} E_k &= 5.00\text{ MeV} = 5.00 \times 10^6\text{ eV} \\ &= 5.00 \times 10^6 \times (1.60 \times 10^{-19}\text{ J}) = 8.00 \times 10^{-13}\text{ J} \end{aligned} $$
Step 2: Determine the nuclear charge variables: The alpha particle has a charge of $Q = 2e$. The gold target nucleus has an atomic number of $Z = 79$, meaning its charge is $q = 79e$.
Step 3: Set up energy conservation and solve for distance $d$: $$E_k = \frac{k(2e)(79e)}{d} = \frac{158 k e^2}{d} \implies d = \frac{158 k e^2}{E_k}$$ Substitute our known physical constants ($k = 8.99 \times 10^9$, $e = 1.60 \times 10^{-19}$): $$ \begin{aligned} d &= \frac{158 \times (8.99 \times 10^9\text{ N m}^2\text{ C}^{-2}) \times (1.60 \times 10^{-19}\text{ C})^2}{8.00 \times 10^{-13}\text{ J}} \\ &= \frac{158 \times (8.99 \times 10^9) \times (2.56 \times 10^{-38})}{8.00 \times 10^{-13}} \\ &= \frac{3.6363 \times 10^{-26}}{8.00 \times 10^{-13}} = \mathbf{4.55 \times 10^{-14}\text{ m}} \quad (\mathbf{45.5\text{ fm}}) \end{aligned} $$ Analysis: Because $45.5\text{ fm}$ is significantly larger than the actual physical radius of a gold nucleus ($\approx 7\text{ fm}$), the alpha particle stops well outside the nuclear core. This confirms that the assumption of pure classical Coulomb scattering is completely valid at this energy.
Example 2: Recoil Correction
Problem: In Example 1, we assumed the gold nucleus remained completely fixed because it is much heavier than an alpha particle. In reality, the gold nucleus recoils slightly due to the electrostatic force, which consumes a small amount of kinetic energy.
Using conservation of momentum and energy, prove that if a light particle of mass $m$ strikes a heavy, free target particle of mass $M$, the threshold distance of closest approach is adjusted by the mass ratio factor. Calculate the true corrected value of $d$ for the collision in Example 1, given $m_{\alpha} = 4.00\text{ u}$ and $M_{\text{gold}} = 197.0\text{ u}$.
Solution:
Step 1: Derivation of Center-of-Mass Energy Correction: Let the incoming alpha particle have mass $m$ and initial speed $v_0$. The initial kinetic energy is $$ E_{k0} = \frac{1}{2}mv_0^2 $$ At the point of closest approach, both particles are moving forward at the exact same velocity ($v_f$) because the relative speed between them drops to zero momentarily. By conservation of linear momentum: $$m v_0 = (m + M) v_f \implies v_f = \frac{m}{m+M} v_0$$ At this instant, the total remaining kinetic energy of the system ($E_{kf}$) is: $$E_{kf} = \frac{1}{2}(m + M)v_f^2 = \frac{1}{2}(m+M)\left(\frac{m v_0}{m+M}\right)^2 = \frac{1}{2}\frac{m^2 v_0^2}{m+M} = E_{k0} \left(\frac{m}{m+M}\right)$$ The remaining kinetic energy cannot be converted into potential energy because it is locked up maintaining system momentum. Therefore, the maximum energy available to convert into electric potential energy ($E_p$) is: $$\Delta E_{\text{available}} = E_{k0} - E_{kf} = E_{k0}\left(1 - \frac{m}{m+M}\right) = E_{k0}\left(\frac{M}{m+M}\right)$$ Equating this available energy to the potential energy at distance $d_{\text{true}}$ gives: $$E_{k0}\left(\frac{M}{m+M}\right) = \frac{2kZe^2}{d_{\text{true}}} \implies d_{\text{true}} = \frac{2kZe^2}{E_{k0}} \left(\frac{m+M}{M}\right) = d_{\text{fixed}} \left(1 + \frac{m}{M}\right)$$
Step 2: Calculate the corrected value using the mass numbers: From Example 1, $d_{\text{fixed}} = 4.545 \times 10^{-14}\text{ m}$. Apply the mass correction factor: $$ \begin{aligned} d_{\text{true}} &= 4.545 \times 10^{-14}\text{ m} \times \left(1 + \frac{4.00}{197.0}\right) \\ &= 4.545 \times 10^{-14} \times (1 + 0.0203) = 4.545 \times 10^{-14} \times 1.0203 \\ &= \mathbf{4.64 \times 10^{-14}\text{ m}} \end{aligned} $$ This university-level correction shows that nuclear recoil causes a small but significant $2.03\%$ increase in the distance of closest approach, preventing the particles from getting quite as close as the simplified fixed-target model predicts.