Atomic Structure: Distance of Closest Approach and Rutherford's Model

Distance of Closest Approach

• Definition: The minimum distance between an alpha particle and a nucleus when the alpha particle is projected directly towards the nucleus.
• Alpha particles are bombarded at an atom during experiments.
  • Most go straight.
  • Some deviate slightly.
  • Some deviate greatly.
  • One goes straight towards the nucleus and deviates 180 degrees.
• The alpha particle going straight towards the nucleus slows down until it stops and returns.
• The distance at which it stops is the distance of closest approach.

Finding the Distance of Closest Approach

• Principle used: Energy conservation.
• Initial state: Alpha particles start from a large distance (approximated as infinity).
• Interaction: As the alpha particle approaches the nucleus (both positively charged), it experiences repulsion, causing it to slow down.
• At infinity:
  • Kinetic energy is maximum: $\frac{1}{2}mv^2$
  • Potential energy is zero.
• At the closest point:
  • Kinetic energy is zero (velocity is zero).
  • Potential energy is maximum.

Energy Conservation Equation

• Kinetic energy + Potential energy at infinity = Kinetic energy + Potential energy at closest distance.
• $KE<em>{\infty} + PE</em>{\infty} = KE<em>{closest} + PE</em>{closest}$
• At infinity: $KE<em>{\infty} = \frac{1}{2}mv^2$, $PE</em>{\infty} = 0$
• At closest approach: $KE<em>{closest} = 0$, $PE</em>{closest} = k \frac{q<em>1 q</em>2}{r}$
  • Where $k = \frac{1}{4 \pi \epsilon_0}$ (electrostatic constant).
• Charge on alpha particle ($q_1$) = +2e (2 times the elementary charge).
• Charge inside the nucleus ($q_2$) = Ze (Z is the atomic number, i.e., number of protons).
• Therefore, $\frac{1}{2}mv^2 = k \frac{(2e)(Ze)}{r<em>0}$, where $r</em>0$ is the distance of closest approach.
• $r<em>0 = \frac{1}{4 \pi \epsilon</em>0} \frac{4Ze^2}{mv^2}$

Numerical Example 1

• Problem: Calculate the distance of closest approach for a proton with energy 3 MeV approaching a gold nucleus (Z = 79).
• Note: The question uses a proton instead of an alpha particle.
  • Charge on proton = +e.
  • Mass of proton = $1.67 \times 10^{-27}$ kg.
• Convert MeV to Joules:
  • 3 MeV = $3 \times 10^6$ eV.
  • 1 eV = $1.6 \times 10^{-19}$ Joules.
  • 3 MeV = $3 \times 10^6 \times 1.6 \times 10^{-19}$ Joules.
• For gold, Z = 79 (79 protons in the nucleus).
• Energy conservation equation:
  • Initial kinetic energy = Final potential energy.
  • $KE = \frac{1}{4 \pi \epsilon<em>0} \frac{q</em>1 q<em>2}{r</em>0}$
  • $KE = \frac{1}{4 \pi \epsilon<em>0} \frac{(Ze)(e)}{r</em>0}$
  • $r<em>0 = \frac{1}{4 \pi \epsilon</em>0} \frac{Ze^2}{KE}$
  • Given: $KE = 4.8 \times 10^{-13} J$, $Z = 79$, and $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 Nm^2/C^2$
  • $r_0 = \frac{(9 \times 10^9)(79)(1.6 \times 10^{-19})^2}{4.8 \times 10^{-13}} \approx 3.8 \times 10^{-14} m$

Numerical Example 2

• Problem: Calculate the energy of an alpha particle whose distance of closest approach to a gold nucleus is 29.5 Fermi.
• 1 Fermi (fm) = $10^{-15}$ meters.
• Given: $r_0 = 29.5 \times 10^{-15} m$, Z for gold = 79.
• Energy conservation:
  • Initial kinetic energy = Final potential energy.
  • $KE = \frac{1}{4 \pi \epsilon<em>0} \frac{q</em>1 q<em>2}{r</em>0}$
  • $KE = \frac{1}{4 \pi \epsilon<em>0} \frac{(2e)(Ze)}{r</em>0}$
  • $KE = \frac{(9 \times 10^9)(2 \times 1.6 \times 10^{-19})(79 \times 1.6 \times 10^{-19})}{29.5 \times 10^{-15}}$
  • $KE \approx 12.34 \times 10^{-13} J$
• Converting Joules to MeV:
  • Divide by $1.6 \times 10^{-13}$
  • $KE \approx 7.7 MeV$

Conversions

• Joules to eV: Divide by $1.6 \times 10^{-19}$.
• eV to MeV: Divide by $10^6$.
• Joules to MeV: Divide by $1.6 \times 10^{-13}$.

Failures of Rutherford Model

• Based on classical electromagnetic theory:
  • A charged particle in accelerated motion must radiate energy.
  • Electrons orbiting the nucleus undergo centripetal acceleration.
  • Therefore, electrons should continuously lose energy.
  • As they lose energy, they should spiral into the nucleus, causing the atom to collapse.
  • This does not happen in reality; atoms are stable.
• Rutherford's model does not specify the orbits in which electrons can revolve.
  • Rutherford stated electrons can revolve in any orbit.
  • Later, Bohr proposed that electrons can only revolve in specific, fixed orbits.
• Line emission spectra:
  • If electrons could revolve in any orbit, they should emit radiation of all frequencies (continuous spectrum).
  • However, experiments showed that atoms emit radiation only at specific frequencies (line spectrum).

Summary of Failures

1. Electrons should spiral into the nucleus due to energy loss from accelerated motion, but atoms are stable.
2. Rutherford's model allows electrons to revolve in any orbit, contradicting the observed line emission spectra, which indicate fixed orbits.