41  Structure of the Atom

41.1 Syllabus inquiry question

  • How is it known that atoms are made of protons, neutrons, and electrons?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 1:

If all scientific knowledge were to be destroyed and only one sentence passed on to future generations, it should be: “All things are made of atoms—little particles that move around in perpetual motion, attracting each other when they are a little distance apart, but repelling upon being squeezed into one another.”

41.2 Learning Objectives

  • Analyse cathode ray experiments as evidence for electrons
  • Apply Thomson’s charge-to-mass ratio: \(q/m = v/(Br)\)
  • Explain Millikan’s oil drop experiment for quantised charge
  • Describe Rutherford’s nuclear model from scattering evidence
  • Explain Chadwick’s discovery of the neutron

41.3 Content

41.3.1 Cathode Ray Experiments

Key observations that proved cathode rays are particles:

Observation Evidence For
Deflection by electric field Rays are charged
Deflection by magnetic field Rays are charged particles with mass
Same properties regardless of cathode material Rays are fundamental (not atoms)
Cause paddle wheel to rotate Rays have momentum (mass × velocity)
Travel in straight lines Rays are particles, not waves

41.3.2 Interactive: Cathode Ray Tube

41.3.3 Thomson’s Charge-to-Mass Experiment (1897)

Thomson measured the charge-to-mass ratio of electrons by balancing electric and magnetic forces:

\[\frac{q}{m} = \frac{v}{Br}\]

where: - \(q/m\) = charge-to-mass ratio (C/kg) - \(v\) = electron velocity (m/s) - \(B\) = magnetic field strength (T) - \(r\) = radius of circular path (m)

Thomson’s Method:

  1. Velocity selector: Balance electric and magnetic forces so only electrons with \(v = E/B\) pass through undeflected
  2. Magnetic deflection: Remove electric field; electrons curve in a circle
  3. Measure radius: From the curvature, calculate \(r\)
  4. Calculate q/m: Use \(q/m = v/(Br)\)
Thomson’s Result

Thomson found \(q/m = 1.76 \times 10^{11}\) C/kg for cathode rays—about 1800 times larger than for hydrogen ions. This meant electrons were either very light or very highly charged (or both). Later work showed electrons are very light.

41.3.4 Interactive: Thomson’s Experiment

41.3.5 Millikan’s Oil Drop Experiment (1909)

Millikan determined the charge of a single electron by suspending charged oil drops in an electric field:

\[q = \frac{mg}{E}\]

where: - \(q\) = charge on droplet (C) - \(m\) = mass of droplet (kg) - \(g\) = 9.8 m/s² - \(E\) = electric field strength (V/m)

Key Finding: All measured charges were integer multiples of a fundamental unit:

\[e = 1.6 \times 10^{-19}\ \text{C}\]

Quantisation of Charge

Millikan found charges like 3.2 × 10⁻¹⁹ C, 4.8 × 10⁻¹⁹ C, 6.4 × 10⁻¹⁹ C—always multiples of e. This proved charge is quantised in discrete units, not continuous.

41.3.6 Interactive: Millikan Oil Drop

41.3.7 Geiger-Marsden Experiment (1909-1911)

Also called the “gold foil experiment” or “alpha scattering experiment”:

Setup: Alpha particles fired at thin gold foil

Observations:

Result Percentage Implication
Passed straight through ~99% Atom is mostly empty space
Deflected at small angles ~1% Some positive charge present
Scattered back (>90°) ~0.01% Concentrated positive charge

41.3.8 Interactive: Rutherford Scattering

41.3.9 Rutherford’s Nuclear Model (1911)

From the scattering data, Rutherford concluded:

  1. Nucleus: Tiny, dense, positively charged centre (~10⁻¹⁵ m)
  2. Electrons: Orbit at relatively large distances (~10⁻¹⁰ m)
  3. Mostly empty space: Atom is ~100,000× larger than nucleus

Closest approach formula (all kinetic energy → potential energy):

\[r_{min} = \frac{kZe \cdot 2e}{KE}\]

where Z is the atomic number and 2e is the alpha particle charge.

41.3.10 Chadwick’s Discovery of the Neutron (1932)

The Problem: Nuclei were heavier than protons alone could explain. Rutherford predicted “neutral particles” in 1920.

Chadwick’s Experiment: 1. Bombard beryllium with alpha particles 2. Observe highly penetrating radiation that wasn’t deflected by fields 3. This radiation could knock protons from paraffin wax 4. From momentum conservation, Chadwick calculated the mass

\[m_n \approx 1.008665\ \text{u} \approx 1.67 \times 10^{-27}\ \text{kg}\]

Why Paraffin?

Paraffin (wax) is rich in hydrogen. When neutrons collide with protons (similar mass), they transfer maximum momentum—like billiard balls. This made detection possible.

41.3.11 Atomic Model Development

41.4 Worked Examples

41.4.1 Example 1: Thomson’s q/m ratio

An electron travelling at 3.0 × 10⁷ m/s enters a magnetic field of 0.020 T and curves with radius 0.0085 m. Calculate q/m.

Solution:

  1. Use \(q/m = v/(Br)\)

  2. \(q/m = (3.0 \times 10^7)/(0.020 \times 0.0085)\)

  3. \(q/m = 1.76 \times 10^{11}\ \text{C/kg}\)

41.4.2 Example 2: Millikan oil drop

An oil drop of mass 4.9 × 10⁻¹⁵ kg is suspended by an electric field of 1.0 × 10⁵ V/m. Find the charge on the drop.

Solution:

  1. At equilibrium: \(qE = mg\)

  2. \(q = mg/E = (4.9 \times 10^{-15} \times 9.8)/(1.0 \times 10^5)\)

  3. \(q = 4.8 \times 10^{-19}\ \text{C} = 3e\) (three electron charges)

41.4.3 Example 3: Closest approach

An alpha particle with KE = 8.0 × 10⁻¹³ J approaches a gold nucleus (Z = 79). Find the closest approach distance.

Solution:

  1. At closest approach, all KE → electric PE

  2. \(KE = \frac{k(2e)(79e)}{r_{min}}\)

  3. \(r_{min} = \frac{8.99 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{8.0 \times 10^{-13}}\)

  4. \(r_{min} = 4.5 \times 10^{-14}\ \text{m}\)

41.4.4 Example 4: Neutron momentum transfer

A neutron (mass 1.67 × 10⁻²⁷ kg) moving at 2.0 × 10⁷ m/s collides head-on with a stationary proton (same mass). Find the maximum velocity transferred to the proton.

Solution:

  1. For equal masses in head-on elastic collision, velocities exchange

  2. Maximum proton velocity = initial neutron velocity

  3. \(v_{proton} = 2.0 \times 10^7\ \text{m/s}\)

41.4.5 Example 5: Electron mass calculation

Given q/m = 1.76 × 10¹¹ C/kg and e = 1.6 × 10⁻¹⁹ C, calculate the electron mass.

Solution:

  1. \(m = q/(q/m) = e/(q/m)\)

  2. \(m = (1.6 \times 10^{-19})/(1.76 \times 10^{11})\)

  3. \(m = 9.1 \times 10^{-31}\ \text{kg}\)

41.5 Common Misconceptions

Common Misconceptions
  • Misconception: Thomson discovered the electron’s charge and mass separately. Correction: Thomson only measured the ratio q/m. Millikan later measured e separately; then m could be calculated.

  • Misconception: Most alpha particles were deflected in the gold foil experiment. Correction: ~99% passed straight through. Only ~0.01% were scattered back—but this tiny fraction was revolutionary evidence for the nuclear model.

  • Misconception: Rutherford’s model explained all atomic behaviour. Correction: Rutherford’s model couldn’t explain why electrons don’t spiral into the nucleus (accelerating charges should radiate). Bohr’s model addressed this.

  • Misconception: Neutrons were discovered by direct detection. Correction: Neutrons are uncharged and don’t interact with detectors. Chadwick inferred their existence from the momentum they transferred to protons.

  • Misconception: The nucleus was discovered because it’s positively charged. Correction: The nucleus was discovered because of its concentrated mass—the back-scattering required a massive target.

41.6 Practice Questions

41.6.1 Easy (2 marks)

Calculate the charge-to-mass ratio for an electron that curves with radius 0.05 m in a 0.015 T field while travelling at 1.3 × 10⁸ m/s.

  • Use \(q/m = v/(Br)\) (1)
  • \(q/m = (1.3 \times 10^8)/(0.015 \times 0.05) = 1.7 \times 10^{11}\) C/kg (1)

Answer: \(q/m = 1.7 \times 10^{11}\) C/kg

41.6.2 Medium (4 marks)

An oil drop in Millikan’s apparatus has mass 8.2 × 10⁻¹⁵ kg. Calculate the electric field needed to suspend it if it carries 5 electron charges.

  • \(q = 5e = 5 \times 1.6 \times 10^{-19} = 8.0 \times 10^{-19}\) C (1)
  • At equilibrium: \(qE = mg\) (1)
  • \(E = mg/q = (8.2 \times 10^{-15} \times 9.8)/(8.0 \times 10^{-19})\) (1)
  • \(E = 1.0 \times 10^5\) V/m (1)

Answer: E = 1.0 × 10⁵ V/m

41.6.3 Hard (5 marks)

Explain how the Geiger-Marsden experiment led to the nuclear model of the atom, and calculate the closest approach for a 5.0 MeV alpha particle to a gold nucleus (Z = 79).

  • Most alphas passed through → atom mostly empty space (1)
  • Large-angle scattering → small, massive, positive nucleus (1)
  • Convert: 5.0 MeV = 5.0 × 10⁶ × 1.6 × 10⁻¹⁹ = 8.0 × 10⁻¹³ J (1)
  • \(r_{min} = kq_\alpha q_{Au}/KE = (8.99 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2)/(8.0 \times 10^{-13})\) (1)
  • \(r_{min} = 4.5 \times 10^{-14}\) m (1)

Answer: The mostly-undeflected alphas showed empty space; back-scattered alphas proved a small, dense, positive nucleus. \(r_{min} = 4.5 \times 10^{-14}\) m.

41.7 Multiple Choice Questions

Test your understanding with these interactive questions:

41.8 Quick Quiz: Atomic Models

41.9 Extended Response Practice

41.10 Summary

Key Takeaways
  • Cathode rays: deflected by E and B fields → charged particles with mass
  • Thomson: \(q/m = v/(Br) = 1.76 \times 10^{11}\) C/kg
  • Millikan: charge is quantised, \(e = 1.6 \times 10^{-19}\) C
  • Geiger-Marsden: most alphas pass through, few scatter back → nuclear model
  • Rutherford: atom has tiny, dense, positive nucleus
  • Chadwick: neutral particle (neutron) with mass ≈ proton mass

41.11 Self-Assessment

Check your understanding:

After studying this section, you should be able to: