42  Quantum Mechanical Nature of the Atom

42.1 Syllabus inquiry question

  • How is it known that classical physics cannot explain the atom?
Feynman Insight

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

Any experiment that can distinguish between indistinguishable paths will destroy the interference pattern. The electron somehow “knows” whether both slits are open. This is the central mystery of quantum mechanics—and we cannot explain it classically.

42.2 Learning Objectives

  • Assess limitations of the Rutherford and Bohr atomic models
  • Apply the Rydberg equation for hydrogen spectral lines
  • Calculate photon energy: \(E = hf = hc/\lambda\)
  • Apply de Broglie wavelength: \(\lambda = h/(mv)\)
  • Analyse Schrödinger’s contribution to atomic theory

42.3 Content

42.3.1 Limitations of Rutherford’s Model

Classical Problem: Accelerating charges radiate energy (Maxwell’s equations)

An orbiting electron is constantly accelerating (centripetal). According to classical physics:

  1. Electron radiates electromagnetic energy
  2. Loses energy → spirals inward
  3. Continuous spectrum emitted (not discrete lines)
  4. Atom collapses in ~10⁻¹¹ seconds

Reality: Atoms are stable and emit discrete spectral lines!

The Classical Catastrophe

Classical physics predicted atoms would collapse almost instantly. The fact that matter is stable proves classical physics is incomplete at atomic scales.

42.3.2 Bohr’s Model (1913)

Bohr’s postulates saved the atom:

  1. Stationary states: Electrons can only orbit at specific radii without radiating
  2. Quantised angular momentum: \(L = mvr = n\hbar\) where \(\hbar = h/(2\pi)\)
  3. Photon emission/absorption: Transitions between states emit/absorb photons with \(E = hf\)

42.3.3 Interactive: Bohr Energy Levels

42.3.4 Energy Levels of Hydrogen

The energy of the nth level:

\[E_n = \frac{-13.6}{n^2}\ \text{eV}\]

Level n Energy (eV)
Ground state 1 -13.6
First excited 2 -3.4
Second excited 3 -1.51
Third excited 4 -0.85
Ionised 0
Negative Energy

Negative energy means the electron is bound. Zero energy means just free. The 13.6 eV ionisation energy is needed to remove the electron completely.

42.3.5 The Balmer Series

Visible hydrogen lines from transitions to n = 2:

\[\frac{1}{\lambda} = R_H \left[\frac{1}{n_f^2} - \frac{1}{n_i^2}\right]\]

where \(R_H = 1.097 \times 10^7\ \text{m}^{-1}\) (Rydberg constant)

Transition Wavelength Colour
3 → 2 (Hα) 656 nm Red
4 → 2 (Hβ) 486 nm Blue-green
5 → 2 (Hγ) 434 nm Violet
6 → 2 (Hδ) 410 nm Deep violet

42.3.6 Interactive: Hydrogen Emission Spectrum

42.3.7 Other Spectral Series

Series Final Level Wavelength Region
Lyman n = 1 Ultraviolet
Balmer n = 2 Visible
Paschen n = 3 Infrared
Brackett n = 4 Far infrared

42.3.8 Photon Energy and Wavelength

When an electron transitions between levels:

\[E_{photon} = E_i - E_f = hf = \frac{hc}{\lambda}\]

where: - \(h = 6.63 \times 10^{-34}\) J·s (Planck’s constant) - \(f\) = frequency (Hz) - \(\lambda\) = wavelength (m) - \(c = 3.0 \times 10^8\) m/s

42.3.9 Interactive: Energy Level Transitions

42.3.10 Limitations of Bohr’s Model

While successful for hydrogen, Bohr’s model fails for:

Limitation Description
Multi-electron atoms Can’t predict spectra of helium or heavier elements
Fine structure Can’t explain splitting of spectral lines in magnetic fields
Chemical bonding No explanation for molecular formation
Intensity Can’t predict relative brightness of spectral lines
Ad hoc Quantisation postulated without deeper explanation

42.3.11 de Broglie’s Matter Waves (1924)

Louis de Broglie proposed that particles have wave properties:

\[\lambda = \frac{h}{mv} = \frac{h}{p}\]

where: - \(\lambda\) = de Broglie wavelength (m) - \(h = 6.63 \times 10^{-34}\) J·s - \(m\) = mass (kg) - \(v\) = velocity (m/s) - \(p\) = momentum (kg·m/s)

Wave-Particle Duality

All matter exhibits wave properties, but the wavelength is only significant for very small masses. An electron at 10⁶ m/s has λ ≈ 7 × 10⁻¹⁰ m (atomic scale). A cricket ball has λ ≈ 10⁻³⁴ m (undetectable).

42.3.12 de Broglie Explains Bohr’s Quantisation

The allowed orbits correspond to standing waves around the nucleus:

\[2\pi r = n\lambda\]

Only integer numbers of wavelengths fit around the orbit, explaining why only certain radii are allowed.

42.3.13 Interactive: de Broglie Standing Waves

42.3.14 Schrödinger’s Wave Mechanics (1926)

Erwin Schrödinger developed a wave equation describing electron behaviour:

Key Concepts:

  1. Wave function (\(\psi\)): Mathematical description of quantum state
  2. Probability density (\(|\psi|^2\)): Probability of finding electron at a location
  3. Orbitals: 3D probability distributions replace definite orbits
  4. Quantum numbers: n, l, m, s describe the state completely
Orbitals vs Orbits

Electrons don’t follow definite paths (orbits). Instead, they exist as probability clouds (orbitals). The “1s orbital” is a spherical cloud where you’re most likely to find the electron.

42.3.15 Interactive: Probability Orbitals

42.3.16 Heisenberg’s Uncertainty Principle

You cannot simultaneously know both position and momentum precisely:

\[\Delta x \cdot \Delta p \geq \frac{h}{4\pi}\]

This is a fundamental limit of nature, not a limitation of measurement technology.

42.4 Worked Examples

42.4.1 Example 1: Rydberg equation

Calculate the wavelength of the Hα line (n = 3 → n = 2).

Solution:

  1. Use \(1/\lambda = R_H[1/n_f^2 - 1/n_i^2]\)

  2. \(1/\lambda = 1.097 \times 10^7 [1/4 - 1/9]\)

  3. \(1/\lambda = 1.097 \times 10^7 \times 0.139 = 1.52 \times 10^6\)

  4. \(\lambda = 6.56 \times 10^{-7}\ \text{m} = 656\ \text{nm}\) (red)

42.4.2 Example 2: Photon energy from transition

Calculate the energy of a photon emitted when hydrogen transitions from n = 4 to n = 2.

Solution:

  1. Energy levels: \(E_4 = -13.6/16 = -0.85\) eV, \(E_2 = -13.6/4 = -3.4\) eV

  2. Energy difference: \(\Delta E = E_4 - E_2 = -0.85 - (-3.4) = 2.55\) eV

  3. Convert: \(E = 2.55 \times 1.6 \times 10^{-19} = 4.08 \times 10^{-19}\) J

42.4.3 Example 3: de Broglie wavelength of electron

Calculate the de Broglie wavelength of an electron travelling at 2.0 × 10⁶ m/s.

Solution:

  1. Use \(\lambda = h/(mv)\)

  2. \(\lambda = (6.63 \times 10^{-34})/(9.11 \times 10^{-31} \times 2.0 \times 10^6)\)

  3. \(\lambda = 3.6 \times 10^{-10}\ \text{m} = 0.36\ \text{nm}\)

42.4.4 Example 4: Wavelength from photon energy

A photon has energy 3.0 × 10⁻¹⁹ J. Find its wavelength.

Solution:

  1. Use \(E = hc/\lambda\), so \(\lambda = hc/E\)

  2. \(\lambda = (6.63 \times 10^{-34} \times 3.0 \times 10^8)/(3.0 \times 10^{-19})\)

  3. \(\lambda = 6.63 \times 10^{-7}\ \text{m} = 663\ \text{nm}\)

42.4.5 Example 5: de Broglie wavelength comparison

Compare the de Broglie wavelength of (a) an electron and (b) a proton, both travelling at 1.0 × 10⁶ m/s.

Solution:

  1. Electron: \(\lambda_e = h/(m_e v) = (6.63 \times 10^{-34})/(9.11 \times 10^{-31} \times 10^6) = 7.3 \times 10^{-10}\) m

  2. Proton: \(\lambda_p = h/(m_p v) = (6.63 \times 10^{-34})/(1.67 \times 10^{-27} \times 10^6) = 4.0 \times 10^{-13}\) m

  3. Ratio: \(\lambda_e/\lambda_p = m_p/m_e \approx 1836\)

The electron wavelength is ~1800 times longer (more wave-like).

42.5 Common Misconceptions

Common Misconceptions
  • Misconception: Electrons orbit the nucleus like planets orbit the Sun. Correction: In quantum mechanics, electrons exist as probability clouds. There’s no definite trajectory.

  • Misconception: The de Broglie wavelength only applies to electrons. Correction: All matter has a de Broglie wavelength. It’s just negligible for macroscopic objects.

  • Misconception: Bohr’s model is correct for all atoms. Correction: Bohr’s model only works for hydrogen (one electron). Multi-electron atoms require quantum mechanics.

  • Misconception: Uncertainty is due to poor measurement techniques. Correction: Heisenberg’s uncertainty is a fundamental property of nature, not a measurement limitation.

  • Misconception: Higher n means higher energy means faster electron. Correction: Higher n actually means the electron is further from the nucleus and moving more slowly. The energy is higher because it’s less bound (less negative).

42.6 Practice Questions

42.6.1 Easy (2 marks)

Calculate the de Broglie wavelength of an electron travelling at 5.0 × 10⁶ m/s.

  • Use \(\lambda = h/(mv)\) (1)
  • \(\lambda = (6.63 \times 10^{-34})/(9.11 \times 10^{-31} \times 5.0 \times 10^6) = 1.5 \times 10^{-10}\) m (1)

Answer: λ = 1.5 × 10⁻¹⁰ m = 0.15 nm

42.6.2 Medium (4 marks)

Use the Rydberg equation to calculate the wavelength of light emitted when hydrogen transitions from n = 5 to n = 2. Identify which spectral series this belongs to.

  • \(1/\lambda = R_H[1/n_f^2 - 1/n_i^2] = 1.097 \times 10^7[1/4 - 1/25]\) (1)
  • \(1/\lambda = 1.097 \times 10^7 \times 0.21 = 2.30 \times 10^6\) m⁻¹ (1)
  • \(\lambda = 4.34 \times 10^{-7}\) m = 434 nm (1)
  • This is the Balmer series (transitions to n = 2, visible light) (1)

Answer: λ = 434 nm; Balmer series (Hγ line)

42.6.3 Hard (5 marks)

Explain how de Broglie’s hypothesis resolved the ad hoc nature of Bohr’s quantisation condition, and calculate the wavelength of an electron in the n = 3 orbit of hydrogen (radius = 4.77 × 10⁻¹⁰ m).

  • Bohr postulated quantisation without explanation (1)
  • de Broglie showed electrons have wavelength; standing waves require integer wavelengths (1)
  • For n = 3 orbit: \(3\lambda = 2\pi r\) (1)
  • \(\lambda = 2\pi r/3 = 2\pi \times 4.77 \times 10^{-10}/3\) (1)
  • \(\lambda = 1.0 \times 10^{-9}\) m = 1.0 nm (1)

Answer: de Broglie explained Bohr’s quantisation as a requirement for standing waves. λ = 1.0 nm for n = 3.

42.7 Multiple Choice Questions

Test your understanding with these interactive questions:

42.8 Summary

Key Takeaways
  • Rutherford model fails: accelerating electrons should radiate and spiral in
  • Bohr’s postulates: stationary states, quantised L, photon emission
  • Hydrogen energy levels: \(E_n = -13.6/n^2\) eV
  • Rydberg equation: \(1/\lambda = R_H[1/n_f^2 - 1/n_i^2]\)
  • de Broglie wavelength: \(\lambda = h/(mv)\)
  • Schrödinger: wave function, probability density, orbitals
  • Uncertainty principle: \(\Delta x \cdot \Delta p \geq h/(4\pi)\)

42.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: