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Year 12 Science

Introduction to Quantum Physics

Step into the strange world of quantum mechanics -- where light behaves as both a wave and a particle, and energy comes in discrete packets called quanta.

The Photoelectric Effect

The photoelectric effect is the emission of electrons from a metal surface when light of sufficient frequency shines upon it. Classical wave theory could not explain this phenomenon -- it was Einstein's explanation using photons (light quanta) in 1905 that earned him the Nobel Prize.

How the Photoelectric Effect Works

Photon

Energy = hf

Metal Surface

Work function = φ

Electron Ejected

KEmax = hf - φ

Einstein's photoelectric equation: KEmax = hf - φ, where h is Planck's constant (6.63 × 10-34 J·s), f is the frequency of light, and φ is the work function of the metal.

Classical Predictions (Wrong)

  • • Brighter light = more KE (wrong)
  • • Any frequency should work (wrong)
  • • Time delay before emission (wrong)

Quantum Explanation (Correct)

  • • Higher frequency = more KE
  • • Threshold frequency f0 required
  • • Instantaneous emission

Wave-Particle Duality

One of the most profound ideas in quantum physics is that all matter and radiation exhibit both wave-like and particle-like behaviour. This was first proposed for light and later extended to all matter by Louis de Broglie in 1924.

Wave Evidence

Diffraction: Light bends around obstacles and through slits, producing interference patterns.

Young's double-slit experiment: Light passing through two slits creates an interference pattern of bright and dark fringes.

This can only be explained by wave behaviour.

Particle Evidence

Photoelectric effect: Light ejects electrons one at a time, with energy dependent on frequency, not intensity.

Compton scattering: X-rays scatter off electrons like billiard balls, with momentum transfer.

This can only be explained by particle behaviour.

De Broglie Wavelength

λ = h / p = h / (mv)

Every particle with momentum p has an associated wavelength λ. For everyday objects, λ is incredibly tiny and undetectable. For electrons, λ is comparable to atomic spacings.

Electron diffraction: When electrons are fired at a thin crystal, they produce a diffraction pattern just like X-rays -- confirming that particles have wave properties. This was demonstrated by Davisson and Germer in 1927.

Photon Energy and Atomic Spectra

Light is quantised into photons, each carrying a discrete amount of energy. When atoms absorb or emit photons, electrons jump between specific energy levels, producing characteristic line spectra.

Photon Energy Equations

E = hf

Energy from frequency

E = hc / λ

Energy from wavelength

Energy Level Transitions

n = 4 (-0.85 eV)
n = 3 (-1.51 eV)
Photon emitted
n = 2 (-3.40 eV)
n = 1 (ground) (-13.6 eV)

Hydrogen atom energy levels (not to scale)

Key relationship: When an electron drops from a higher level (Ei) to a lower level (Ef), a photon is emitted with energy Ephoton = Ei - Ef = hf. This explains why each element has a unique emission spectrum.

Key Vocabulary

Photon

A quantum (discrete packet) of electromagnetic radiation with energy E = hf. Photons have zero rest mass and always travel at the speed of light.

Work Function (φ)

The minimum energy required to liberate an electron from the surface of a metal. Different metals have different work functions.

Threshold Frequency (f0)

The minimum frequency of light needed to eject electrons from a metal surface. Below this frequency, no electrons are emitted regardless of intensity.

De Broglie Wavelength

The wavelength associated with a moving particle, given by λ = h/p. Demonstrates that matter has wave-like properties.

Worked Examples

1

Calculate the energy of a photon of ultraviolet light with frequency 1.5 × 1015 Hz.

Step 1: Use E = hf, where h = 6.63 × 10-34 J·s.

Step 2: E = 6.63 × 10-34 × 1.5 × 1015.

Step 3: E = 9.945 × 10-19 J ≈ 9.95 × 10-19 J.

Answer: The photon energy is approximately 9.95 × 10-19 J (or about 6.2 eV).

2

Light of frequency 8.0 × 1014 Hz strikes a metal with work function 3.0 × 10-19 J. Find the maximum kinetic energy of the emitted electrons.

Step 1: Find photon energy: E = hf = 6.63 × 10-34 × 8.0 × 1014 = 5.304 × 10-19 J.

Step 2: Apply Einstein's equation: KEmax = hf - φ.

Step 3: KEmax = 5.304 × 10-19 - 3.0 × 10-19 = 2.3 × 10-19 J.

Answer: The maximum KE of emitted electrons is 2.3 × 10-19 J (about 1.4 eV).

3

Calculate the de Broglie wavelength of an electron (mass 9.11 × 10-31 kg) moving at 2.0 × 106 m/s.

Step 1: Calculate momentum: p = mv = 9.11 × 10-31 × 2.0 × 106 = 1.822 × 10-24 kg·m/s.

Step 2: Apply de Broglie: λ = h/p = 6.63 × 10-34 / 1.822 × 10-24.

Step 3: λ = 3.64 × 10-10 m = 0.364 nm.

Answer: The de Broglie wavelength is 3.64 × 10-10 m -- comparable to atomic spacing, so electron diffraction is observable.

Knowledge Check

Select the correct answer for each question. Click "Check Answer" to see if you are right.

Question 1

In the photoelectric effect, increasing the intensity of light above the threshold frequency will:

Question 2

The de Broglie wavelength of a particle will decrease if:

Question 3

A photon has a wavelength of 500 nm. Its energy is closest to:

Question 4

Which experiment provided the first direct evidence for the particle nature of light?

Question 5

If the frequency of incident light on a metal surface is doubled (while remaining above the threshold), the maximum kinetic energy of emitted electrons:

Key Concepts Summary

Year 12: Special Relativity Year 12: Electrochemistry