37  Light: The Quantum Model

37.1 Syllabus inquiry question

  • What evidence supports the particle model of light?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 37:

The photoelectric effect reveals that light comes in lumps—discrete packets of energy we call photons. This was Einstein’s revolutionary insight: light has both wave and particle properties, depending on how you look at it.

37.2 Learning Objectives

  • Explain black body radiation and the ultraviolet catastrophe
  • Apply Wien’s displacement law: \(\lambda_{max} = b/T\)
  • Describe evidence for the photoelectric effect
  • Apply the photoelectric equation: \(K_{max} = hf - \phi\)
  • Explain the concept of photon energy: \(E = hf\)

37.3 Content

37.3.1 Black Body Radiation

A black body is an ideal absorber and emitter of electromagnetic radiation. It emits a continuous spectrum that depends only on temperature.

Key observations: - Intensity distribution has a characteristic peak - Peak wavelength shifts to shorter wavelengths as temperature increases - Total power radiated increases dramatically with temperature

37.3.2 The Ultraviolet Catastrophe

Classical physics predicted that black body radiation intensity should increase without limit at short wavelengths—the ultraviolet catastrophe.

Planck’s Solution (1900): Energy is quantised in discrete packets:

\[E = nhf\]

where n is an integer, h is Planck’s constant, and f is frequency.

This was the birth of quantum physics.

37.3.3 Interactive: Black Body Spectrum

37.3.4 Wien’s Displacement Law

\[\lambda_{max} = \frac{b}{T}\]

where: - \(\lambda_{max}\) = peak wavelength (m) - \(b = 2.90 \times 10^{-3}\) m·K (Wien’s constant) - \(T\) = absolute temperature (K)

Temperature and Colour

Higher temperature → shorter peak wavelength → bluer colour Lower temperature → longer peak wavelength → redder colour

37.3.5 The Photoelectric Effect

When light shines on a metal surface, electrons may be ejected. Key observations:

Observation Wave Prediction Actual Result
Threshold frequency Any frequency works Below threshold, no emission
Time delay Delay expected Emission is instantaneous
Maximum KE Depends on intensity Depends on frequency
Number of electrons Depends on frequency Depends on intensity

37.3.6 Interactive: Photoelectric Effect

37.3.7 Einstein’s Photoelectric Equation

Einstein explained the photoelectric effect using photons—discrete packets of light energy:

\[K_{max} = hf - \phi\]

where: - \(K_{max}\) = maximum kinetic energy of ejected electron (J) - \(h = 6.63 \times 10^{-34}\) J·s (Planck’s constant) - \(f\) = frequency of incident light (Hz) - \(\phi\) = work function of metal (J)

Threshold frequency: \(f_0 = \phi/h\) (minimum frequency for emission)

37.3.8 Photon Energy

Each photon carries energy:

\[E = hf = \frac{hc}{\lambda}\]

Electron-Volt (eV)

A convenient energy unit: 1 eV = \(1.6 \times 10^{-19}\) J

This is the energy gained by an electron accelerating through 1 V.

37.3.9 Stopping Voltage

The stopping voltage (\(V_s\)) is the potential needed to stop the most energetic photoelectrons:

\[eV_s = K_{max} = hf - \phi\]

Graphing \(V_s\) vs \(f\) gives: - Slope = \(h/e\) - y-intercept = \(-\phi/e\)

37.4 Worked Examples

37.4.1 Example 1: Wien’s law

A star has surface temperature 12,000 K. Find its peak wavelength.

Solution:

  1. Use \(\lambda_{max} = b/T\)

  2. \(\lambda_{max} = (2.90 \times 10^{-3})/12000\)

  3. \(\lambda_{max} = 2.4 \times 10^{-7}\ \text{m} = 240\ \text{nm}\) (UV)

37.4.2 Example 2: Photon energy

Calculate the energy of a photon with wavelength 500 nm.

Solution:

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

  2. \(E = (6.63 \times 10^{-34} \times 3.0 \times 10^8)/(5.0 \times 10^{-7})\)

  3. \(E = 4.0 \times 10^{-19}\ \text{J} = 2.5\ \text{eV}\)

37.4.3 Example 3: Photoelectric equation

Light of frequency \(8.0 \times 10^{14}\) Hz shines on a metal with work function \(3.0 \times 10^{-19}\) J. Find the maximum kinetic energy of emitted electrons.

Solution:

  1. Use \(K_{max} = hf - \phi\)

  2. \(K_{max} = (6.63 \times 10^{-34} \times 8.0 \times 10^{14}) - 3.0 \times 10^{-19}\)

  3. \(K_{max} = 5.3 \times 10^{-19} - 3.0 \times 10^{-19} = 2.3 \times 10^{-19}\ \text{J}\)

37.4.4 Example 4: Threshold frequency

A metal has work function 2.0 eV. Find the threshold frequency.

Solution:

  1. Convert: \(\phi = 2.0 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}\ \text{J}\)

  2. \(f_0 = \phi/h = (3.2 \times 10^{-19})/(6.63 \times 10^{-34})\)

  3. \(f_0 = 4.8 \times 10^{14}\ \text{Hz}\)

37.4.5 Example 5: Stopping voltage

Find the stopping voltage for electrons with \(K_{max} = 1.6 \times 10^{-19}\) J.

Solution:

  1. Use \(eV_s = K_{max}\)

  2. \(V_s = K_{max}/e = (1.6 \times 10^{-19})/(1.6 \times 10^{-19})\)

  3. \(V_s = 1.0\ \text{V}\)

37.5 Common Misconceptions

Common Misconceptions

  • Misconception: Brighter light produces faster electrons. Correction: Intensity affects the number of photoelectrons, not their maximum energy. Only frequency determines \(K_{max}\).

  • Misconception: Below the threshold frequency, intense light eventually ejects electrons. Correction: No matter how intense, light below threshold frequency cannot eject electrons. Each photon must individually have enough energy.

  • Misconception: Work function is the energy of the ejected electron. Correction: Work function is the minimum energy needed to remove an electron. \(K_{max}\) is the remaining energy after this is overcome.

  • Misconception: Higher frequency light is always brighter. Correction: Brightness (intensity) depends on the number of photons, not their energy. UV light can be dim; red light can be intense.

  • Misconception: The photoelectric effect disproves wave theory. Correction: It shows wave theory is incomplete. Light exhibits both wave and particle properties—wave-particle duality.

37.6 Practice Questions

37.6.1 Easy (2 marks)

Calculate the energy of a photon with frequency \(6.0 \times 10^{14}\) Hz.

  • Use \(E = hf\) (1)
  • \(E = 6.63 \times 10^{-34} \times 6.0 \times 10^{14} = 4.0 \times 10^{-19}\) J (1)

Answer: \(4.0 \times 10^{-19}\) J (or 2.5 eV)

37.6.2 Medium (4 marks)

Light of wavelength 400 nm shines on a metal with work function 2.5 eV. Calculate the maximum kinetic energy of photoelectrons and the stopping voltage.

  • Photon energy: \(E = hc/\lambda = (6.63 \times 10^{-34} \times 3.0 \times 10^8)/(4.0 \times 10^{-7}) = 5.0 \times 10^{-19}\) J = 3.1 eV (1)
  • \(K_{max} = E - \phi = 3.1 - 2.5 = 0.6\) eV (1)
  • Convert: \(K_{max} = 0.6 \times 1.6 \times 10^{-19} = 9.6 \times 10^{-20}\) J (1)
  • \(V_s = K_{max}/e = 0.6\) V (1)

Answer: \(K_{max} = 0.6\) eV (or \(9.6 \times 10^{-20}\) J), \(V_s = 0.6\) V

37.6.3 Hard (5 marks)

A graph of stopping voltage vs frequency for a metal gives a gradient of \(4.1 \times 10^{-15}\) V/Hz and a y-intercept of -2.0 V. Calculate Planck’s constant and the work function.

  • Gradient = h/e, so h = gradient × e (1)
  • \(h = 4.1 \times 10^{-15} \times 1.6 \times 10^{-19} = 6.6 \times 10^{-34}\) J·s (1)
  • y-intercept = \(-\phi/e\), so \(\phi = -\)intercept × e (1)
  • \(\phi = 2.0 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}\) J = 2.0 eV (1)
  • Comment on agreement with accepted values (1)

Answer: \(h = 6.6 \times 10^{-34}\) J·s, \(\phi = 2.0\) eV (or \(3.2 \times 10^{-19}\) J)

37.7 Multiple Choice Questions

Test your understanding with these interactive questions:

37.8 Summary

Key Takeaways
  • Black body radiation led to quantisation: \(E = nhf\)
  • Wien’s displacement law: \(\lambda_{max} = b/T\)
  • Photon energy: \(E = hf = hc/\lambda\)
  • Photoelectric equation: \(K_{max} = hf - \phi\)
  • Threshold frequency: \(f_0 = \phi/h\)
  • Stopping voltage: \(eV_s = K_{max}\)
  • Wave-particle duality: light behaves as both wave and particle

37.9 Self-Assessment

Check your understanding:

After studying this section, you should be able to: