The wave model and the photon model of electromagnetic radiation, and how scientists moved from one to the other

Photon energy, E = hf, and the electronvolt

The photoelectric effect: threshold frequency, work function and Einstein's equation hf = φ+ ½ mv²ₘₐₓ

Stopping potential as a way to measure the maximum kinetic energy of photoelectrons

Why the photoelectric effect is evidence for photons, and where the wave model fails

Emission and absorption line spectra, discrete energy levels, hf = E₁ - E₂ and energy level diagrams in eV

A model is a simplified picture that explains observations and makes testable predictions. Light has had two, and the scientific community changed its mind as new evidence arrived.

Today neither model alone is complete. Light shows wave behaviour as it travels (diffraction, interference, polarisation) and particle behaviour when it interacts with matter (photoelectric effect, line spectra): wave–particle duality. The history shows how science works: a model is kept while it explains the evidence, and replaced or extended when new, reproducible evidence contradicts it.

Here E is the photon energy (J), h the Planck constant, 6.63 ×10⁻³⁴ J s, and f the frequency (Hz). E = hf is in the formulae list, and so is c = fλ. Combining them gives E = hc/λ, which is not given, so derive it.

Higher frequency (shorter wavelength) means more energy per photon: an ultraviolet photon carries more energy than a red one.

The intensity of a beam depends on how many photons arrive per second, not on the energy of each one.

For a beam of power P: number of photons per second = P ÷hf.

WORKED EXAMPLE · QUESTION

A green laser pointer emits light of wavelength 532 nm with an output power of 5.0 mW. (a) Calculate the frequency of the light. (b) Calculate the energy of one photon. (c) Calculate the number of photons emitted each second.

SOLUTION

(a) f = c/λ= 3.00 ×10⁸ ÷532 ×10⁻⁹ = 5.64 × 10¹⁴ Hz.

(b) E = hf = 6.63 ×10⁻³⁴ ×5.64 ×10¹⁴ = 3.74 × 10⁻¹⁹ J.

(c) Energy per second ÷ energy per photon: N = 5.0 ×10⁻³ ÷3.74 ×10⁻¹⁹ = 1.3 × 10¹⁶ s⁻¹ (1.34 × 10¹⁶ s⁻¹).

Even a weak beam carries an enormous number of photons each second, which is why light usually looks continuous.

From W = QV: 1 eV = 1.60 ×10⁻¹⁹ C ×1 V = 1.60 ×10⁻¹⁹ J. To go from eV to J, multiply by 1.60 ×10⁻¹⁹; from J to eV, divide. An electron accelerated from rest through a p.d. V gains kinetic energy eV: through 1500 V it gains 1500 eV. The eV is a convenient size for photon energies, work functions and atomic energy levels, which are all a few eV; 1 keV = 10³ eV and 1 MeV = 10⁶ eV.

WORKED EXAMPLE · QUESTION

(a) Express the photon energy from worked example 1 in electronvolts. (b) An electron is accelerated from rest through a potential difference of 1500 V. Calculate its kinetic energy in joules and its final speed. Mass of electron = 9.11 ×10⁻³¹ kg.

SOLUTION

(a) 3.74 ×10⁻¹⁹ ÷1.60 ×10⁻¹⁹ = 2.34 eV.

(b) Eₖ = 1500 eV = 1500 ×1.60 ×10⁻¹⁹ = 2.4 ×10⁻¹⁶ J. From Eₖ = ½ mv²: v = √(2 ×2.4 ×10⁻¹⁶ ÷9.11 ×10⁻³¹) = 2.4 × 10⁻¹⁶ J; 2.30 × 10⁷ m s⁻¹.

A classic demonstration: a clean zinc plate on a negatively charged electroscope discharges at once in ultraviolet light, but not at all in bright visible light. One photon interacts with one electron and gives it all its energy, hf; the photon ceases to exist. The electron needs a minimum energy to escape the attraction of the metal's positive ions.

The work function φ of a metal is the minimum energy needed to remove an electron from the surface of the metal.

The threshold frequency f₀ is the minimum frequency of radiation that will release electrons from the surface of a metal.

Einstein's photoelectric equation is in the formulae list. It is an energy balance: photon energy in = energy to escape + kinetic energy left over. It says "max" because electrons from below the surface lose some energy on the way out; only those at the surface leave with all of hf - φ. At threshold, hf₀ = φ, so f₀ = φ/h; derive this, as it is not given.

A graph of ½ mv²ₘₐₓ against f is a straight line: the gradient is h for every metal, the f-intercept is f₀, and the line extrapolates to -φ on the energy axis.

WORKED EXAMPLE · QUESTION

Ultraviolet radiation of wavelength 220 nm falls on a clean zinc surface. The work function of zinc is 4.3 eV. Mass of electron = 9.11 ×10⁻³¹ kg. (a) Calculate the photon energy in eV. (b) Calculate the maximum speed of the photoelectrons. (c) Calculate the threshold frequency for zinc.

SOLUTION

(a) E = hc/λ= 6.63 ×10⁻³⁴ ×3.00 ×10⁸ ÷220 ×10⁻⁹ = 9.04 ×10⁻¹⁹ J = 5.65 eV.

(b) Maximum kinetic energy = 5.65 - 4.3 = 1.35 eV = 1.35 ×1.60 ×10⁻¹⁹ = 2.16 ×10⁻¹⁹ J. Then vₘₐₓ = √(2 ×2.16 ×10⁻¹⁹ ÷9.11 ×10⁻³¹) = 6.9 × 10⁵ m s⁻¹ (6.89 × 10⁵ m s⁻¹).

(c) f₀ = φ/h = 4.3 ×1.60 ×10⁻¹⁹ ÷6.63 ×10⁻³⁴ = 1.0 × 10¹⁵ Hz (1.04 × 10¹⁵ Hz).

The work function is given to 2 s.f., so the speed and frequency are given to 2 s.f.

Light falls on the metal cathode of a vacuum photocell; photoelectrons cross to the anode and a sensitive ammeter registers a photocurrent. Now make the anode negative relative to the cathode. Electrons are repelled and must do work against the field to reach it, so only those with enough kinetic energy arrive. Increase this reverse p.d. until the current just falls to zero. That p.d. is the stopping potential Vₛ: even the fastest electrons are just turned back, so

A stopping potential of 1.35 V means a maximum kinetic energy of 1.35 eV. This relation is not in the formulae list; learn it.

Combining eVₛ = ½ mv²ₘₐₓ with Einstein's equation gives eVₛ = hf - φ, so Vₛ = (h/e)f - φ/e. A graph of Vₛ against f is a straight line of gradient h/e, a way to measure the Planck constant.

WORKED EXAMPLE · QUESTION

A student shines light of five different frequencies on the same photocell and measures the stopping potential each time.

(a) Use the graph above to determine a value for the Planck constant. (b) Determine the work function of the cathode, in eV.

SOLUTION

(a) Gradient of the best-fit line, from a large triangle: 1.66 V ÷4.0 ×10¹⁴ Hz = 4.15 ×10⁻¹⁵ V s. Gradient = h/e, so h = 1.60 ×10⁻¹⁹ ×4.15 ×10⁻¹⁵ = 6.64 × 10⁻³⁴ J s.

(b) From the f-intercept, f₀ ≈5.3 ×10¹⁴ Hz. φ= hf₀ = 6.64 ×10⁻³⁴ ×5.3 ×10¹⁴ = 3.5 ×10⁻¹⁹ J = 2.2 eV.

Use points on your best-fit line, not individual data points, for the gradient. Equivalent method: the line extrapolates to -φ/e = -2.2 V at f = 0.

In the wave model, the energy of a light wave is spread continuously across the wavefront and grows with intensity, so electrons could soak up energy gradually until they had enough to escape. The observations contradict this.

WORKED EXAMPLE · QUESTION

Explain how observations of the photoelectric effect support a particle (photon) model of light rather than a wave model. Refer to threshold frequency, the time before emission begins and the energy of the emitted electrons.

SOLUTION

A full answer links each observation to the model:

Light consists of photons, each of energy E = hf; one photon is absorbed by one electron.

An electron needs at least the work function φ to escape, so emission needs hf ≥φ: below f₀ = φ/h there is no emission, whatever the intensity. In the wave model energy could build up, so any frequency should eventually work; this is not observed.

Emission is instantaneous because each photon delivers its energy in a single interaction; the wave model predicts a delay in dim light.

The maximum kinetic energy is hf - φ, so it rises with frequency and does not depend on intensity. Greater intensity means more photons per second, so more electrons per second, not faster ones; the wave model predicts faster electrons in brighter light.

Threshold, instant emission and frequency-dependent energy all need one-photon-one-electron energy packets.

Electrons in an atom can only have certain discrete energies, called energy levels. The lowest is the ground state; higher ones are excited states.

Levels are given negative values: zero is the energy of an electron that has just escaped (ionisation). The ground state value, without its sign, is the ionisation energy.

Emission: an excited electron drops to a lower level and emits one photon carrying the energy difference. In a gas discharge tube, electrons accelerated by the p.d. collide with atoms and excite them; the atoms then de-excite.

Absorption: an electron absorbs a photon only if the photon energy exactly equals the difference between its level and a higher one. Other photons pass straight through.

Because the levels are discrete, only certain photon energies, and so only certain frequencies, are involved. Each element has its own set of levels, so its line spectrum identifies it.

Here E₁ is the energy of the higher level and E₂ of the lower level (J), and f the photon frequency (Hz). This is not in the formulae list, so learn it; it is E = hf applied to the gap between two levels. Diagrams are usually in eV: convert the difference to joules before using h.

Emission and absorption spectra

An emission line spectrum is bright coloured lines on a dark background, from a hot, low-pressure gas.

An absorption line spectrum is dark lines across a continuous spectrum, seen when white light passes through a cooler gas. The atoms absorb exactly the photon energies that match their level differences, then re-emit in random directions, so less light at those wavelengths continues towards the observer. The dark lines of an element are at the same wavelengths as some of its bright emission lines, because the same energy gaps are involved.

WORKED EXAMPLE · QUESTION

The energy level diagram above has levels at −7.80 eV (n = 1), −4.20 eV (n = 2), −2.40 eV (n = 3) and −1.30 eV (n = 4). (a) State the ionisation energy of the atom in joules. (b) Calculate the frequency of the highest-energy photon that can be emitted. (c) Show that the longest wavelength emitted lies in the infrared. (d) A photon of energy 5.00 eV and, separately, a free electron with 5.00 eV of kinetic energy meet atoms in the ground state. Explain what can happen in each case.

SOLUTION

(a) Energy to lift an electron from the ground state to 0 eV: 7.80 ×1.60 ×10⁻¹⁹ = 1.25 × 10⁻¹⁸ J.

(b) Largest gap, n = 4 → n = 1: ΔE = -1.30 - (-7.80) = 6.50 eV, so f = 6.50 ×1.60 ×10⁻¹⁹ ÷6.63 ×10⁻³⁴ = 1.57 × 10¹⁵ Hz.

(c) Longest wavelength = smallest gap, n = 4 → n = 3: λ= 6.63 ×10⁻³⁴ ×3.00 ×10⁸ ÷(1.10 ×1.60 ×10⁻¹⁹) = 1.13 × 10⁻⁶ m, longer than about 7 ×10⁻⁷ m, so infrared.

(d) Photon: all or nothing. No gap from the ground state equals 5.00 eV (they are 3.60, 5.40 and 6.50 eV), so the photon is not absorbed. Electron: it can give up part of its energy, so it can excite an atom to n = 2 (3.60 eV) and carry on with 1.40 eV; it cannot reach n = 3 (5.40 eV).

1–2.4 Waves

c = fλ links the wavelength in a spectrum to the photon frequency; diffraction and interference are the evidence for the wave model, and a diffraction grating (2.4) is how line spectra are measured precisely.

6 onwards: electric circuits

The electronvolt comes from W = QV; the stopping potential is the p.d. that does enough work on an electron to remove all its kinetic energy.

Units 4 and 5

Particles accelerated through a p.d. gain energy eV, and particle energies are quoted in MeV and GeV (Unit 4). Absorption lines in starlight identify the elements in stars, and black-body radiation, the problem that started the quantum idea, is used to find stellar temperatures (Unit 5).

1. Define the work function of a metal.

2. Calculate the energy, in eV, of a photon of frequency 1.0 × 10¹⁵ Hz.

3. Light below the threshold frequency falls on a metal. Its intensity is doubled. What happens?

4. The stopping potential for a photocell is 0.80 V. What is the maximum kinetic energy of the photoelectrons in joules?

5. Why do the dark lines in the absorption spectrum of a gas appear at the same wavelengths as bright lines in its emission spectrum?

6. Light of wavelength 450 nm falls on a metal of work function 2.2 eV. Calculate the stopping potential.

1. The minimum energy needed to remove an electron from the surface of the metal.

2. E = 6.63 ×10⁻³⁴ ×1.0 ×10¹⁵ = 6.63 ×10⁻¹⁹ J; ÷1.60 ×10⁻¹⁹ = 4.14 eV.

3. Still no electrons are emitted. Doubling the intensity doubles the number of photons per second, but each photon still has less energy than the work function, and an electron absorbs only one photon.

4. ½ mv²ₘₐₓ = eVₛ = 1.60 ×10⁻¹⁹ ×0.80 = 1.28 × 10⁻¹⁹ J (0.80 eV).

5. Both involve the same pairs of energy levels. A photon is absorbed only if its energy equals a gap E₁ - E₂, and a photon of exactly that energy is emitted when an electron drops back across the same gap, so the frequencies are the same.

6. E = hc/λ= 6.63 ×10⁻³⁴ ×3.00 ×10⁸ ÷450 ×10⁻⁹ = 4.42 ×10⁻¹⁹ J = 2.76 eV. Maximum kinetic energy = 2.76 - 2.2 = 0.56 eV, so Vs = 0.56 V.