2.5 Photons, the photoelectric effect and line spectra
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AS content for Unit 2 (Waves and Electricity). Interference and diffraction of light are on 2.2 and 2.4; electron diffraction and the de Broglie wavelength are on 2.4.
What this page covers
6 points
- 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 \(hf = \phi + \tfrac12 mv^2_\text{max}\)
- 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, \(hf = E_1 - E_2\) and energy level diagrams in eV
In one minute
5 points
- Light travels as a wave (diffraction, interference) but is emitted and absorbed in photons, packets of energy \(E = hf = hc/\lambda\).
- 1 eV is the energy transferred when an electron moves through a p.d. of 1 V, \(1.60 \times 10^{-19}\) J.
- One photon gives all its energy to one electron. If \(hf\) is at least the work function \(\phi\), the electron can escape: \(hf = \phi + \tfrac12 mv^2_\text{max}\).
- Below the threshold frequency \(f_0 = \phi/h\) no electrons are emitted, however bright the light. Brighter light above \(f_0\) gives more electrons, not faster ones.
- Electrons in atoms occupy discrete energy levels. A transition emits or absorbs one photon with \(hf = E_1 - E_2\), so each element has its own line spectrum.
Contents
13 sections
- How did the wave and photon models of light develop?
- What is a photon, and how much energy does it carry?
- What is the electronvolt?
- What is the photoelectric effect?
- How do you measure the maximum kinetic energy of photoelectrons?
- Why is the photoelectric effect evidence for photons?
- What are line spectra, and what do they tell us about atoms?
- Exam tips for 2.5
- Common misconceptions
- How this topic connects
- Key equations
- Check your understanding
- Specification checklist
How did the wave and photon models of light develop?
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.
| When | Idea or evidence | Which model it supported |
|---|---|---|
| late 1600s | Newton pictures light as a stream of tiny "corpuscles"; Huygens describes light as a wave spreading from each point on a wavefront | both proposed; Newton's reputation keeps the particle view dominant for a century |
| 1801 | Young's double-slit experiment shows interference fringes | wave: particles cannot cancel each other out |
| 1860s | Maxwell shows that light is an electromagnetic wave travelling at \(3.00 \times 10^{8}\) m s⁻¹ | wave |
| 1887–1902 | Hertz notices that ultraviolet light helps sparks jump; later work shows light ejects electrons from metals only above a threshold frequency | a puzzle the wave model cannot explain |
| 1900 | Planck explains the spectrum of hot objects by assuming energy is emitted in packets \(hf\) | the first quantum idea |
| 1905 | Einstein proposes that light itself travels as quanta (photons) and explains the photoelectric effect | photon |
| 1916 | Millikan's careful measurements confirm Einstein's equation and give a value of \(h\) | photon: accepted after the prediction was tested |
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.
What is a photon, and how much energy does it carry?
Here \(E\) is the photon energy (J), \(h\) the Planck constant, \(6.63 \times 10^{-34}\) J s, and \(f\) the frequency (Hz). \(E = hf\) is in the formulae list, and so is \(c = f\lambda\). Combining them gives \(E = hc/\lambda\), 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 \(= P \div hf\).
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.
(a) \(f = c/\lambda = 3.00 \times 10^{8} \div 532 \times 10^{-9} =\) 5.64 × 10¹⁴ Hz.
(b) \(E = hf = 6.63 \times 10^{-34} \times 5.64 \times 10^{14} =\) 3.74 × 10⁻¹⁹ J.
(c) Energy per second ÷ energy per photon: \(N = 5.0 \times 10^{-3} \div 3.74 \times 10^{-19} =\) 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.
What is the electronvolt?
From \(W = QV\): \(1\ \text{eV} = 1.60 \times 10^{-19}\ \text{C} \times 1\ \text{V} = 1.60 \times 10^{-19}\) J. To go from eV to J, multiply by \(1.60 \times 10^{-19}\); 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.
(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 \(= 9.11 \times 10^{-31}\) kg.
(a) \(3.74 \times 10^{-19} \div 1.60 \times 10^{-19} =\) 2.34 eV.
(b) \(E_k = 1500\ \text{eV} = 1500 \times 1.60 \times 10^{-19} = 2.4 \times 10^{-16}\) J. From \(E_k = \tfrac12 mv^2\): \(v = \sqrt{2 \times 2.4 \times 10^{-16} \div 9.11 \times 10^{-31}} =\) 2.4 × 10⁻¹⁶ J; 2.30 × 10⁷ m s⁻¹.
Putting an energy in eV straight into \(\tfrac12 mv^2\) or \(E = hf\). Whenever \(h\), a mass in kg or a speed in m s⁻¹ appears in the same equation, the energy must be in joules.
What is the photoelectric effect?
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 \(\phi\) of a metal is the minimum energy needed to remove an electron from the surface of the metal.
- The threshold frequency \(f_0\) 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 \(hf - \phi\). At threshold, \(hf_0 = \phi\), so \(f_0 = \phi/h\); derive this, as it is not given.
A graph of \(\tfrac12 mv^2_\text{max}\) against \(f\) is a straight line: the gradient is \(h\) for every metal, the f-intercept is \(f_0\), and the line extrapolates to \(-\phi\) on the energy axis.
Ultraviolet radiation of wavelength 220 nm falls on a clean zinc surface. The work function of zinc is 4.3 eV. Mass of electron \(= 9.11 \times 10^{-31}\) kg. (a) Calculate the photon energy in eV. (b) Calculate the maximum speed of the photoelectrons. (c) Calculate the threshold frequency for zinc.
(a) \(E = hc/\lambda = 6.63 \times 10^{-34} \times 3.00 \times 10^{8} \div 220 \times 10^{-9} = 9.04 \times 10^{-19}\ \text{J} =\) 5.65 eV.
(b) Maximum kinetic energy \(= 5.65 - 4.3 = 1.35\ \text{eV} = 1.35 \times 1.60 \times 10^{-19} = 2.16 \times 10^{-19}\) J. Then \(v_\text{max} = \sqrt{2 \times 2.16 \times 10^{-19} \div 9.11 \times 10^{-31}} =\) 6.9 × 10⁵ m s⁻¹ (6.89 × 10⁵ m s⁻¹).
(c) \(f_0 = \phi/h = 4.3 \times 1.60 \times 10^{-19} \div 6.63 \times 10^{-34} =\) 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.
Subtracting 4.3 (eV) from \(9.04 \times 10^{-19}\) (J). Decide on one energy unit before you subtract: convert \(hf\) to eV, or convert \(\phi\) to joules.
How do you measure the maximum kinetic energy of photoelectrons?
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 \(V_s\): even the fastest electrons are just turned back, so
\[eV_s = \tfrac12 mv^2_\text{max}\]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 \(eV_s = \tfrac12 mv^2_\text{max}\) with Einstein's equation gives \(eV_s = hf - \phi\), so \(V_s = (h/e)f - \phi/e\). A graph of \(V_s\) against \(f\) is a straight line of gradient \(h/e\), a way to measure the Planck constant.
A student shines light of five different frequencies on the same photocell and measures the stopping potential each time.
| f / 10¹⁴ Hz | 5.8 | 6.9 | 7.9 | 9.1 | 10.2 |
|---|---|---|---|---|---|
| Vs / V | 0.21 | 0.66 | 1.08 | 1.58 | 2.03 |
(a) Use the graph above to determine a value for the Planck constant. (b) Determine the work function of the cathode, in eV.
(a) Gradient of the best-fit line, from a large triangle: \(1.66\ \text{V} \div 4.0 \times 10^{14}\ \text{Hz} = 4.15 \times 10^{-15}\) V s. Gradient \(= h/e\), so \(h = 1.60 \times 10^{-19} \times 4.15 \times 10^{-15} =\) 6.64 × 10⁻³⁴ J s.
(b) From the f-intercept, \(f_0 \approx 5.3 \times 10^{14}\) Hz. \(\phi = hf_0 = 6.64 \times 10^{-34} \times 5.3 \times 10^{14} = 3.5 \times 10^{-19}\ \text{J} =\) 2.2 eV.
Use points on your best-fit line, not individual data points, for the gradient. Equivalent method: the line extrapolates to \(-\phi/e = -2.2\) V at \(f = 0\).
Saying the gradient of \(V_s\) against \(f\) is \(h\). It is \(h/e\), because the vertical axis is a p.d., not an energy. If the graph is of maximum kinetic energy in joules against \(f\), then the gradient is \(h\).
Why is the photoelectric effect evidence for photons?
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.
| Observation | Wave model predicts | Photon model explains |
|---|---|---|
| Below a threshold frequency no electrons are emitted, however intense the light. | Any frequency should work if the light is bright enough or shines for long enough. | Each electron absorbs one photon. If \(hf < \phi\), no single photon has enough energy; more photons only means more electrons that each fail to escape. |
| Emission begins instantly, even in very dim light (above \(f_0\)). | In dim light an electron would need time to build up enough energy. | One photon delivers \(hf\) to one electron at once. |
| The maximum kinetic energy depends on frequency, not intensity. | Brighter light carries more energy, so electrons should leave faster. | \(\tfrac12 mv^2_\text{max} = hf - \phi\) depends only on \(f\). Intensity sets the number of photons per second. |
| Photocurrent is proportional to intensity (above \(f_0\)). | Consistent, but cannot explain the rows above. | More photons per second → more photoelectrons per second. |
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.
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 \(\phi\) to escape, so emission needs \(hf \ge \phi\): below \(f_0 = \phi/h\) 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 \(hf - \phi\), 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.
What are line spectra, and what do they tell us about atoms?
- 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 \(E_1\) is the energy of the higher level and \(E_2\) 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.
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.
(a) Energy to lift an electron from the ground state to 0 eV: \(7.80 \times 1.60 \times 10^{-19} =\) 1.25 × 10⁻¹⁸ J.
(b) Largest gap, n = 4 → n = 1: \(\Delta E = -1.30 - (-7.80) = 6.50\) eV, so \(f = 6.50 \times 1.60 \times 10^{-19} \div 6.63 \times 10^{-34} =\) 1.57 × 10¹⁵ Hz.
(c) Longest wavelength = smallest gap, n = 4 → n = 3: \(\lambda = 6.63 \times 10^{-34} \times 3.00 \times 10^{8} \div (1.10 \times 1.60 \times 10^{-19}) =\) 1.13 × 10⁻⁶ m, longer than about \(7 \times 10^{-7}\) 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).
Using one level's value instead of the difference, or adding the two negative values. The photon energy is always the gap: subtract the lower (more negative) level from the higher one, so the answer is positive.
Exam tips for 2.5
Wavelengths in nm need \(\times 10^{-9}\); energies in eV need \(\times 1.60 \times 10^{-19}\) before they meet \(h\), \(c\) or a mass. Write the conversion as a separate line so it is visible.
Write "photon energy = work function + maximum kinetic energy" before substituting. Always say maximum kinetic energy: photoelectrons leave with a range of energies, because those below the surface lose energy on the way out.
Frequency controls the energy of each photon, and so whether emission happens and the maximum kinetic energy. Intensity controls the number of photons per second, and so the number of electrons per second (the current).
Pair each observation with its explanation, and say what the wave model would have predicted. Use the words one photon, one electron, work function and intensity = number of photons per second.
From the n-th level, the number of different photon energies possible is \(n(n - 1)/2\): six from n = 4. Check by drawing every downward arrow between pairs of levels.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| Brighter light gives photoelectrons more kinetic energy. | Intensity changes the number of photons, not the energy of each. | Brighter light (above \(f_0\)) gives more electrons per second; \(\tfrac12 mv^2_\text{max}\) depends only on \(f\). |
| Very bright red light will eventually release electrons from zinc. | Each red photon has less energy than the work function, and electrons cannot store energy from several photons. | Below \(f_0\) there is no emission at any intensity. |
| All photoelectrons leave with the same kinetic energy. | Electrons below the surface lose energy escaping. | \(hf - \phi\) is the maximum kinetic energy. |
| The work function is the energy of the photon. | It is a property of the metal surface. | The work function is the minimum energy to remove an electron from the metal. |
| An atom absorbs any photon with more energy than the gap. | A photon is absorbed whole or not at all. | Only photons whose energy exactly matches a gap are absorbed (a free electron can give up part of its energy). |
| Dark lines in an absorption spectrum mean the light there is completely removed. | Absorbed energy is re-emitted, but in all directions. | Those wavelengths are much fainter in the direction of the original beam, so the lines look dark by contrast. |
How this topic connects
2.1–2.4 Waves
\(c = f\lambda\) 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.
2.6 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).
Key equations
| Equation | Use | In the formulae list? |
|---|---|---|
| \(E = hf\) | energy of one photon; E in J, f in Hz | Yes |
| \(c = f\lambda\) | frequency from wavelength | Yes |
| \(E = hc/\lambda\) | photon energy from wavelength | No — derive it from the two above |
| 1 eV \(= 1.60 \times 10^{-19}\) J | converting energies; \(E_k = eV\) after acceleration through V | No — learn it (\(e\) is on the data list) |
| \(hf = \phi + \tfrac12 mv^2_\text{max}\) | photoelectric effect: energy balance for one photon and one electron | Yes |
| \(f_0 = \phi/h\) | threshold frequency | No — derive it (set ½mv²max = 0) |
| \(eV_s = \tfrac12 mv^2_\text{max}\) | maximum kinetic energy from stopping potential | No — learn it |
| \(hf = E_1 - E_2\) | photon emitted or absorbed in a transition between energy levels | No — learn it (it is E = hf for the gap) |
| number of photons per second \(= P/hf\) | photon rate in a beam of power P | No — derive it |
Check your understanding
1. Define the work function of a metal.
The minimum energy needed to remove an electron from the surface of the metal.
2. Calculate the energy, in eV, of a photon of frequency 1.0 × 10¹⁵ Hz.
\(E = 6.63 \times 10^{-34} \times 1.0 \times 10^{15} = 6.63 \times 10^{-19}\) J; \(\div 1.60 \times 10^{-19} =\) 4.14 eV.
3. Light below the threshold frequency falls on a metal. Its intensity is doubled. What happens?
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. The stopping potential for a photocell is 0.80 V. What is the maximum kinetic energy of the photoelectrons in joules?
\(\tfrac12 mv^2_\text{max} = eV_s = 1.60 \times 10^{-19} \times 0.80 =\) 1.28 × 10⁻¹⁹ J (0.80 eV).
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?
Both involve the same pairs of energy levels. A photon is absorbed only if its energy equals a gap \(E_1 - E_2\), 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. Light of wavelength 450 nm falls on a metal of work function 2.2 eV. Calculate the stopping potential.
\(E = hc/\lambda = 6.63 \times 10^{-34} \times 3.00 \times 10^{8} \div 450 \times 10^{-9} = 4.42 \times 10^{-19}\ \text{J} = 2.76\) eV. Maximum kinetic energy \(= 2.76 - 2.2 = 0.56\) eV, so Vs = 0.56 V.
Specification checklist
Numbers refer to the specification points.
- (57) I can describe the wave model and the photon model of electromagnetic radiation, and outline how the evidence led scientists from one to the other.
- (58) I can calculate the energy of a photon from its frequency or wavelength using \(E = hf\).
- (59) I can explain that a photoelectron is released when an electron at a metal surface absorbs a single photon.
- (60) I can define threshold frequency and work function, and use \(hf = \phi + \tfrac12 mv^2_\text{max}\), including with the stopping potential.
- (61) I can define the electronvolt and convert between electronvolts and joules.
- (62) I can explain why the photoelectric effect is evidence for the particle nature of light, and why the wave model cannot account for it.
- (63) I can explain emission and absorption line spectra in terms of transitions between discrete energy levels, and calculate the frequency of the photon involved using \(hf = E_1 - E_2\) from an energy level diagram in eV.