The Photoelectric Effect
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You should be familiar with the quantisation of energy from the previous lecture on blackbody radiation and understand that electromagnetic radiation carries energy in discrete packets called photons.
Why this matters
The photoelectric effect provided the decisive experimental evidence that light behaves as a particle, cementing the quantum hypothesis.
Classical wave theory predicted that increasing the intensity of light should eventually eject electrons from a metal surface, regardless of frequency, and that there should be a time delay between illumination and emission. Experiments contradicted both predictions. Instead, electrons are ejected instantaneously only if the light frequency exceeds a specific threshold, and their kinetic energy depends linearly on frequency, not intensity. This behaviour is perfectly explained by treating light as a stream of photons, each carrying energy \(E = hf\), where \(h\) is Planck’s constant and \(f\) is the frequency. This lecture explores the mechanism of photoemission, the concept of the work function, and the implications for atomic ionisation.
The Particle Nature of Light
Imagine shining a stream of marbles at a pile of sand. If each marble is heavy enough, it will knock a grain of sand out immediately. If the marbles are too light, no amount of increasing the number of marbles (intensity) will eject a grain, because each individual impact lacks the necessary energy. The photoelectric effect works similarly. Light is not a continuous wave of energy spread over the surface; it is a collection of discrete quanta, or photons. Each photon interacts with a single electron in the metal. If the photon’s energy is sufficient to overcome the binding force holding the electron in the metal lattice, the electron is ejected. If the photon’s energy is too low, the electron remains bound, regardless of how many photons strike the surface per second. This one-to-one interaction explains why there is no time delay in emission and why intensity affects the number of ejected electrons, not their energy.
The Work Function and Threshold Frequency
Electrons in a metal are bound by the collective electrostatic forces of the lattice. The minimum energy required to remove an electron from the surface of a specific material is called the work function, denoted by \(\phi\). This value is characteristic of the material; for example, it is lower for alkali metals like caesium than for noble metals like platinum. Because photon energy is given by \(E = hf\), there is a minimum frequency, known as the threshold frequency \(f_0\), below which no photoemission occurs. This threshold is defined by the condition where the photon energy exactly equals the work function. If the incident light has a frequency \(f\) lower than \(f_0\), the photons do not carry enough energy to liberate the electrons, and the current remains zero.
It is crucial to note that the work function is a property of the material, not the light source. Changing the light source to a different frequency changes the photon energy, but the material’s requirement for energy to release an electron remains constant.
Energy Conservation in Photoemission
When a photon of frequency \(f\) strikes an electron in the metal, the photon is absorbed entirely. The energy of the photon is used for two purposes: first, to overcome the work function \(\phi\) and free the electron from the surface; second, to provide the electron with kinetic energy. According to the principle of conservation of energy, the maximum kinetic energy \(KE_{\max}\) of the ejected electron is the difference between the photon energy and the work function. Electrons deeper within the metal may require more energy to reach the surface, so they will have less kinetic energy upon exit, but the maximum kinetic energy corresponds to electrons at the surface that require the minimum energy to escape.
Since \(KE_{\max} = \frac{1}{2}m_e v_{\max}^2\), where \(m_e\) is the electron mass and \(v_{\max}\) is the maximum velocity, we can relate the stopping potential \(V_s\) (the voltage required to stop the most energetic electrons) to the frequency. The work done by the electric field is \(eV_s\), so \(eV_s = hf - \phi\). This linear relationship between stopping potential and frequency is a hallmark of the quantum model.
Photoionisation of Atoms
The same principles apply to isolated atoms in the gas phase. Here, the "work function" is replaced by the first ionisation energy \(I_1\), which is the energy required to remove the most loosely bound electron from a neutral atom. Photoionisation occurs only if the incident photon energy is greater than or equal to \(I_1\). If \(hf \ge I_1\), the electron is ejected, and any excess energy appears as the kinetic energy of the ionised electron. This process is fundamental in mass spectrometry and atmospheric chemistry, where solar ultraviolet radiation ionises molecules in the upper atmosphere.
Unlike metals, where the work function is a bulk property, the ionisation energy is a discrete atomic property determined by the nuclear charge and electron shielding. This connects the photoelectric effect directly to the periodic trends in ionisation energies discussed in later lectures.
The result
Ionisation and the photoelectric effect
When light of frequency \(f\) strikes a material with work function \(\phi\) (the minimum energy needed to remove an electron), the maximum kinetic energy of any ejected electron is \(KE_{\max}=hf-\phi\), and no electrons are ejected at all if \(f\) is below the threshold frequency \(f_0=\phi/h\), no matter how intense the light. For an isolated gas-phase atom, the same relation holds with the work function replaced by the atom's first ionisation energy \(I_1\): photoionisation occurs only for \(hf\ge I_1\).
A common misconception is that increasing the intensity of light below the threshold frequency will eventually eject electrons because the energy accumulates over time. This is incorrect. In the quantum model, energy transfer is discrete and instantaneous per photon-electron interaction. An electron cannot "save up" energy from multiple low-energy photons in a single step to overcome the work function. Therefore, no emission occurs below \(f_0\), regardless of intensity or duration. C
Worked example
Problem. Ultraviolet light with a wavelength of \(250\ \mathrm{nm}\) is incident on a sodium surface. The work function of sodium is \(2.28\ \mathrm{eV}\). Calculate the maximum kinetic energy of the ejected electrons in electronvolts (\(\mathrm{eV}\)) and in joules (\(\mathrm{J}\)).
- Convert the wavelength to frequency using \(f = c/\lambda\), where \(c = 3.00 \times 10^8\ \mathrm{m\,s^{-1}}\) and \(\lambda = 250 \times 10^{-9}\ \mathrm{m}\). \[ f = \frac{3.00 \times 10^8}{250 \times 10^{-9}} = 1.20 \times 10^{15}\ \mathrm{Hz} \]
- Calculate the photon energy in joules using \(E = hf\), where \(h = 6.626 \times 10^{-34}\ \mathrm{J\,s}\). \[ E = (6.626 \times 10^{-34})(1.20 \times 10^{15}) = 7.951 \times 10^{-19}\ \mathrm{J} \]
- Convert the photon energy to electronvolts. Since \(1\ \mathrm{eV} = 1.602 \times 10^{-19}\ \mathrm{J}\): \[ E_{\mathrm{eV}} = \frac{7.951 \times 10^{-19}}{1.602 \times 10^{-19}} = 4.96\ \mathrm{eV} \]
- Subtract the work function \(\phi = 2.28\ \mathrm{eV}\) to find the maximum kinetic energy. \[ KE_{\max} = 4.96\ \mathrm{eV} - 2.28\ \mathrm{eV} = 2.68\ \mathrm{eV} \]
- Convert the kinetic energy back to joules for the final answer. \[ KE_{\max} = 2.68 \times 1.602 \times 10^{-19} = 4.29 \times 10^{-19}\ \mathrm{J} \]
Units check. Energy is expressed in both electronvolts and joules, consistent with the input units for the work function and the SI unit for energy.
Three quick tests
- If the frequency of incident light is doubled, what happens to the maximum kinetic energy of the ejected electrons?
Answer
It increases by an amount equal to \(hf\), assuming the initial frequency was above the threshold. The relationship is linear, not proportional to the square of the frequency. - Why does increasing the intensity of light below the threshold frequency not result in photoemission?
Answer
Because each photon individually lacks sufficient energy to overcome the work function. Energy transfer is a one-photon-to-one-electron process, so accumulation of energy from multiple low-energy photons does not occur in the standard photoelectric effect. - How does the photoelectric effect relate to the ionisation energy of an atom?
Answer
For an isolated atom, the work function is replaced by the first ionisation energy \(I_1\). Photoionisation occurs only if \(hf \ge I_1\), and the excess energy becomes the kinetic energy of the ejected electron.
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Every lecture and every result in this unit, with prerequisites and threads.
Lecture notes drafted with a local model and checked line by line by the editor; every number in the worked example was recomputed.