Derivation

The Photoelectric Effect and Einstein's Light-Quantum Equation

D-430 Home PU-202 Threads light · matter · energy Depends on The Energy-Momentum Four-Vector, The Four-Wavevector and Phase Invariance

Statement

When light of frequency \(f\) strikes a metal surface with work function \(\phi\), the maximum kinetic energy of the emitted photoelectrons is given by Einstein's light-quantum equation \(K_{\max} = hf - \phi\), where \(h\) is Planck's constant. Electrons are emitted only if \(f\) exceeds the threshold frequency \(f_0 = \phi/h\); the stopping potential \(V_s\) required to halt the fastest electrons satisfies \(eV_s = K_{\max}\), yielding the linear relation \(V_s = \frac{h}{e}f - \frac{\phi}{e}\) with slope \(h/e\) and intercept \(-\phi/e\).

Why it matters

The photoelectric effect was the decisive experimental evidence that forced the abandonment of the classical wave theory of light. Classical electromagnetism predicted that electron emission should depend on light intensity rather than frequency, and that there should be a measurable time delay while electrons accumulate energy. Neither prediction held. The observed threshold frequency, the linear dependence of maximum kinetic energy on frequency, and the immediate emission at any intensity above threshold could only be explained if light energy arrived in discrete packets, or quanta, each carrying energy \(E = hf\).

Einstein's 1905 explanation introduced the photon concept and established the particle-like nature of light. This duality—wave-like propagation with particle-like energy exchange—became a cornerstone of quantum mechanics. The photoelectric equation also provided the first precise experimental method for measuring Planck's constant, later refined by Millikan's stopping-potential measurements, and it underpins modern technologies including photodiodes, solar cells, and photomultiplier tubes.

Assumptions

One photon ejects one electron.The energy transfer is a single quantum event: one photon of energy \(hf\) gives all its energy to one electron. If multiple photons were required to eject one electron (multiphoton absorption), the linear \(K_{\max}\) vs \(f\) relation would break down at low intensities, and the threshold frequency would depend on intensity, contradicting experiment.

The work function \(\phi\) is constant for a given metal surface.\(\phi\) is the minimum energy needed to extract an electron from the Fermi level to the vacuum level. If \(\phi\) varied significantly across the surface (due to impurities, oxidation, or crystal orientation), the stopping potential would not be sharp, and the linear \(V_s\) vs \(f\) plot would show a spread rather than a single slope \(h/e\).

The electron starts from rest at the Fermi level.The maximum kinetic energy \(K_{\max}\) corresponds to electrons at the Fermi level, which have the least binding energy. Electrons deeper in the band require more energy to escape, so they emerge with less kinetic energy. The equation \(K_{\max} = hf - \phi\) applies only to the fastest electrons; the full energy distribution is broader.

The photon energy \(hf\) is fully absorbed by a single electron in a single interaction.In classical wave theory, energy is spread over the wavefront and accumulated gradually. The assumption of a single, instantaneous quantum jump is essential for explaining the absence of a time delay. If energy were accumulated over time, the emission rate would depend on intensity, and low-intensity light would show a delay, which is not observed.

The metal is treated as a free-electron gas near the surface.The derivation assumes that the electron behaves as a free particle after absorption, with no further interactions that would dissipate energy before emission. In real metals, electron-phonon scattering and surface scattering can reduce the observed kinetic energy, but the maximum energy still obeys Einstein's equation.

Derivation

\[ E_{\text{photon}} = hf \]

Einstein's light-quantum hypothesis: light of frequency \(f\) consists of quanta (photons) each carrying energy \(hf\), where \(h = 6.626 \times 10^{-34}\ \mathrm{J\,s}\) is Planck's constant. A

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

Energy conservation for a single photon-electron interaction: the photon energy is used to overcome the work function \(\phi\) (the binding energy of the electron to the metal) and the remainder appears as the maximum kinetic energy of the emitted electron. B

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

Rearrange Step 2 to solve for the maximum kinetic energy. This is Einstein's photoelectric equation. A

\[ K_{\max} = eV_s \]

The stopping potential \(V_s\) is the retarding voltage that just prevents the fastest electrons from reaching the collector. The work done by the electric field on an electron of charge \(e\) is \(eV_s\), which must equal \(K_{\max}\). A

\[ eV_s = hf - \phi \]

Combine Steps 3 and 4 to eliminate \(K_{\max}\). This relates the measurable stopping potential to the frequency and work function. A

\[ V_s = \frac{h}{e}f - \frac{\phi}{e} \]

Divide by \(e\) to solve for \(V_s\). This is a linear equation in \(f\) with slope \(h/e\) and intercept \(-\phi/e\). B

\[ f_0 = \frac{\phi}{h} \]

The threshold frequency \(f_0\) is the minimum frequency for which \(K_{\max} = 0\). Set \(K_{\max} = 0\) in Step 3: \(hf_0 = \phi\), so \(f_0 = \phi/h\). Below this frequency, no electrons are emitted regardless of intensity. C

\[ \lambda_0 = \frac{hc}{\phi} \]

The threshold wavelength \(\lambda_0\) is the maximum wavelength for which emission occurs. Since \(f_0 = c/\lambda_0\), substitute into Step 7: \(\lambda_0 = c/f_0 = hc/\phi\). A

\[ \text{Slope of } V_s \text{ vs } f \text{ plot} = \frac{h}{e} \]

From Step 6, a plot of \(V_s\) against \(f\) is a straight line with slope \(h/e\). Measuring this slope allows determination of \(h\) if \(e\) is known, or \(e\) if \(h\) is known. A

\[ \text{Intercept} = -\frac{\phi}{e} \]

The \(y\)-intercept of the \(V_s\) vs \(f\) line is \(-\phi/e\). Measuring the intercept gives the work function \(\phi = -e \times \text{intercept}\). A

\[ K_{\max} = \frac{1}{2}m_e v_{\max}^2 \]

The maximum kinetic energy is related to the maximum speed \(v_{\max}\) of the emitted electrons. This connects the photon energy equation to the measurable electron velocity. B

\[ hf = \phi + \frac{1}{2}m_e v_{\max}^2 \]

Combine Steps 2 and 11 to express the photoelectric equation in terms of the maximum electron speed. A

\[ I \propto \text{number of photons per unit time} \]

Light intensity \(I\) is proportional to the number of photons incident per unit area per unit time. Increasing intensity increases the number of emitted electrons (current) but not their maximum kinetic energy, since each photon still has energy \(hf\). B

\[ \text{No time delay} \]

Since each photon transfers its full energy \(hf\) to a single electron in a single quantum event, emission is immediate. There is no accumulation time, consistent with the experimental observation that emission begins as soon as light above threshold frequency strikes the surface. A

Result

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

Reading. The maximum kinetic energy of photoelectrons is the photon energy \(hf\) minus the work function \(\phi\). If \(hf < \phi\), no electrons are emitted. The stopping potential \(V_s\) is directly proportional to \(f\), with slope \(h/e\) and intercept \(-\phi/e\). The threshold frequency is \(f_0 = \phi/h\), and the threshold wavelength is \(\lambda_0 = hc/\phi\).

Units check. \([hf] = \mathrm{J\,s \cdot s^{-1} = J}\). \([\phi] = \mathrm{J}\). \([K_{\max}] = \mathrm{J}\). \([eV_s] = \mathrm{C \cdot V = C \cdot J/C = J}\). All terms have units of joules. Consistent.

Limiting cases

Threshold frequency. When \(f = f_0 = \phi/h\), \(K_{\max} = 0\) and \(V_s = 0\). Electrons are just barely emitted with zero kinetic energy. For \(f < f_0\), no emission occurs regardless of intensity.
High frequency, low intensity. Even at very low intensity, if \(f > f_0\), electrons are emitted immediately with \(K_{\max} = hf - \phi\). The current is small (few photons per second), but the maximum kinetic energy is unchanged.
Low frequency, high intensity. If \(f < f_0\), no electrons are emitted even at extremely high intensity. The classical prediction that high intensity should eventually eject electrons is contradicted by experiment.
Very high frequency (X-rays). The photon can now free inner-shell electrons as well, each with \(K = hf - E_{\text{binding}}\); the conduction-band electrons still give the fastest electrons, \(K_{\max} = hf - \phi\). Only at much higher energies does Compton scattering take over.
Multiphoton regime. At extremely high intensities (e.g., laser pulses), multiple photons can be absorbed simultaneously by a single electron. The threshold frequency is then lower, and the emission depends on intensity, violating the single-photon assumption.

Breaks when

Multiphoton absorption. At very high intensities (e.g., femtosecond laser pulses), a single electron can absorb two or more photons simultaneously. The threshold frequency is then \(f_0 = \phi/(nh)\) for \(n\)-photon absorption, and the emission rate depends on intensity to the \(n\)-th power, contradicting the linear single-photon model.
X-ray energies. Photons of keV energies mostly eject tightly bound inner-shell electrons; the equation still holds with that shell’s binding energy in place of \(\phi\) (the basis of X-ray photoelectron spectroscopy). At still higher energies (hundreds of keV to a few MeV in light elements) Compton scattering, in which the photon is scattered rather than absorbed, becomes the dominant interaction.
Surface variations. If the work function \(\phi\) varies across the surface (due to impurities, oxidation, or crystal orientation), the stopping potential is not sharp, and the \(V_s\) vs \(f\) plot shows a spread rather than a single slope. The measured slope may still be \(h/e\), but the intercept is ambiguous.
Finite temperature. At \(T > 0\) a few electrons occupy states slightly above the Fermi level, so the cut-off at \(K_{\max} = hf - \phi\) is smeared by a few \(kT\) (about \(0.03\ \mathrm{eV}\) at room temperature), and a weak photocurrent survives just below \(f_0\).

Failure modes

Confusing intensity with frequency. Assuming that increasing light intensity increases the maximum kinetic energy of photoelectrons. In fact, intensity affects the number of emitted electrons (current), not their maximum energy. Only frequency affects \(K_{\max}\).
Ignoring the work function. Writing \(K_{\max} = hf\) instead of \(K_{\max} = hf - \phi\). The work function is the energy required to extract the electron from the metal, and it must be subtracted from the photon energy.
Misinterpreting the threshold frequency. Thinking that below the threshold frequency, electrons are emitted with zero kinetic energy. In fact, no electrons are emitted at all if \(f < f_0\), regardless of intensity.
Confusing stopping potential with accelerating potential. The stopping potential \(V_s\) is the retarding voltage that just prevents the fastest electrons from reaching the collector. It is not the voltage needed to accelerate electrons; it is the voltage that stops them.

Discussion

The photoelectric effect was the first clear experimental evidence that light has a particle-like nature. Classical wave theory predicted that the energy of a light wave is proportional to its intensity (amplitude squared), and that electrons should accumulate energy over time until they have enough to escape the metal. This predicted a time delay at low intensities and a dependence of electron energy on intensity, neither of which was observed. Einstein's photon hypothesis resolved these contradictions by proposing that light energy is quantized, with each photon carrying energy \(hf\). The one-photon-one-electron interaction explains the threshold frequency, the linear dependence of \(K_{\max}\) on \(f\), and the immediate emission at any intensity above threshold. The classical delay is easy to estimate: light of \(1\ \mathrm{W/m^2}\) falling on one sodium atom (cross-section about \(10^{-19}\ \mathrm{m^2}\)) delivers \(10^{-19}\ \mathrm{W}\), so collecting \(\phi = 2.3\ \mathrm{eV} = 3.7\times10^{-19}\ \mathrm{J}\) would take about \(4\ \mathrm{s}\), yet emission is observed within nanoseconds even in far weaker light.

The experimental verification of Einstein's equation was provided by Millikan's stopping-potential measurements. By measuring \(V_s\) for various frequencies \(f\), Millikan obtained a straight line with slope \(h/e\) and intercept \(-\phi/e\). The slope gave a value of \(h\) consistent with other measurements, and the intercept gave the work function \(\phi\). This was a precise test of the photon hypothesis, and it confirmed Einstein's prediction to within experimental error.

The photoelectric effect also illustrates the principle of energy conservation at the quantum level. The photon energy \(hf\) is fully absorbed by a single electron, and the energy is partitioned between overcoming the work function and providing kinetic energy. There is no partial absorption or gradual accumulation; the interaction is all-or-nothing. This discreteness is a hallmark of quantum mechanics and is in sharp contrast to the continuous energy exchange predicted by classical wave theory.

At a deeper level, two subtleties. First, a free electron cannot absorb a photon at all: energy and momentum cannot both be conserved. The effect needs electrons bound to a lattice or atom, which takes up the photon’s momentum. Second, the threshold and the linear rise of \(K_{\max}\) with \(f\) can also be reproduced by treating the light as a classical wave and only the atoms quantum-mechanically (Lamb and Scully, 1969), because Fermi’s golden rule then forces energy \(hf\) to be absorbed in one step. Einstein’s equation is therefore exactly right, but the photoelectric effect alone does not prove that the field is quantised; that proof came later from experiments such as photon antibunching. Historically, however, it was Einstein’s light-quantum reading that opened the road to quantum theory.

Common misconceptions. One common misconception is that the photoelectric effect proves that light is a particle. In fact, light exhibits both wave-like and particle-like properties, and the photoelectric effect is evidence of the particle-like aspect. Another misconception is that the work function \(\phi\) is a property of the photon. In fact, \(\phi\) is a property of the metal surface, and it depends on the material and its surface condition. A third misconception is that the stopping potential \(V_s\) is the voltage needed to accelerate electrons. In fact, \(V_s\) is the retarding voltage that just stops the fastest electrons, and it is a measure of their maximum kinetic energy.

Worked examples

Example 1. Light of frequency \(f = 1.0 \times 10^{15}\ \mathrm{Hz}\) strikes a sodium surface with work function \(\phi = 2.3\ \mathrm{eV}\). Find the maximum kinetic energy of the emitted photoelectrons and the stopping potential. Use \(h = 6.626 \times 10^{-34}\ \mathrm{J\,s}\) and \(e = 1.602 \times 10^{-19}\ \mathrm{C}\).

\[ E_{\text{photon}} = hf = (6.626 \times 10^{-34}\ \mathrm{J\,s})(1.0 \times 10^{15}\ \mathrm{Hz}) = 6.626 \times 10^{-19}\ \mathrm{J} \]

Calculate the photon energy using \(E = hf\). A

\[ E_{\text{photon}} = \frac{6.626 \times 10^{-19}\ \mathrm{J}}{1.602 \times 10^{-19}\ \mathrm{J/eV}} = 4.14\ \mathrm{eV} \]

Convert the photon energy from joules to electron-volts. A

\[ K_{\max} = hf - \phi = 4.14\ \mathrm{eV} - 2.3\ \mathrm{eV} = 1.84\ \mathrm{eV} \]

Apply Einstein's photoelectric equation to find the maximum kinetic energy. B

\[ V_s = \frac{K_{\max}}{e} = \frac{1.84\ \mathrm{eV}}{1\ \mathrm{eV}} = 1.84\ \mathrm{V} \]

The stopping potential is the maximum kinetic energy in electron-volts, since \(eV_s = K_{\max}\). A

\[ K_{\max} = 1.84\ \mathrm{eV}, \qquad V_s = 1.84\ \mathrm{V} \]

Reading. The photon energy is 4.14 eV, and 2.3 eV is used to overcome the work function, leaving 1.84 eV as the maximum kinetic energy. The stopping potential is 1.84 V.

Units check. \([hf] = \mathrm{J\,s \cdot s^{-1} = J}\). \([\phi] = \mathrm{eV}\). \([K_{\max}] = \mathrm{eV}\). \([V_s] = \mathrm{V}\). Consistent.

Example 2. The threshold wavelength for a certain metal is \(\lambda_0 = 500\ \mathrm{nm}\). Find the work function \(\phi\) of the metal and the maximum kinetic energy of photoelectrons emitted when light of wavelength \(\lambda = 400\ \mathrm{nm}\) is incident on the surface. Use \(hc = 1240\ \mathrm{eV\,nm}\).

\[ \phi = \frac{hc}{\lambda_0} = \frac{1240\ \mathrm{eV\,nm}}{500\ \mathrm{nm}} = 2.48\ \mathrm{eV} \]

The work function is the energy of a photon with wavelength equal to the threshold wavelength. B

\[ E_{\text{photon}} = \frac{hc}{\lambda} = \frac{1240\ \mathrm{eV\,nm}}{400\ \mathrm{nm}} = 3.10\ \mathrm{eV} \]

Calculate the photon energy for the incident light. A

\[ K_{\max} = hf - \phi = \frac{hc}{\lambda} - \phi = 3.10\ \mathrm{eV} - 2.48\ \mathrm{eV} = 0.62\ \mathrm{eV} \]

Apply Einstein's photoelectric equation to find the maximum kinetic energy. B

\[ V_s = \frac{K_{\max}}{e} = \frac{0.62\ \mathrm{eV}}{1\ \mathrm{eV}} = 0.62\ \mathrm{V} \]

The stopping potential is the maximum kinetic energy in electron-volts. A

\[ \phi = 2.48\ \mathrm{eV}, \qquad K_{\max} = 0.62\ \mathrm{eV}, \qquad V_s = 0.62\ \mathrm{V} \]

Reading. The work function is 2.48 eV, and the photon energy is 3.10 eV, leaving 0.62 eV as the maximum kinetic energy. The stopping potential is 0.62 V.

Units check. \([hc/\lambda] = \mathrm{eV\,nm / nm = eV}\). \([\phi] = \mathrm{eV}\). \([K_{\max}] = \mathrm{eV}\). \([V_s] = \mathrm{V}\). Consistent.

Problems

Light of frequency \(f = 2.0 \times 10^{15}\ \mathrm{Hz}\) strikes a metal surface with work function \(\phi = 3.0\ \mathrm{eV}\). Find the maximum kinetic energy of the emitted photoelectrons and the stopping potential. Use \(h = 6.626 \times 10^{-34}\ \mathrm{J\,s}\) and \(e = 1.602 \times 10^{-19}\ \mathrm{C}\).
Solution
\(E_{\text{photon}} = hf = (6.626 \times 10^{-34})(2.0 \times 10^{15}) = 1.325 \times 10^{-18}\ \mathrm{J}\). Convert to eV: \(E_{\text{photon}} = 1.325 \times 10^{-18} / 1.602 \times 10^{-19} = 8.27\ \mathrm{eV}\). \(K_{\max} = 8.27 - 3.0 = 5.27\ \mathrm{eV}\). \(V_s = 5.27\ \mathrm{V}\).
The threshold wavelength for a metal is \(\lambda_0 = 600\ \mathrm{nm}\). Find the work function \(\phi\) of the metal. Use \(hc = 1240\ \mathrm{eV\,nm}\).
Solution
\(\phi = hc/\lambda_0 = 1240 / 600 = 2.07\ \mathrm{eV}\).
Light of wavelength \(\lambda = 500\ \mathrm{nm}\) strikes a metal surface with work function \(\phi = 2.0\ \mathrm{eV}\). Find the maximum kinetic energy of the emitted photoelectrons and the stopping potential. Use \(hc = 1240\ \mathrm{eV\,nm}\).
Solution
\(E_{\text{photon}} = hc/\lambda = 1240 / 500 = 2.48\ \mathrm{eV}\). \(K_{\max} = 2.48 - 2.0 = 0.48\ \mathrm{eV}\). \(V_s = 0.48\ \mathrm{V}\).
A plot of stopping potential \(V_s\) versus frequency \(f\) for a certain metal has a slope of \(4.14 \times 10^{-15}\ \mathrm{V\,s}\) and an intercept of \(-2.0\ \mathrm{V}\). Find Planck's constant \(h\) and the work function \(\phi\) of the metal. Use \(e = 1.602 \times 10^{-19}\ \mathrm{C}\).
Solution
Slope \(= h/e\), so \(h = e \times \text{slope} = (1.602 \times 10^{-19})(4.14 \times 10^{-15}) = 6.63 \times 10^{-34}\ \mathrm{J\,s}\). Intercept \(= -\phi/e\), so \(\phi = -e \times \text{intercept} = -(1.602 \times 10^{-19})(-2.0) = 3.20 \times 10^{-19}\ \mathrm{J}\). Convert to eV: \(\phi = 3.20 \times 10^{-19} / 1.602 \times 10^{-19} = 2.00\ \mathrm{eV}\).
Light of intensity \(I = 100\ \mathrm{W/m^2}\) and frequency \(f = 1.5 \times 10^{15}\ \mathrm{Hz}\) strikes a metal surface with work function \(\phi = 2.5\ \mathrm{eV}\). Find the number of photons incident per unit area per unit time and the maximum kinetic energy of the emitted photoelectrons. Use \(h = 6.626 \times 10^{-34}\ \mathrm{J\,s}\).
Solution
Energy per photon: \(E_{\text{photon}} = hf = (6.626 \times 10^{-34})(1.5 \times 10^{15}) = 9.94 \times 10^{-19}\ \mathrm{J}\). Number of photons per unit area per unit time: \(N = I / E_{\text{photon}} = 100 / 9.94 \times 10^{-19} = 1.01 \times 10^{20}\ \mathrm{photons/(m^2\,s)}\). \(K_{\max} = hf - \phi = 9.94 \times 10^{-19}\ \mathrm{J} - 2.5 \times 1.602 \times 10^{-19}\ \mathrm{J} = 9.94 \times 10^{-19} - 4.01 \times 10^{-19} = 5.93 \times 10^{-19}\ \mathrm{J}\). Convert to eV: \(K_{\max} = 5.93 \times 10^{-19} / 1.602 \times 10^{-19} = 3.70\ \mathrm{eV}\).