The Photoelectric Equation Revision Notes | A Level AQA Physics

1. The Central Question: Where Does the Photon's Energy Go?

When a photon strikes a metal surface, it hands all of its energy to a single electron. That energy has two jobs to do: first, it must pay the cost of freeing the electron from the metal, and second, whatever is left over becomes the electron's kinetic energy as it flies away. The photoelectric equation is an energy accounting statement for this transaction.

This is where the photon model of light earns its credibility. A wave model cannot explain why emission depends on frequency rather than intensity — but the photon model does, precisely because energy arrives in discrete packets, one photon per electron. The equation you are about to use is the mathematical expression of that idea.

2. The Photoelectric Equation

Here is the energy balance written as an equation. Every term has a physical meaning you need to be able to state precisely in an exam.

Read the equation left to right: the photon arrives carrying energy $hf$. It spends $\phi$ to pull the electron free from the metal's surface. The remainder, $hf - \phi$, becomes kinetic energy. Rearranged: $E_{k(max)} = hf - \phi$. This rearrangement is the form you will use in almost every calculation.

The work function is the energy cost of the escape itself. An electron sitting right at the surface pays exactly $\phi$ and leaves with zero kinetic energy. An electron deeper inside the metal has to do extra work fighting through the lattice, so it arrives at the surface with less energy to spare — which is why most electrons leave with less than the maximum kinetic energy.

3. Why $E_{k(max)}$ Is a Maximum, Not a Fixed Value

This is one of the most important conceptual points in the whole topic, and it is frequently tested with 'explain' questions. The subscript 'max' is not decorative — it carries real physical meaning.

Electrons in a metal are not all sitting at the same depth. Surface electrons need only the bare minimum energy $\phi$ to escape, so they leave with the full surplus as kinetic energy. Electrons deeper in the lattice must do additional work fighting intermolecular forces on their way out, consuming some of the photon's energy before they even reach the surface. Those electrons leave with less kinetic energy than the surface electrons.

The result is a spread of kinetic energies among emitted electrons, ranging from zero up to $E_{k(max)}$. The equation $hf = \phi + E_{k(max)}$ describes only the most energetic electrons — those that started right at the surface and lost no energy to internal collisions. Every other electron has kinetic energy somewhere below this maximum.

4. Threshold Frequency and the Condition for Emission

Emission only occurs when the photon's energy is at least equal to the work function. At the threshold frequency $f_0$, the photon has just enough energy to free a surface electron, which then leaves with zero kinetic energy. Setting $E_{k(max)} = 0$ in the photoelectric equation gives:

Below $f_0$, each photon carries insufficient energy to overcome $\phi$, so no electron is released — not even if the intensity is extremely high. Increasing intensity only increases the number of photons arriving per second; it does not increase the energy of any individual photon. This is a direct consequence of the quantum model and cannot be explained by classical wave theory.

5. Worked Calculations Using the Photoelectric Equation

The equation $hf = \phi + E_{k(max)}$ can be rearranged to find any one of the three quantities if the other two are known. The worked examples below show the method for the two most common exam question types: finding $E_{k(max)}$ and finding the threshold frequency.

Worked example: Finding Maximum Kinetic Energy of Photoelectrons

Ultraviolet radiation of frequency $1.50 \times 10^{15}$ Hz is incident on a sodium surface. The work function of sodium is $3.65 \times 10^{-19}$ J. Calculate the maximum kinetic energy of the emitted photoelectrons. ($h = 6.63 \times 10^{-34}$ J s) [3 marks]

1. $hf = \phi + E_{k(max)}$ — [M1] Stating or implying the correct equation earns the method mark.
2. $hf = 6.63 \times 10^{-34} \times 1.50 \times 10^{15}$ $hf = 9.945 \times 10^{-19}$ J — Multiply the Planck constant by the frequency. Keep the full value for now to avoid rounding errors mid-calculation. [M1] Correct substitution.
3. $E_{k(max)} = hf - \phi$ $E_{k(max)} = 9.945 \times 10^{-19} - 3.65 \times 10^{-19}$ $E_{k(max)} = 6.295 \times 10^{-19}$ J — Subtract the work function from the photon energy. Both values are in joules, so no unit conversion is needed.
4. $E_{k(max)} = 6.30 \times 10^{-19}$ J — [A1] Final answer with correct unit. Three significant figures match the data given. Always include the unit — omitting it costs the accuracy mark.

Answer: $E_{k(max)} = 6.30 \times 10^{-19}$ J

Worked example: Finding the Threshold Frequency from the Work Function

The work function of zinc is $6.89 \times 10^{-19}$ J. Calculate the threshold frequency for zinc. ($h = 6.63 \times 10^{-34}$ J s) [2 marks]

1. $hf_0 = \phi + 0$ $hf_0 = \phi$ — At the threshold frequency, the photon has just enough energy to free a surface electron, which leaves with zero kinetic energy. [M1] Recognising this condition.
2. $f_0 = \frac{\phi}{h}$ — Divide both sides by $h$.
3. $f_0 = \frac{6.89 \times 10^{-19}}{6.63 \times 10^{-34}}$ $f_0 = 1.039 \times 10^{15}$ Hz — Divide the numerator by the denominator. Check the power of ten: $10^{-19} \div 10^{-34} = 10^{-19-(-34)} = 10^{15}$. This is in the ultraviolet range, which is physically reasonable for zinc.
4. $f_0 = 1.04 \times 10^{15}$ Hz — [A1] Correct answer with unit. Three significant figures are appropriate here.

Answer: $f_0 = 1.04 \times 10^{15}$ Hz

6. Working with Electron-Volts

Work function values are frequently quoted in electron-volts (eV) in data books and exam questions. One electron-volt is the energy gained by an electron when it is accelerated through a potential difference of one volt: $1 \text{ eV} = 1.60 \times 10^{-19}$ J. You must be comfortable converting between eV and joules before substituting into the photoelectric equation.

The conversion is straightforward: to convert from eV to joules, multiply by $1.60 \times 10^{-19}$. To convert from joules to eV, divide by $1.60 \times 10^{-19}$. The equation itself always works in SI units (joules), so convert first, then substitute.

Worked example: Using the Photoelectric Equation with eV Data

Light of wavelength $2.50 \times 10^{-7}$ m is incident on a metal surface with a work function of $4.50$ eV. Calculate the maximum kinetic energy of the emitted photoelectrons in joules. ($h = 6.63 \times 10^{-34}$ J s, $c = 3.00 \times 10^{8}$ m s$^{-1}$, $1$ eV $= 1.60 \times 10^{-19}$ J) [4 marks]

1. $\phi = 4.50 \times 1.60 \times 10^{-19}$ $\phi = 7.20 \times 10^{-19}$ J — [M1] Converting eV to joules before substitution. Failing to convert is one of the most common errors in this type of question.
2. $E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{2.50 \times 10^{-7}}$ — The question gives wavelength, not frequency. Use $c = f\lambda$ rearranged to $f = c/\lambda$, then substitute into $E = hf$ to get $E = hc/\lambda$. [M1] Correct method for finding photon energy from wavelength.
3. Numerator: $6.63 \times 10^{-34} \times 3.00 \times 10^{8} = 1.989 \times 10^{-25}$ J m $E = \frac{1.989 \times 10^{-25}}{2.50 \times 10^{-7}} = 7.956 \times 10^{-19}$ J — Power of ten check: $10^{-25} \div 10^{-7} = 10^{-25-(-7)} = 10^{-18}$. Then $1.989 \div 2.50 = 0.7956$, giving $7.956 \times 10^{-19}$ J.
4. $E_{k(max)} = hf - \phi = 7.956 \times 10^{-19} - 7.20 \times 10^{-19}$ $E_{k(max)} = 7.56 \times 10^{-20}$ J — [A1] Correct subtraction. Note that $7.956 - 7.200 = 0.756$, so $E_{k(max)} = 0.756 \times 10^{-19} = 7.56 \times 10^{-20}$ J. [A1] Correct unit.

Answer: $E_{k(max)} = 7.56 \times 10^{-20}$ J

7. The Graph of $E_{k(max)}$ Against Frequency

Rearranging the photoelectric equation as $E_{k(max)} = hf - \phi$ reveals that a graph of $E_{k(max)}$ on the y-axis against frequency $f$ on the x-axis is a straight line. This graph is a powerful tool for extracting both the Planck constant and the work function from experimental data.

The gradient of this graph equals the Planck constant $h$. The x-intercept gives the threshold frequency $f_0$. The y-intercept, if extrapolated, gives $-\phi$ (a negative value, since at $f = 0$ the equation gives $E_{k(max)} = -\phi$). Examiners regularly ask you to interpret this graph, so knowing what each feature represents is essential.

8. Exam Technique and Common Errors

The photoelectric equation appears in calculation questions, 'explain' questions, and graph interpretation questions. Each type rewards a slightly different approach, and the marks are distributed across method, substitution, and final answer. Knowing where marks are allocated helps you avoid losing them unnecessarily.

The photoelectric equation is ultimately a statement of energy conservation applied to a single photon-electron interaction. Every term has a precise physical meaning, and every exam question on this topic — whether a calculation, a graph, or an explanation — tests whether you can apply that meaning correctly. Keeping the energy accounting clear in your mind is the key to full marks.