Threshold Frequency and Work Function Revision Notes | A Level AQA Physics

1. The Puzzle That Broke Classical Physics

Shine ultraviolet light on a zinc plate and electrons fly off instantly. Shine red light — no matter how bright — and nothing happens. Classical wave theory predicted that any frequency of light, given enough intensity, should eventually release electrons. The universe disagreed.

This is the photoelectric effect, and it forced physicists to accept something radical: light arrives in discrete packets of energy called photons. Each photon carries energy $E = hf$, where $h$ is Planck's constant ($6.63 \times 10^{-34}$ J s) and $f$ is frequency. The wave model cannot explain what you are about to learn.

2. Threshold Frequency — Why There Is a Cut-Off

Picture a metal surface as a crowd of electrons sitting in a shallow pit. Each electron needs a minimum amount of energy to escape that pit and reach the surface. That minimum energy is called the work function, symbol $\phi$.

Now here is the key insight: one photon interacts with one electron — it is a one-to-one exchange. If that single photon carries energy less than $\phi$, the electron cannot escape, full stop. No accumulation of energy from multiple photons occurs. This is why frequency matters, not intensity.

The threshold frequency $f_0$ is the exact frequency at which a photon carries just enough energy to equal the work function. At this frequency, the electron escapes with zero kinetic energy — it barely makes it out.

Different metals have different work functions, which is why ultraviolet light releases electrons from zinc but visible light does not — zinc's work function is too high for visible photons to overcome. Caesium, with a much lower work function, releases electrons under visible light, which is why it is used in photoelectric cells.

3. The Photon Explanation of Threshold Frequency

The photon model gives a precise, mechanistic explanation for the threshold frequency — something the wave model cannot do. When a photon strikes the metal surface, it transfers all of its energy $hf$ to a single electron in a single interaction. There is no partial transfer and no waiting for energy to build up.

If $hf < \phi$: the photon energy is insufficient to free the electron. The electron absorbs the energy but cannot escape — it re-emits it as heat. Increasing the intensity of this sub-threshold light only means more photons per second, each still individually too weak. No emission occurs.

If $hf \geq \phi$: the photon has enough energy to liberate the electron. Any energy above $\phi$ becomes kinetic energy of the emitted photoelectron. This is why emission is instantaneous — there is no delay waiting for energy to accumulate.

4. The Photoelectric Equation and Maximum Kinetic Energy

When a photon of frequency $f > f_0$ strikes the surface, the electron uses $\phi$ worth of energy to escape, and the remainder becomes kinetic energy. The most energetic photoelectrons are those at the very surface — they need to do the least work to escape. Electrons deeper in the metal lose additional energy through collisions before reaching the surface, so they emerge with less kinetic energy.

The photoelectric equation captures this energy conservation precisely. It links the incoming photon energy to the work function and the maximum kinetic energy of emitted electrons.

Notice that $E_{k(\text{max})}$ depends only on the frequency of the light, not its intensity. Increasing intensity increases the number of photons per second, which increases the number of photoelectrons emitted per second (the photocurrent), but it does not change how much kinetic energy each photoelectron carries.

Worked example: Calculating Maximum Kinetic Energy of Photoelectrons

Ultraviolet radiation of frequency $1.50 \times 10^{15}$ Hz is incident on a metal surface with a work function of $3.60 \times 10^{-19}$ J. Calculate the maximum kinetic energy of the emitted photoelectrons.

1. $E_{k(\text{max})} = hf - \phi$ — Energy conservation: photon energy minus the energy needed to escape equals the kinetic energy left over. [M1]
2. $hf = 6.63 \times 10^{-34} \times 1.50 \times 10^{15}$ $hf = 6.63 \times 1.50 \times 10^{-34+15}$ $6.63 \times 1.50 = 9.945$ $hf = 9.945 \times 10^{-19}$ J — Multiply the coefficient and add the exponents separately to avoid errors. [M1]
3. $E_{k(\text{max})} = 9.945 \times 10^{-19} - 3.60 \times 10^{-19}$ $E_{k(\text{max})} = (9.945 - 3.60) \times 10^{-19}$ $E_{k(\text{max})} = 6.345 \times 10^{-19}$ J — Subtract the work function from the photon energy. Both values are in joules, so no unit conversion is needed here. [M1]
4. $E_{k(\text{max})} \approx 6.35 \times 10^{-19}$ J — The data is given to 3 significant figures, so the answer should match. Always state the unit. [A1]
5. $hf = 9.95 \times 10^{-19}$ J $> \phi = 3.60 \times 10^{-19}$ J — Photon energy exceeds the work function, so emission is expected and $E_{k(\text{max})}$ must be positive. A negative result would indicate an error.

Answer: $E_{k(\text{max})} \approx 6.35 \times 10^{-19}$ J

5. Stopping Potential — Measuring Kinetic Energy Electrically

You cannot directly measure the kinetic energy of a photoelectron with a ruler. Instead, physicists use an elegant electrical method: apply a reverse voltage across the photoelectric cell to decelerate the electrons. The voltage at which even the fastest electrons are brought to a halt is called the stopping potential, $V_s$.

The connection to kinetic energy is direct. The work done by the electric field on the electron as it decelerates equals the electron's initial kinetic energy. Using the definition of work done by a charge moving through a potential difference:

Combining this with the photoelectric equation gives a complete picture: $hf = \phi + eV_s$. This form is particularly useful in experiments because $V_s$ is directly measurable with a voltmeter, allowing you to determine either the work function or Planck's constant from experimental data.

Worked example: Using Stopping Potential to Find Work Function

Light of frequency $8.00 \times 10^{14}$ Hz is shone on a metal surface. The stopping potential is measured to be $1.20$ V. Calculate the work function of the metal in joules.

1. $hf = \phi + eV_s$ $\phi = hf - eV_s$ — Rearrange to make $\phi$ the subject. [M1]
2. $hf = 6.63 \times 10^{-34} \times 8.00 \times 10^{14}$ $= 6.63 \times 8.00 \times 10^{-34+14}$ $6.63 \times 8.00 = 53.04$ $hf = 53.04 \times 10^{-20} = 5.304 \times 10^{-19}$ J — Multiply coefficients and add exponents. Rewrite in standard form. [M1]
3. $eV_s = 1.60 \times 10^{-19} \times 1.20$ $= 1.60 \times 1.20 \times 10^{-19}$ $1.60 \times 1.20 = 1.92$ $eV_s = 1.92 \times 10^{-19}$ J — This is the maximum kinetic energy of the photoelectrons, determined from the stopping potential. [M1]
4. $\phi = 5.304 \times 10^{-19} - 1.92 \times 10^{-19}$ $= (5.304 - 1.92) \times 10^{-19}$ $= 3.384 \times 10^{-19}$ J — Subtract to find the energy used to overcome the surface binding. [A1]
5. $\phi \approx 3.38 \times 10^{-19}$ J — Three significant figures matches the precision of the given data. Always include units.

Answer: $\phi \approx 3.38 \times 10^{-19}$ J

6. Graphical Analysis — The $E_{k(\text{max})}$ Against $f$ Graph

The photoelectric equation $E_{k(\text{max})} = hf - \phi$ is linear in $f$. Plotting $E_{k(\text{max})}$ on the y-axis against frequency $f$ on the x-axis produces a straight line — and every feature of that line carries physical meaning.

• Gradient of the line = Planck's constant $h$ (same for all metals)
• x-intercept = threshold frequency $f_0$ (different for each metal)
• y-intercept = $-\phi$ (the negative of the work function)
• The line does not extend below the x-axis — $E_{k(\text{max})}$ cannot be negative

This graph is a favourite in AQA exam questions. You may be asked to read off $f_0$, calculate $h$ from the gradient, or determine $\phi$ from the y-intercept. Practise extracting all three pieces of information from a single graph.

8. Exam Technique and Common Mistakes

AQA questions on this topic test both calculation and explanation. The calculation marks are straightforward if you apply the photoelectric equation correctly. The explanation marks are where students most often drop points — examiners want precise, mechanistic language, not vague references to 'energy'.

For 'explain' questions, the mark scheme typically awards one mark for stating that a photon interacts with one electron, one mark for comparing $hf$ with $\phi$, and one mark for the consequence (emission or no emission). Structure your answer around these three beats every time.