# Photoelectric effect

> Physics · CIE A-Level 9702
> Source: https://www.owlsprep.com/study/cie-9702-u26-photoelectric-effect/

This sub-topic covers the photoelectric effect experiment, its unexpected observations that disproved the classical wave model of light, Einstein's photon explanation, and core calculation skills for CIE A-Level Physics.

**Prerequisites:** Classical wave model of electromagnetic radiation; Photon energy equation; Energy units including electronvolts

## Learning objectives

- Explain the photoelectric effect and its role as evidence for the photon model of light
- Recall and correctly apply Einstein's photoelectric equation to solve calculation problems
- Describe key experimental observations and explain why classical wave theory cannot explain them
- Interpret graphs of stopping potential against frequency to find Planck's constant and work function

## Observations and Failure of Wave Theory

**Photoelectric effect** — Emission of electrons from a metal surface when electromagnetic radiation of sufficiently high frequency is absorbed by the surface.

*Example:* Photoemission from a zinc plate illuminated by ultraviolet radiation

Early 20th century experiments produced four key observations that could not be explained by the classical wave model of light, which assumes energy is spread evenly across a wavefront:

1. Photoelectrons are only emitted if incident frequency is above a metal-specific **threshold frequency** $f_0$
2. Emission is instantaneous, even at very low intensity
3. Maximum kinetic energy of photoelectrons depends only on frequency, not intensity
4. Increasing intensity increases the number of photoelectrons emitted per second, not their maximum kinetic energy

> **warning**
>
> Wave theory predicts emission should occur at any frequency (given enough time for energy to build up) and that higher intensity should produce higher kinetic energy. Neither prediction matches experimental results.

**Worked example:** A student observes instantaneous photoelectron emission when low-intensity blue light shines on a sodium surface. They then shine high-intensity low-frequency red light on the same surface, but detect no emission. Explain why no photoelectrons are released.

1. First, recall that photoelectric emission only occurs when incident frequency is above the metal's threshold frequency.
2. Red light has a frequency below sodium's threshold frequency. The classical wave model incorrectly predicts that high intensity would eventually provide enough energy for emission.
3. In the photon model, one photon interacts with one electron. Each photon of red light has energy below sodium's work function, so no electron can gain enough energy to escape, regardless of how many photons (how high the intensity) hit the surface.

## Einstein's Photoelectric Equation

**Work function** — Minimum energy required to release one electron from the surface of a metal, dependent on the metal type.

*Notation:* $\Phi$

**Threshold frequency** — Minimum frequency of incident radiation that can cause photoelectric emission, related to work function by $\Phi = hf_0$.

*Notation:* $f_0$

Einstein explained the observations by assuming light behaves as discrete packets called photons, each with energy $E = hf$. One photon interacts with one electron at the metal surface. The electron uses at least the work function energy to escape the metal, and any remaining energy becomes its kinetic energy. This gives the core photoelectric equation:

$$hf = \Phi + KE_{max}$$

The maximum kinetic energy can also be written in terms of stopping potential $V_s$, the reverse potential needed to stop the most energetic photoelectrons: $KE_{max} = eV_s$.

**Worked example:** The work function of caesium is 2.14 eV. Calculate (a) the threshold frequency, (b) the stopping potential for incident radiation of frequency $6.5 \times 10^{14}$ Hz. Take $h = 6.63 \times 10^{-34} Js$, $e = 1.60 \times 10^{-19} C$.

1. Convert work function from eV to joules:
2. $$\Phi = 2.14 \times 1.60 \times 10^{-19} = 3.424 \times 10^{-19} J$$
3. (a) Calculate threshold frequency from $f_0 = \frac{\Phi}{h}$:
4. $$f_0 = \frac{3.424 \times 10^{-19}}{6.63 \times 10^{-34}} = 5.16 \times 10^{14} Hz$$
5. (b) Substitute into Einstein's equation rearranged for stopping potential, where $eV_s = hf - \Phi$:
6. $$hf = 6.63 \times 10^{-34} \times 6.5 \times 10^{14} = 4.31 \times 10^{-19} J$$
7. $$V_s = \frac{4.31 \times 10^{-19} - 3.424 \times 10^{-19}}{1.60 \times 10^{-19}} = 0.55 V$$

## Graphical Interpretation

Rearranging Einstein's equation gives a linear relationship between maximum kinetic energy (or stopping potential) and incident frequency. This linear relationship confirms the photon model and allows experimental measurement of Planck's constant and work function.

| Graph Feature | Physical Meaning |
| --- | --- |
| Gradient of $KE_{max}$ vs $f$ | $h$ (Planck constant) |
| Gradient of $V_s$ vs $f$ | $h/e$ |
| Y-intercept of $KE_{max}$ vs $f$ | $-\Phi$ |
| X-intercept (where $KE_{max}=0$) | Threshold frequency $f_0$ |

> **tip**
>
> CIE exams regularly ask you to calculate Planck's constant from a $V_s$ vs $f$ graph. Always multiply the gradient by $e$ (elementary charge) to get $h$.

**Worked example:** A graph of $V_s$ (V) against $f$ (Hz) for an unknown metal has a gradient of $4.1 \times 10^{-15} V/Hz$ and an x-intercept at $5.0 \times 10^{14} Hz$. Calculate $h$ and the work function of the metal.

1. From $V_s = \frac{h}{e}f - \frac{\Phi}{e}$, gradient equals $\frac{h}{e}$:
2. $$h = \text{gradient} \times e = 4.1 \times 10^{-15} \times 1.60 \times 10^{-19} = 6.56 \times 10^{-34} Js$$
3. The x-intercept is threshold frequency $f_0$, so $\Phi = hf_0$:
4. $$\Phi = 6.56 \times 10^{-34} \times 5.0 \times 10^{14} = 3.28 \times 10^{-19} J = 2.05 eV$$

## Common pitfalls

- **Wrong:** Forgetting to convert work function from eV to joules before calculating threshold frequency.
  - Why it fails: Planck's constant is usually given in $Js$, so mixing energy units gives a result that is orders of magnitude wrong.
  - Correct: Always convert all energy values to joules before substitution, or use $h$ in $eVs$ for unit consistency.
- **Wrong:** Claiming higher intensity of incident light increases the maximum kinetic energy of photoelectrons.
  - Why it fails: Intensity measures the number of photons per second, not the energy per photon. Maximum kinetic energy depends only on frequency.
  - Correct: State that higher intensity increases the rate of photoelectron emission, not their maximum kinetic energy.
- **Wrong:** Thinking emission can occur below threshold frequency if you wait long enough at high intensity.
  - Why it fails: This is a wrong prediction from classical wave theory. The photon model requires each individual photon to have enough energy.
  - Correct: No emission can ever occur below threshold frequency, regardless of intensity or exposure time.
- **Wrong:** Taking the gradient of a $V_s$ vs $f$ graph to be equal to $h$ directly.
  - Why it fails: Confusing the $KE_{max}$ vs $f$ relationship with the $V_s$ vs $f$ relationship.
  - Correct: Gradient equals $h$ for $KE_{max}$ vs $f$, and $h/e$ for $V_s$ vs $f$.
- **Wrong:** Assuming all photoelectrons have kinetic energy equal to $KE_{max}$.
  - Why it fails: $KE_{max}$ describes only electrons emitted from the metal surface. Electrons from deeper inside lose extra energy escaping.
  - Correct: Only use $KE_{max}$ in Einstein's equation, as it refers to the most energetic emitted electrons.

## Cheatsheet

| Concept | Key Formula/Relationship |
| --- | --- |
| Photon energy | $E = hf = \frac{hc}{\lambda}$ |
| Work function - threshold frequency | $\Phi = hf_0$ |
| Einstein's photoelectric equation | $hf = \Phi + KE_{max}$ |
| Stopping potential relation | $KE_{max} = eV_s$ |
| Gradient of $KE_{max}$ vs $f$ | $= h$ |
| Gradient of $V_s$ vs $f$ | $= h/e$ |
| Core principle | One photon interacts with one electron |

## What's next

The photoelectric effect provided the first definitive experimental evidence for the particle nature of light, forming the foundation of all modern quantum physics. Mastering its principles and calculations is critical for all subsequent quantum topics in the CIE 9702 syllabus. The photon model introduced here is extended to matter particles in wave-particle duality, and applied to atomic energy levels to explain line spectra, both core exam topics.

- [Wave-particle duality](https://www.owlsprep.com/study/cie-9702-u26-wave-particle-duality/)
- [Electron energy levels](https://www.owlsprep.com/study/cie-9702-u26-electron-energy-levels/)
- [Line spectra](https://www.owlsprep.com/study/cie-9702-u26-line-spectra/)

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