Photons and the Photoelectric Effect

AP Physics 2· AP Physics 2 CED — Quantum, Atomic, and Nuclear Physics· 14 min read

1. Introduction to the Photoelectric Effect★★☆☆☆⏱ 2 min

The photoelectric effect is the phenomenon where electrons are ejected from a material (typically a metal) when light of sufficient frequency is incident on its surface. Early 20th century experiments showed results that could not be explained by the classical wave model of light: the kinetic energy of ejected electrons depends only on light frequency, not intensity, and no electrons are ejected below a threshold frequency regardless of intensity.

Albert Einstein explained this in 1905 by proposing that light is made of discrete energy packets called photons, rather than a continuous wave. This topic accounts for 10-16% of your total AP Physics 2 exam score, and appears regularly in both multiple-choice and free-response questions, often testing conceptual understanding of the photon model versus classical wave theory.

2. Photon Energy and Quantization of Light★★☆☆☆⏱ 4 min

📘 Definition

Quantization of Light

Light energy is carried in discrete, indivisible packets called photons, rather than distributed continuously as predicted by classical wave theory.

Example:

All photons of light with the same frequency have identical energy, regardless of light intensity.

Every photon of light with frequency $f$ has energy proportional to its frequency, given by:

E = hf = \frac{hc}{\lambda}

where $h = 6.626 \times 1 0^{- 34} J \cdotp s$ is Planck's constant, $c = 3 \times 1 0^{8} m/s$ is the speed of light, and $λ$ is the wavelength of the light. A very useful approximation for AP problems converts $h c$ to electron-volt nanometer units: $h c \approx 1240 eV \cdotp nm$ . This eliminates unit conversions when working with wavelength in nanometers, the most common unit for visible/UV light.

Intensity of light in the photon model is the number of photons incident per unit area per unit time, not the energy per photon. Higher intensity means more photons, not more energetic photons, which is the key difference from the classical wave model.

📐 Worked Example

Calculate the energy of a photon of red light with wavelength 650 nm, give your answer in electron volts.

1
Use the simplified photon energy formula for electron volts and nanometers: $E = \frac{h c}{λ}$
2
Substitute the known values: $h c = 1240 eV \cdotp nm$ , $λ = 650 nm$
3
$E = \frac{1240}{650} \approx 1.91 \text{ eV}$
4
Confirm with SI units to verify the result:
5
$E = \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{650 \times 10^{-9}} \approx 3.06 \times 10^{-19} \text{ J} = \frac{3.06 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.91 \text{ eV}$

Exam tip:

Always memorize $h c = 1240 eV \cdotp nm$ for the AP exam. It cuts down calculation time for photon energy problems by 75% and eliminates unit conversion errors.

3. Einstein's Photoelectric Effect Equation★★★☆☆⏱ 4 min

📘 Definition

Work Function

\Phi

The minimum energy required for an electron to break free of the electrostatic attraction of a metal surface. It is an intrinsic property of the metal, different for every element.

Example:

Cesium has a work function of 2.14 eV, much lower than platinum's work function of ~6.35 eV.

The minimum frequency of photon that can eject an electron is called the threshold frequency $f_{0}$ , where $h f_{0} = Φ$ , so $f_{0} = \frac{Φ}{h}$ . The corresponding maximum wavelength that can eject an electron is threshold wavelength $λ_{0} = \frac{h c}{Φ}$ .

By conservation of energy, the energy of the incoming photon goes into escaping the metal plus the kinetic energy of the ejected electron. For the most loosely bound electrons (which have the highest kinetic energy after ejection), this gives Einstein's photoelectric equation:

hf = \Phi + K_{max}

A key result: increasing the intensity of light (adding more photons) does not change $K_{ma x}$ , it only increases the number of electrons ejected. Only increasing frequency (increasing energy per photon) increases $K_{ma x}$ .

📐 Worked Example

A cesium metal surface has a work function of 2.14 eV. Light of wavelength 400 nm is incident on the surface. Calculate (a) the maximum kinetic energy of ejected electrons, and (b) the threshold wavelength of cesium.

1
First calculate the incident photon energy using the $h c$ shortcut:
2
$E = \frac{1240}{400} = 3.10 \text{ eV}$
3
Use Einstein's equation to find $K_{ma x}$ :
4
$K_{max} = E - \Phi = 3.10 \text{ eV} - 2.14 \text{ eV} = 0.96 \text{ eV}$
5
Calculate threshold wavelength using $λ_{0} = \frac{h c}{Φ}$ :
6
$\lambda_0 = \frac{1240}{2.14} \approx 579 \text{ nm}$
7
Check logic: any wavelength longer than 579 nm has energy less than 2.14 eV, so no electrons are ejected, which matches our 400 nm being shorter than threshold.

Exam tip:

When asked to explain why no electrons are ejected by high-intensity light below threshold frequency, always explicitly state that intensity corresponds to number of photons, not energy per photon; each individual photon still has energy below the work function.

4. Stopping Potential and Graphical Analysis★★★☆☆⏱ 4 min

In the classic photoelectric effect experiment, $K_{ma x}$ is measured experimentally using a reverse potential difference (called the stopping potential $V_{s}$ ) between the metal emitter and a collector plate. The stopping potential is the minimum voltage that stops the most energetic electrons from reaching the collector, so all of the maximum kinetic energy is converted to electric potential energy:

K_{max} = eV_s

Substituting into Einstein's equation gives a linear relationship between $V_{s}$ and incident frequency $f$ , which is used to measure Planck's constant experimentally:

V_s = \left(\frac{h}{e}\right)f - \frac{\Phi}{e}

This is a straight line with slope equal to $\frac{h}{e}$ , which is the same for all metals, and x-intercept equal to the threshold frequency $f_{0}$ . The y-intercept is $- \frac{Φ}{e}$ , so you can calculate the work function directly from the graph.

📐 Worked Example

A student plots stopping potential vs incident frequency for an unknown metal, and finds the line of best fit has a slope of $4.1 \times 1 0^{- 15} V \cdotp s$ and x-intercept at $9.0 \times 1 0^{13} Hz$ . Calculate the work function of the metal from this data.

1
Recall that slope $m = \frac{h}{e}$ , so rearrange to solve for $h$ :
2
$h = m e = (4.1 \times 10^{-15} \text{ V·s})(1.6 \times 10^{-19} \text{ C}) = 6.56 \times 10^{-34} \text{ J·s}$
3
The x-intercept is threshold frequency $f_{0} = 9.0 \times 1 0^{13} Hz$ , so $Φ = h f_{0}$ :
4
$\Phi = (6.56 \times 10^{-34})(9.0 \times 10^{13}) = 5.904 \times 10^{-20} \text{ J}$
5
Convert to electron volts:
6
$\Phi = \frac{5.904 \times 10^{-20}}{1.6 \times 10^{-19}} \approx 0.37 \text{ eV}$

Exam tip:

If a question shows a $V_{s}$ vs $f$ graph for two different metals, the lines are always parallel (same slope = same $h / e$ for all metals); any answer option claiming different slopes for different metals is automatically wrong.

5. Concept Check (AP Style)★★★☆☆⏱ 2 min

✓ Quick check

Test your understanding of core concepts with these AP-style questions:

Blue light photons can eject electrons from a given metal, but green light photons cannot. When the intensity of green light is tripled, which of the following outcomes is correct?
- Three times as many electrons are ejected, each with the same maximum kinetic energy
- No electrons are ejected from the metal
- Electrons are ejected, with maximum kinetic energy one-third that from blue light
- The work function of the metal decreases, allowing electrons to be ejected
Reveal answer
1 —
Ejection only occurs if an individual photon has energy at least equal to the work function. Green photons have energy below the work function, and tripling intensity only increases the number of photons, not energy per photon. Work function is an intrinsic metal property that does not change with incident light.
A student investigates the photoelectric effect with potassium (work function 2.30 eV). (a) Calculate threshold frequency in hertz. (b) Stopping potential is 0.70 V, find $K_{ma x}$ in eV. (c) Calculate incident wavelength in nm.

6. Common Pitfalls

Wrong move:

Calculating maximum kinetic energy by adding the work function to photon energy instead of subtracting.

Why:

Students mix up energy flow, incorrectly thinking the electron receives both the photon energy and the work function to escape.

Correct move:

Always write the full energy conservation statement $E_{p h o t o n} = Energy to escape + K_{ma x}$ before rearranging to solve for $K_{ma x}$ .

Wrong move:

Stating that increasing light intensity increases the maximum kinetic energy of ejected electrons.

Why:

Confuses classical wave theory predictions with the photon model, mixing up intensity (number of photons) and energy per photon.

Correct move:

Remember $K_{ma x}$ depends only on incident light frequency, not intensity; increasing intensity only increases the number of ejected electrons.

Wrong move:

Using nanometers for wavelength directly in SI unit calculations without converting to meters.

Why:

The $h c = 1240 eV \cdotp nm$ shortcut works for nanometers, but students accidentally use nanometers when calculating energy in joules.

Correct move:

If working in SI units, always convert wavelength from nanometers to meters by multiplying by $1 0^{- 9}$ ; use the 1240 eV·nm shortcut for electron volt calculations.

Wrong move:

Taking the y-intercept of a $V_{s}$ vs $f$ graph as the work function directly.

Why:

Forgets the $1/ e$ factor in the linear relation between $V_{s}$ and $f$ .

Correct move:

The y-intercept is $- Φ/ e$ , so multiply the absolute value of the y-intercept by $e$ to get the work function.

Wrong move:

Claiming photons have no mass so they have no energy.

Why:

Misapplies classical mass-energy relations to relativistic photons.

Correct move:

Photons have zero rest mass but carry discrete energy $E = h f$ , which is experimentally confirmed by the photoelectric effect.

7. Quick Reference Cheatsheet

Category	Formula	Notes
Photon Energy	$E = h f = \frac{h c}{λ}$	$h = 6.626 \times 1 0^{- 34} J \cdotp s$ ; $h c \approx 1240 eV \cdotp nm$ for $λ$ in nm
Threshold Frequency	$f_{0} = \frac{Φ}{h}$	Minimum frequency that can eject electrons from a metal
Threshold Wavelength	$λ_{0} = \frac{h c}{Φ}$	Maximum wavelength that can eject electrons from a metal
Einstein Photoelectric Equation	$h f = Φ + K_{ma x}$	Conservation of energy; $K_{ma x}$ = maximum kinetic energy of ejected electrons
Stopping Potential Relation	$K_{ma x} = e V_{s}$	$V_{s}$ is reverse voltage that stops all electrons from reaching the collector
Linear Graph Relation	$V_{s} = (\frac{h}{e}) f - \frac{Φ}{e}$	Slope = $h / e$ (same for all metals); x-intercept = $f_{0}$ ; y-intercept = $- Φ/ e$
Energy Unit Conversion	$1 eV = 1.6 \times 1 0^{- 19} J$	Use to convert between joules and electron volts

8. Frequently Asked

Why can't high intensity low frequency light eject electrons?

Each photon interacts with one individual electron; intensity only increases the number of photons, not the energy per photon. If each individual photon has energy less than the work function, no electron can gain enough energy to escape, no matter how many photons hit the surface.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · AP Physics 2
Photoelectric effect graph interpretation
2022 · AP Physics 2
Calculate stopping potential for unknown metal

Going deeper

unit overviewUnit 7 Quantum, Atomic, and Nuclear Physics Overview

What's Next

This topic is the foundational introduction to quantum mechanics for AP Physics 2, and all subsequent topics in Unit 7 build on the core idea of photon energy quantization. Next you will apply this concept to wave-particle duality, extending the relations between energy, frequency, and wavelength to matter waves. Without mastering the photon model and photoelectric effect energy relations, you will not be able to correctly solve problems involving atomic energy level transitions or Compton scattering, common AP exam questions. This topic also establishes the quantum framework that underpins all later nuclear physics topics.