Photons and the Photoelectric Effect — AP Physics 2 Study Guide

For: AP Physics 2 candidates sitting AP Physics 2.

Covers: Photon energy quantization, Planck's relation, work function, Einstein's photoelectric effect equation, threshold frequency/wavelength, stopping potential, and graphical analysis of photoelectric effect data.

You should already know: Wave properties of light including frequency and wavelength relationships. Conservation of energy for closed systems. Electric potential energy and potential difference.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics 2 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Photons and the Photoelectric Effect?

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 is part of Unit 7: Quantum, Atomic, and Nuclear Physics, which accounts for 10-16% of the total AP Physics 2 exam score. It appears regularly in both multiple-choice questions (MCQ) and as a short multi-part section of free-response questions (FRQ), and often tests conceptual understanding of the photon model versus classical wave theory.

2. Photon Energy and Quantization of Light

Quantization of light means that light energy is carried in discrete, indivisible packets called photons, rather than distributed continuously as in the classical wave model. Every photon of light with frequency $f$ has exactly the same amount of energy, proportional to its frequency: $$E=hf=\frac{hc}{\lambda }$$ where $h=6.626\times 10^{-34}\text{ J}\cdot \text{s}$ is Planck's constant, $c=3\times 10^8\text{ m/s}$ is the speed of light, and $\lambda$ is the wavelength of the light. A very useful approximation for AP problems converts $hc$ to electron-volt nanometer units: $hc\approx 1240\text{ eV}\cdot \text{nm}$. 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: $E=\frac{hc}{\lambda }$.
2. Substitute the known values: $hc=1240\text{ eV}\cdot \text{nm}$, $\lambda =650\text{ nm}$.
3. Calculate: $E=\frac{1240}{650}\approx 1.91\text{ eV}$.
4. Confirm with SI units: $\lambda =650\times 10^{-9}\text{ m}$, $E=\frac{(6.626\times 10^{-34})(3\times 10^8)}{650\times 10^{-9}}\approx 3.06\times 10^{-19}\text{ J}=\frac{3.06\times 10^{-19}}{1.6\times 10^{-19}}\approx 1.91\text{ eV}$, which matches.

Exam tip: Always memorize $hc=1240\text{ eV}\cdot \text{nm}$ for the AP exam. It cuts down calculation time for photon energy problems by 75% and eliminates unit conversion errors.

3. Key Definitions and Einstein's Photoelectric Equation

When a photon hits a metal surface, it is fully absorbed by a single electron. The electron needs a minimum amount of energy to break free of the metal's electrostatic attraction; this minimum energy is called the work function $\Phi$, which is an intrinsic property of the metal, different for every element. The minimum frequency of photon that can eject an electron is called the threshold frequency $f_0$, where $h{f}_0=\Phi$, so $f_0=\frac{\Phi }{h}$. The corresponding maximum wavelength that can eject an electron is threshold wavelength ${\lambda }_0=\frac{hc}{\Phi }$.

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: $$hf=\Phi +K_{max}$$ where $K_{max}$ is the maximum kinetic energy of ejected electrons. Any electron bound more tightly will lose more energy escaping, so their kinetic energy will be lower than $K_{max}$. A key result: increasing the intensity of light (adding more photons) does not change $K_{max}$, it only increases the number of electrons ejected. Only increasing frequency (increasing energy per photon) increases $K_{max}$.

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: $E=\frac{1240}{400}=3.10\text{ eV}$.
2. Use Einstein's equation to find $K_{max}$: $K_{max}=E-\Phi =3.10\text{ eV}-2.14\text{ eV}=0.96\text{ eV}$.
3. Calculate threshold wavelength: ${\lambda }_0=\frac{hc}{\Phi }=\frac{1240}{2.14}\approx 579\text{ nm}$.
4. 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

In the classic photoelectric effect experiment, $K_{max}$ 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: $$K_{max}=e{V}_s$$ where $e$ is the elementary charge ($1.6\times 10^{-19}\text{ C}$).

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 $-\frac{\Phi }{e}$, 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 $4.1\times 10^{-15}\text{ V}\cdot \text{s}$ and x-intercept at $9.0\times 10^{13}\text{ Hz}$. Calculate the work function of the metal from this data.

1. Recall that slope $m=\frac{h}{e}$, so $h=me=(4.1\times 10^{-15}\text{ V}\cdot \text{s})(1.6\times 10^{-19}\text{ C})=6.56\times 10^{-34}\text{ J}\cdot \text{s}$.
2. The x-intercept is threshold frequency $f_0=9.0\times 10^{13}\text{ Hz}$, so $\Phi =h{f}_0$.
3. Calculate $\Phi =(6.56\times 10^{-34})(9.0\times 10^{13})=5.904\times 10^{-20}\text{ J}$.
4. Convert to electron volts: $\Phi =\frac{5.904\times 10^{-20}}{1.6\times 10^{-19}}\approx 0.37\text{ eV}$.

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.

Common Pitfalls (and how to avoid them)

• 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 $E_{photon}=Energytoescape+K_{max}$ before rearranging to solve for $K_{max}$.
• 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 $K_{max}$ 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 $hc=1240eV\cdot nm$ 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 $10^{-9}$; 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 $-\Phi /e$, 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 $E=hf$, which is experimentally confirmed by the photoelectric effect.

Practice Questions (AP Physics 2 Style)

Question 1 (Multiple Choice)

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? A) Three times as many electrons are ejected, each with the same maximum kinetic energy B) No electrons are ejected from the metal C) Electrons are ejected, with maximum kinetic energy one-third that from blue light D) The work function of the metal decreases, allowing electrons to be ejected

Worked Solution: Ejection of electrons only occurs if the energy of an individual photon is at least equal to the work function of the metal. Green light photons have frequency lower than the threshold frequency for this metal, so every individual green photon has energy less than $\Phi$, regardless of how many green photons are present. Tripling intensity only triples the number of photons per second, not the energy per photon, and the work function is an intrinsic property of the metal that does not change with incident light. Therefore, no electrons can be ejected. Correct answer: B.

Question 2 (Free Response)

A student investigates the photoelectric effect with a clean potassium metal surface, which has a work function of 2.30 eV. (a) Calculate the threshold frequency of potassium, in hertz. (b) When monochromatic light is incident on the surface, the stopping potential is measured to be 0.70 V. What is the maximum kinetic energy of ejected electrons, in electron volts? (c) Calculate the wavelength of the incident light, in nanometers.

Worked Solution: (a) Threshold frequency follows $h{f}_0=\Phi$, so rearrange to get $f_0=\frac{\Phi }{h}$. Convert $\Phi$ to joules: $\Phi =2.30\text{ eV}\times 1.6\times 10^{-19}\text{ J/eV}=3.68\times 10^{-19}\text{ J}$. Substitute values: $f_0=\frac{3.68\times 10^{-19}}{6.626\times 10^{-34}}\approx 5.55\times 10^{14}\text{ Hz}$.

(b) Stopping potential relates to maximum kinetic energy as $K_{max}=e{V}_s$. For $V_s=0.70\text{ V}$, $K_{max}=e(0.70\text{ V})=0.70\text{ eV}$, no unit conversion needed.

(c) Use Einstein's equation: $\frac{hc}{\lambda }=\Phi +K_{max}$. The right-hand side is $2.30\text{ eV}+0.70\text{ eV}=3.00\text{ eV}$. Use $hc=1240\text{ eV}\cdot \text{nm}$, so $\lambda =\frac{1240}{3.00}\approx 413\text{ nm}$.

Question 3 (Application / Real-World Style)

Digital camera sensors use silicon photodiodes, which generate an electrical signal when a photon with energy greater than silicon's 1.12 eV band gap is absorbed. What is the longest wavelength of light that can trigger a signal in a silicon photodiode? The human eye can detect light from ~400 nm (violet) to ~700 nm (red). Is all visible light detectable by a silicon sensor? Explain your answer in context.

Worked Solution: The longest detectable wavelength corresponds to the minimum energy equal to the band gap, so ${\lambda }_{max}=\frac{hc}{E_{gap}}=\frac{1240\text{ eV}\cdot \text{nm}}{1.12\text{ eV}}\approx 1107\text{ nm}$. All visible light wavelengths fall between 400 nm and 700 nm, which are both shorter than 1107 nm. Shorter wavelength means higher photon energy, so all visible light photons have energy greater than 1.12 eV. Therefore, the entire visible spectrum is detectable by a silicon photodiode, which is why silicon is the standard material for consumer digital camera sensors.

Quick Reference Cheatsheet

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 (de Broglie wavelength). 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, which are common AP exam questions. This topic also establishes the quantum framework that underpins all nuclear physics topics later in the unit.

• Wave-Particle Duality of Light and Matter
• Atomic Energy Levels and Spectra
• Nuclear Reactions and Binding Energy