Topic 15.5: The Photoelectric Effect

What Topic 15.5 requires

Topic 15.5 belongs to Unit 15, Modern Physics, which the CED weights at 12-15% of the multiple-choice section and estimates at about 14 to 22 class periods. The unit opener says Unit 15 "lays the groundwork for the study of modern physics by resolving the conflicts and unanswered questions from Units 13 and 14", and it names this topic by name: students "will also revisit the wave-particle duality of light through their investigations of phenomena such as the photoelectric effect."

15.5.A, describe an interaction between photons and matter using the photoelectric effect. One learning objective, three numbered essential knowledge statements, five sub-statements.

• 15.5.A.1 The photoelectric effect is the emission of electrons when electromagnetic radiation is incident upon a photoactive material.
• 15.5.A.2 That emission requires a minimum frequency of incident light, called the threshold frequency.
• 15.5.A.2.i Light at or above the threshold frequency induces electron emission "regardless of the number of photons that strike the material".
• 15.5.A.2.ii The energy of the emitted electrons does not depend on the number of incident photons, "which provides evidence that light is a collection of discrete, quantized energy packets called photons".
• 15.5.A.3 The maximum kinetic energy of an emitted electron is related to the frequency of the incident light and to the work function of the material, $\phi$.
• 15.5.A.3.i The work function is the minimum energy required to emit an electron from atoms in the material.
• 15.5.A.3.ii prints the equation $K_{\text{max}} = hf - \phi$.
• 15.5.A.3.iii describes the apparatus: two metal plates in a vacuum chamber connected to a variable source of potential difference, one plate illuminated by monochromatic light so that electrons are ejected, and the potential difference "adjusted until no current is measured in the circuit".

The boundary statement, whole: "Where applicable, work functions for materials will be provided on the exam; students are not expected to know values of work functions or variables of a material that influence the magnitude of its work function."

Both halves matter. The first says a number will be handed to you, so a memorised table of work functions buys you nothing. The second fences off why one metal holds its electrons more tightly than another, so no question can ask you to rank two metals except from data it supplies.

The CED lists five suggested skills here. Six of the unit's eight topics list four; only 15.5 and 15.7 list five.

• 1.B Create quantitative graphs with appropriate scales and units, including plotting data.
• 2.A Derive a symbolic expression from known quantities by selecting and following a logical mathematical pathway.
• 2.D Predict new values or factors of change of physical quantities using functional dependence between variables.
• 3.A Create experimental procedures that are appropriate for a given scientific question.
• 3.C Justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws.

Read that list as the topic's exam shape: 1.B says a graph, 2.A says a symbolic derivation, 2.D says a question with no numbers in it, 3.A says design the measurement, 3.C says defend the answer. The unit's own Building the Science Practices note uses this topic as its worked illustration, suggesting students "describe conceptually what happens to the maximum kinetic energy of ejected electrons from a metal plate if the plate is replaced by a plate with a higher work function, and then justify what impact that change will have on the required stopping potential.".

The classical prediction, and where it breaks

This is the topic where a good model fails in public, so the contrast is the lesson. The CED does not print the classical prediction anywhere in 15.5. It prints the two results that kill it, 15.5.A.2.i and 15.5.A.2.ii, and leaves you to supply the model they contradict.

Treat light as a classical wave. Its energy is spread continuously across the wavefront, and brightness means more energy arriving per second on each square metre. An electron sitting in the metal is held there by some binding energy, and the wave pours energy in until it has enough to escape. Two predictions follow immediately.

1. Any colour should work. Frequency sets how fast the field oscillates, not how much energy arrives. Turn a dim red lamp up far enough, or leave it on long enough, and electrons should eventually come off.
2. Brighter light should mean faster electrons. More energy per second per electron means more energy left over after escaping, so the ejected electrons should speed up as you turn up the lamp.

Both are wrong, and the CED states the corrections as required content.

One change of model fixes all three rows. Light does not deliver its energy as a spread-out wave; it delivers it in discrete packets, each carrying $E = hf$, and one packet is absorbed by one electron. Then the arithmetic is forced. Whether an electron escapes depends on what a single packet carries, which is fixed by frequency alone. How many escape depends on how many packets arrive, which is what brightness controls. The two questions come apart because energy and count are carried by different variables, and this is exactly what 15.5.A.2.ii means when it calls the observation "evidence that light is a collection of discrete, quantized energy packets called photons".

A third classical failure is often taught alongside these, the absence of any measurable delay while a dim beam accumulates enough energy. It is real physics, but it is not in the CED's essential knowledge for this topic, so do not cite it as required content.

Work function, threshold frequency, and why the maximum is a maximum

Three quantities, one equation, printed in 15.5.A.3.ii and again in the Modern Physics group of the equation sheet:

The work function $\phi$ is defined by 15.5.A.3.i as the minimum energy required to emit an electron from atoms in the material. It is a property of the material, it is a positive energy, and questions quote it in electron volts.

The threshold frequency $f_0$ is not on the sheet and has no equation of its own. You produce it by asking when the equation stops having a physical solution. Set $K_{\text{max}} = 0$:

Below $f_0$ the formula returns a negative kinetic energy, which is not a slow electron but no electron. The graph section below returns to this, because reading the extrapolated part of a line as data is one of the standard ways to lose a mark here.

The threshold wavelength follows from the sheet's $\lambda = c/f$:

and this is where the constants block earns its keep. The sheet prints $hc = 1240 \, \text{eV} \cdot \text{nm}$, so a work function in electron volts divided into 1240 gives a wavelength in nanometres with no unit conversion at all. Note the direction: a larger work function means a shorter threshold wavelength, because a tightly bound electron needs a more energetic photon and more energetic photons are the short ones.

Why "maximum". The CED's own wording chains the two words together. 15.5.A.3.i calls $\phi$ the minimum energy required to get an electron out, so an electron that costs exactly the minimum keeps the largest possible remainder. Any electron that costs more than the minimum keeps less. So $K_{\text{max}} = hf - \phi$ is a ceiling on a spread of ejected electron energies, not a single value they all share, and that is the whole reason the stopping-potential measurement in 15.5.A.3.iii works the way it does: you raise the retarding voltage until even the fastest electron fails to make the crossing.

Number of photons changes the current, never the maximum kinetic energy

Notice a wording choice in the CED. Both 15.5.A.2.i and 15.5.A.2.ii are written in terms of the number of photons, not in terms of intensity. That is deliberate and it is easier to reason with, because it separates two things that "brightness" runs together: how much energy each arriving packet carries, and how many packets arrive per second.

Hold that split and every functional-dependence question in this topic answers itself.

Row one is the headline result of the topic. More photons means more absorption events means more electrons per second, so the ammeter reading climbs. Each electron still absorbed one photon of energy $hf$ and still paid at least $\phi$ to escape, so the fastest electron is no faster than it was. Nothing in $K_{\text{max}} = hf - \phi$ contains a photon count.

Row two is the same equation read the other way. Raise $f$ and every packet carries more, so every electron keeps more. This is the only way to make the electrons faster.

Row three is the one students soften into "the electrons come off slowly". They do not come off. 15.5.A.2 makes the threshold a condition for emission, not a condition for fast emission.

Row four is the CED's own illustration, taken from the Unit 15 science-practices note: swap in a plate with a higher work function and the same light now leaves less over. If the new work function exceeds the photon energy entirely, the current stops.

A useful sanity check on any answer you write here: the words "bright", "intense" and "how many" belong in the current column, and the words "colour", "frequency" and "wavelength" belong in the energy column. A justification that puts one in the other column is wrong however good the algebra beside it looks.

The stopping potential, and why it is not on the equation sheet

Essential knowledge 15.5.A.3.iii describes a real measurement rather than a thought experiment: two metal plates in a vacuum chamber, wired to a variable source of potential difference, one of them lit by monochromatic light that ejects electrons, and the potential difference "adjusted until no current is measured in the circuit". The voltage at that cutoff is the stopping potential.

Convention, declared once and held to the end of this page: $V_{\text{stop}}$ is a positive magnitude, the size of the retarding potential difference that just kills the current. The sign of the plate that the electrons are climbing toward is not something this page tracks, and the CED does not ask you to.

Now the part worth knowing: there is no stopping-potential equation on the AP Physics 2 equation sheet. The Modern Physics group has ten entries, and counting through them, not one mentions a potential difference. You build the relation yourself, which is what suggested skill 2.A is for, and the piece you need is in the Electricity group instead: $\Delta U_E = q \Delta V$.

An electron carries charge of magnitude $e$. Crossing a retarding potential difference of magnitude $V_{\text{stop}}$ costs it $e V_{\text{stop}}$ of kinetic energy. At the cutoff, the fastest electron arrives with nothing left, so the energy it started with is exactly the energy the field took away:

Divide through by $e$ and the graph form appears:

The electron-volt shortcut. The sheet prints $1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J}$ and $e = 1.60 \times 10^{-19} \, \text{C}$, the same digits, because that is how the electron volt is defined. So if you carry $K_{\text{max}}$ in electron volts, the stopping potential in volts is the same number. A maximum kinetic energy of $1.56 \, \text{eV}$ is stopped by $1.56 \, \text{V}$. Nothing to compute, and it is a fast way to check that a joules-and-coulombs answer came out right.

That shortcut is also the reason to keep photoelectric work in electron volts from the start. Work functions arrive in electron volts, $hc = 1240 \, \text{eV} \cdot \text{nm}$ produces photon energies in electron volts, and the stopping potential falls out with no conversion. Convert to joules only when a question asks for a speed, since $K = \tfrac{1}{2} m v^2$ needs kilograms and joules. Speed and kinetic energy in AP Physics 1 terms are covered in Topic 3.1, and the potential-difference idea this derivation leans on is Topic 10.5.

Reading the graph: the slope is Planck's constant

Suggested skill 1.B asks for quantitative graphs with scales and plotted data, and this topic has exactly one graph worth knowing, in two dresses.

Plot $K_{\text{max}}$ against $f$. Compare $K_{\text{max}} = hf - \phi$ with $y = mx + b$:

• slope $= h$, Planck's constant, in $\text{J} \cdot \text{s}$ or $\text{eV} \cdot \text{s}$ depending on the axis units,
• vertical intercept $= -\phi$, the negative of the work function,
• horizontal intercept $= f_0$, the threshold frequency.

Plot $V_{\text{stop}}$ against $f$. Same line divided by $e$:

• slope $= h/e$, which is $4.14 \times 10^{-15} \, \text{V} \cdot \text{s}$ using the sheet's $h = 4.14 \times 10^{-15} \, \text{eV} \cdot \text{s}$,
• vertical intercept $= -\phi/e$,
• horizontal intercept $= f_0$ again, unchanged, because dividing by a constant cannot move a zero.

Four readings are worth carrying into an exam.

The slope does not depend on the material. Every metal gives a line of the same slope, so two metals plot as two parallel lines and the one further right has the larger work function. Skill 2.C questions about two plates are usually this picture in words. If your fitted slopes come out different for two metals, the data or the fit is wrong, not the physics.

The vertical intercept is $-\phi$, not $\phi$. Read the sign. A work function is a positive energy and the intercept is below the axis.

The line below $f_0$ is extrapolation, not data. No electrons come off there, so there is nothing to plot. Real data starts at $f_0$ and runs right. The negative-$K$ segment exists only to locate the intercept.

The horizontal intercept is the cleanest way to read $\phi$. Multiply $f_0$ by the slope and you have the work function without needing to trust a distant $y$-intercept read off a cramped scale.

The CED builds a task on precisely this. Its skill 2.B sample activity hands students the finished arithmetic

and asks them to work backwards to another representation, naming "a graph of stopping potential as a function of frequency" and "an energy bar chart for the ejected electrons" among the options. The numbers in it are the ones the first worked example below uses.

What the equation sheet actually gives you

The AP Physics 2 sheet is a different sheet from the AP Physics 1 one, and its Modern Physics group is where this topic lives. Counting the group rather than remembering it: it holds ten entries, and three of them belong to 15.5.

• $E = hf$, the photon energy.
• $K_{\text{max}} = hf - \phi$, the photoelectric equation itself.
• $\lambda = c/f$, which is how a wavelength in the question becomes a frequency in the equation.

The constants block supplies the rest: $h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s} = 4.14 \times 10^{-15} \, \text{eV} \cdot \text{s}$, $hc = 1.99 \times 10^{-25} \, \text{J} \cdot \text{m} = 1240 \, \text{eV} \cdot \text{nm}$, $c = 3.00 \times 10^8 \, \text{m/s}$, $m_e = 9.11 \times 10^{-31} \, \text{kg}$, $e = 1.60 \times 10^{-19} \, \text{C}$ and $1 \, \text{eV} = 1.60 \times 10^{-19} \, \text{J}$. The symbol table beside that column defines $\phi$ as the work function.

Not printed anywhere on the sheet: a stopping-potential relation, a threshold-frequency formula, and any value of any work function. The first you derive from $\Delta U_E = q \Delta V$, the second you derive by setting $K_{\text{max}}$ to zero, and the third the exam hands you, which the 15.5 boundary statement puts in writing. The CED says the same thing in general terms in its course-framework overview: not all equations in the framework appear on the equation sheet provided during the exam.

One precision warning, because modern-physics problems are where rounding shows. The sheet's constants are each rounded on their own, so two routes to the same quantity disagree in the third digit. Multiplying the printed $h = 4.14 \times 10^{-15} \, \text{eV} \cdot \text{s}$ by the printed $c = 3.00 \times 10^8 \, \text{m/s}$ gives $1242 \, \text{eV} \cdot \text{nm}$, while the sheet prints $hc = 1240 \, \text{eV} \cdot \text{nm}$. The printed $hc$ line is the better number, so when a wavelength is given, use it directly rather than rebuilding it. Two threshold frequencies computed the two ways will differ by about 0.2%, and that is rounding, not an error to hunt down.

Traps, and how 15.5 gets tested

The CED's sample free-response Question 4 is a photoelectric question. In the Exam Information section it reads: "Monochromatic light of wavelength $\lambda_0$ is incident on a metal with work function $\phi$. Electrons are ejected from the metal and have a maximum speed $v_0$." Part A asks whether longer-wavelength light gives a smaller, equal or greater maximum speed, with justification. Part B asks for a symbolic derivation of $v_0$. Part C asks whether the two agree. Its scoring guidelines are printed too, worth 8 points, listing learning objectives 15.1.A and 15.5.A.

Three details in that rubric are worth copying into your own habits.

• A point is awarded "For using $K_{\text{max}} = hf - \phi$ with an attempt to substitute for wavelength", before any answer exists. Writing the starting equation is scored.
• A separate point is awarded "For a correct substitution of $f = \frac{c}{\lambda}$". The unit-conversion step is paid for on its own.
• Parts A and C both demand "conceptual reasoning beyond algebraic solutions". A correct formula with no sentence attached does not collect those points.

Note also what the question hands over: $\phi$ is given, exactly as the boundary statement promises, and $\lambda_0$ is given as a symbol rather than a number. The whole question is symbolic. Question 4 on the AP Physics 2 exam is the Qualitative/Quantitative Translation question, and the exam runs 3 hours with 42 multiple-choice questions and 4 free-response questions, split 50/50 by weight.

Now the errors, each paired with the check that catches it.

• "Brighter light gives faster electrons." The single most direct contradiction of 15.5.A.2.ii. Check: does your reasoning use a photon count to change an energy? If so it is wrong.
• "Below threshold the electrons just come off slowly." They do not come off. A negative $K_{\text{max}}$ from the formula means no emission, not a slow electron.
• Mixing joules and electron volts. $hf$ in joules minus $\phi$ in electron volts is the most common arithmetic wreck in this topic. Pick electron volts, keep everything there, and convert once at the end only if a speed is wanted.
• Reading the intercept as $+\phi$. The vertical intercept of $K_{\text{max}}$ against $f$ is $-\phi$.
• Letting the slope depend on the metal. It cannot. It is $h$.
• Reaching for a remembered work function. The boundary statement says the exam provides them.
• Answering a 2.D question with numbers. "What happens to the stopping potential if the plate is replaced by one with a higher work function" wants a direction and a reason, and the reason is the shape of $K_{\text{max}} = hf - \phi$.

For the photon model this topic rests on, see Topic 15.1. For the other experiment the CED uses as evidence for photons, this time with momentum rather than energy doing the work, see Topic 15.6. Photon absorption and emission by atoms, where the energy goes into a level rather than a free electron, is Topic 15.3. The Unit 15 overview puts the eight topics in order, and the AP Physics 2 equation sheet shows the ten Modern Physics entries you work from.