AP Physics 2 · Topic 15.4

Topic 15.4: Blackbody Radiation

Unit 15: Modern Physics12-15% of the multiple-choice section

Every object above absolute zero radiates, and the spectrum it radiates depends only on its temperature. Two printed equations run the topic: Wien's law puts the peak wavelength at b over T, and the Stefan-Boltzmann law puts the total emitted power at area times sigma times T to the fourth.

AP Physics: Unit 15 (topics 15.4 Blackbody Radiation). AP Physics 2 Unit 15, Topic 15.4. One learning objective, 15.4.A, describe the electromagnetic radiation emitted by an object due to its temperature. Three essential knowledge statements: 15.4.A.1 (matter will spontaneously convert some of its internal thermal energy into electromagnetic energy); 15.4.A.2 (a blackbody is an idealized model of matter that absorbs all radiation that falls on the body, and if the body is in equilibrium at a constant temperature then it must in turn emit energy); and 15.4.A.3 (a blackbody will emit a continuous spectrum that only depends on the body's temperature, often modeled by plotting intensity per unit wavelength as a function of wavelength), with 15.4.A.3.i (the distribution cannot be modeled using only classical physics concepts; a blackbody's spectrum is described by Planck's law, which assumes that the energy of light is quantized), 15.4.A.3.ii (the peak wavelength decreases with increasing temperature, as described by Wien's law; relevant equation lambda_max = b/T) and 15.4.A.3.iii (the rate at which energy is emitted is proportional to surface area and to temperature to the fourth power, as described by the Stefan-Boltzmann law; relevant equation P = A sigma T^4). The topic has NO boundary statement. Suggested skills: 1.C, 2.C, 3.B, 3.C. 1.C is listed for no other topic in Unit 15. Two of the ten equations in the Modern Physics group of the AP Physics 2 equation sheet belong here, together with the printed constants b = 2.90e-3 m K and sigma = 5.67e-8 W/(m^2 K^4). Planck's law itself is named in 15.4.A.3.i but is printed neither in the CED nor on the equation sheet. The phrase ultraviolet catastrophe appears only in an essential question on the unit opener and in no essential knowledge statement. The CED's sample multiple-choice question 11 is aligned to 15.4.A, 15.4.A.3 and skill 2.D, and asks for the new emitted power when a blackbody's peak wavelength is halved; the published answer is sixteen times the original.

What Topic 15.4 requires

Topic 15.4 carries a single learning objective, 15.4.A: describe the electromagnetic radiation emitted by an object due to its temperature. Three numbered essential knowledge statements sit under it, and the topic prints no boundary statement.

15.4.A.1 Matter will spontaneously convert some of its internal thermal energy into electromagnetic energy.
15.4.A.2 A blackbody is an idealized model of matter that absorbs all radiation that falls on the body. If the body is in equilibrium at a constant temperature, then it must in turn emit energy.
15.4.A.3 A blackbody will emit a continuous spectrum that only depends on the body's temperature. The radiation emitted by a blackbody is often modeled by plotting intensity per unit wavelength as a function of wavelength.
15.4.A.3.i The distribution of the intensity of a blackbody's spectrum as a function of temperature cannot be modeled using only classical physics concepts. A blackbody's spectrum is described by Planck's law, which assumes that the energy of light is quantized.
15.4.A.3.ii The peak wavelength emitted by a blackbody (the wavelength at which the blackbody emits the greatest amount of radiation per unit wavelength) decreases with increasing temperature, as described by Wien's law. Relevant equation: $\lambda_{max} = \frac{b}{T}$ .
15.4.A.3.iii The rate at which energy is emitted (power) by a blackbody is proportional to the surface area of the body and to the temperature of the body raised to the fourth power, as described by the Stefan-Boltzmann law. Relevant equation: $P = A \sigma T^4$ .

Suggested skills: 1.C (create qualitative sketches of graphs that represent features of a model or the behavior of a physical system), 2.C (compare physical quantities between two or more scenarios or at different times and locations in a single scenario), 3.B (apply an appropriate law, definition, theoretical relationship, or model to make a claim), and 3.C (justify or support a claim using evidence from experimental data, physical representations, or physical principles or laws).

Two things about that skill list are worth noticing. 1.C is unique to this topic in Unit 15: no other topic in the unit lists it. That is a strong hint that the intensity-against-wavelength curve is something you may be asked to sketch. And the list contains no 2.B, so the emphasis is on comparison and justification rather than on grinding out a number, though the two printed equations obviously support calculation too.

A blackbody is defined by what it absorbs

15.4.A.2 defines the object: a blackbody is an idealized model of matter that absorbs all radiation that falls on the body. If the body is in equilibrium at a constant temperature, then it must in turn emit energy.

The definition is about absorption, and the emission is a deduction. Follow the second sentence carefully, because it is the whole argument:

The body absorbs every bit of radiation that reaches it. Energy is going in.
It is in equilibrium at a constant temperature, so its internal energy is not changing.
Therefore energy must be leaving at the same rate it arrives.
The only channel available is radiation, so the body must emit.

That is a conservation-of-energy argument, of exactly the kind Unit 9 trained you on, and it is a ready-made skill 3.B answer to "why must a perfect absorber also be a perfect emitter".

The word idealized matters too. No real object absorbs everything; the blackbody is a model, in the same sense that an ideal gas or a frictionless surface is a model. The CED never asks you to correct for a real surface, and no emissivity factor appears in the printed Stefan-Boltzmann equation.

15.4.A.1 is broader than the blackbody itself: matter will spontaneously convert some of its internal thermal energy into electromagnetic energy. Not just hot matter, and not just idealized matter. Everything with a nonzero temperature radiates, all the time, without being told to. That is thermal radiation, one of the three transfer processes Topic 9.3 lists alongside conduction and convection, and it is the one that needs no medium at all, because electromagnetic waves do not need one (14.4.A.2 in Topic 14.4).

The curve, and what the axes mean

15.4.A.3 sets up the representation the whole topic is built on: a blackbody will emit a continuous spectrum that only depends on the body's temperature, and that radiation is often modeled by plotting intensity per unit wavelength as a function of wavelength.

Three claims are packed in there.

Continuous. Not lines. This is the direct contrast with Topic 15.3, where atomic transitions produce a photon of a single wavelength each and the spectrum is a set of discrete lines. A blackbody puts out some radiation at every wavelength. If you are shown a spectrum and asked to identify what produced it, lines mean atomic transitions and a smooth hump means thermal emission.

Only depends on temperature. Not on what the body is made of, not on its shape, not on its history. Two blackbodies at the same temperature have identical spectra. That is a strong statement and it is what makes the peak wavelength usable as a thermometer.

Intensity per unit wavelength on the vertical axis. Not intensity, not power, not energy. The quantity plotted is how much intensity falls in each small slice of wavelength, which is why the peak has a location at all. Getting the axis label right is part of skill 1.C.

The shape to be able to draw, and the features that carry the marks:

The curve starts at zero at very short wavelength, rises to a single peak, and falls away with a long tail toward long wavelengths. It is not symmetric: the short-wavelength side is steep, the long-wavelength side is a slow decline.
The peak sits at $\lambda_{max} = b/T$ , so hotter means the peak is further left.
A hotter body's curve is drawn above the cooler body's at every wavelength, so that the area under it is larger. That is consistent with both printed equations at once: the peak has moved left, per Wien, and the total emitted power has gone up as $T^4$ , per Stefan-Boltzmann.
Since the vertical axis is intensity per unit wavelength, the area under the curve corresponds to the total emitted intensity. The CED does not put it in those words, and AP Physics 2 is algebra-based so no integral is ever required, but that reading is what connects the shape of the graph to $P = A \sigma T^4$ .

A sketch that gets the peak in the right place and the hotter curve entirely above the cooler one has said everything 1.C can reasonably ask for.

Why classical physics fails here (15.4.A.3.i)

This is the statement that puts blackbody radiation in a modern physics unit rather than in the thermodynamics unit, and it is worth having whole: the distribution of the intensity of a blackbody's spectrum as a function of temperature cannot be modeled using only classical physics concepts. A blackbody's spectrum is described by Planck's law, which assumes that the energy of light is quantized.

Note what the CED commits to and what it does not.

It commits to the failure of classical physics for this distribution, to the existence of Planck's law as the correct description, and to the reason Planck's law works, namely that it assumes the energy of light is quantized.

It does not print Planck's law. The formula appears nowhere in the CED and nowhere on the equation sheet. You are not expected to write it, use it, or recognise it algebraically. What you are expected to know is that a quantization assumption is what fixes the classical failure.

That single sentence is the load-bearing one for the whole unit. It is the first place quantization is introduced as the answer to a measurement, and everything in Topic 15.1 about photons is downstream of it. It is also why Topic 15.5 lands so hard: once a quantization assumption has already rescued one experiment, a second experiment demanding the same assumption stops looking like a coincidence.

One term to handle carefully. The unit opener lists, among its four essential questions, "how does the ultraviolet catastrophe link thermodynamics and modern physics?" That phrase appears in the essential question and nowhere else in the AP Physics 2 CED. There is no essential knowledge statement defining it, and no boundary statement mentioning it. So it is fair to meet the phrase in class and unwise to build an exam answer on a definition the CED never gives. If you need to describe the failure, use the CED's own wording from 15.4.A.3.i: classical physics concepts alone cannot model the distribution, and Planck's law, which assumes quantized light energy, can.

Wien's law: where the peak goes

15.4.A.3.ii states it with an unusually careful parenthesis, and the parenthesis is the part to keep: the peak wavelength emitted by a blackbody (the wavelength at which the blackbody emits the greatest amount of radiation per unit wavelength) decreases with increasing temperature, as described by Wien's law.

\lambda_{max} = \frac{b}{T}

The constant is printed in the Table of Information as Wien's constant, $b = 2.90 \times 10^{-3}$ m K. Two practical points about it:

The units are metres kelvin. Divide by a temperature in kelvin and you get a wavelength in metres, not nanometres. Converting the answer afterwards is where the factor of $10^9$ tends to go missing.
$T$ must be absolute temperature. The symbol key beside the Modern Physics group of the sheet spells this out: $T$ is absolute temperature. A Celsius value in this equation is simply wrong, and unlike some errors it does not even scale sensibly.

The relationship is inverse, so the reasoning is quick once you trust it:

Change	Effect on $\lambda_{max}$
Double the absolute temperature	Peak wavelength halves
Triple it	Peak wavelength falls to a third
Cool the body	Peak moves to longer wavelength

That direction is why heated metal glows red before it glows white: as it warms, the peak slides down from the infrared toward the visible, and the ordering in 14.4.A.3.ii puts red at the long-wavelength end of the visible band, so red is the first colour to arrive.

One caution about colour language, taken from the Topic 14.4 boundary statement: AP Physics 2 expects students to know the ordering of the electromagnetic spectrum, including visible light, but students will not be expected to define exact wavelength ranges within the electromagnetic spectrum. So argue with ordering. "The peak moved to a shorter wavelength, so it moved toward the blue end" is safe. "The peak is now at 480 nm, which is blue" leans on a boundary the CED declines to draw.

Stefan-Boltzmann: how much, in total

15.4.A.3.iii covers the other axis of the problem: the rate at which energy is emitted (power) by a blackbody is proportional to the surface area of the body and to the temperature of the body raised to the fourth power, as described by the Stefan-Boltzmann law.

P = A \sigma T^4

The Stefan-Boltzmann constant is printed as $\sigma = 5.67 \times 10^{-8}$ W/(m $^2$ K $^4$ ). The symbol $P$ is power, in watts, which is energy per unit time, so this is a rate and not an amount. If a question asks for energy rather than power, multiply by a time.

The fourth power is what makes this equation worth respecting. A modest temperature change produces a large change in output:

Temperature factor	Power factor
$\times 2$	$\times 16$
$\times 3$	$\times 81$
$\times 10$	$\times 10\,000$
$\times 1.1$	$\times 1.46$

Area enters only to the first power, which sets up a clean 2.C comparison the exam can make: doubling the radius of a sphere multiplies its surface area by four and therefore its power by four, while doubling its temperature multiplies its power by sixteen. Temperature wins.

Two places this goes wrong in practice:

Getting the area wrong. For a sphere the emitting surface is $A = 4\pi r^2$ , printed in the Geometry and Trigonometry table on the same equation sheet. For a cube of side $s$ it is $6s^2$ . The equation says surface area, not cross-sectional area and not volume.
Using Celsius. Same trap as Wien's law, and worse here, because raising the wrong number to the fourth power amplifies the error enormously.

The two equations of this topic answer different questions and the exam likes to ask them together. Wien's law says where the emission peaks. Stefan-Boltzmann says how much comes out in total. Both are driven by the single variable $T$ , which is why one measurement of the peak wavelength fixes the whole spectrum.

Comparisons and factors of change (skills 2.C and 2.D)

The CED's own sample multiple-choice question 11 is a Unit 15 blackbody question, aligned to learning objective 15.4.A, essential knowledge 15.4.A.3 and skill 2.D, predict new values or factors of change of physical quantities using functional dependence between variables. It gives a blackbody at some temperature with a stated peak wavelength and a stated emitted power, halves the peak wavelength, and asks what the new power is. The published answer is that the power becomes sixteen times as large.

That is the archetype for this topic, and it is worth being able to run it in your head in two steps rather than three:

Peak wavelength to temperature, through Wien. $\lambda_{max} \propto 1/T$ , so halving the peak wavelength doubles the absolute temperature. No constants needed.
Temperature to power, through Stefan-Boltzmann. $P \propto T^4$ , so doubling the temperature multiplies the power by $2^4 = 16$ .

Combining the two gives a shortcut worth deriving once and keeping. Substituting $T = b/\lambda_{max}$ into $P = A\sigma T^4$ :

P = A\sigma \left(\frac{b}{\lambda_{max}}\right)^4 \qquad \text{so} \qquad P \propto \frac{1}{\lambda_{max}^{4}}

for a body of fixed area. Divide the peak wavelength by any factor and the power goes up by that factor to the fourth. That is not printed anywhere, and deriving it is a one-line skill 2.A move.

The general habit for a 2.D question: work in ratios, not in numbers. Write the two situations as a quotient so every constant cancels.

\frac{P_2}{P_1} = \frac{A_2}{A_1}\left(\frac{T_2}{T_1}\right)^4 \qquad \frac{\lambda_{max,2}}{\lambda_{max,1}} = \frac{T_1}{T_2}

With those two lines you do not need $\sigma$ or $b$ at all for a comparison question, which removes the two constants and with them the easiest place to slip.

Skill 2.C, comparing quantities between scenarios, is also on this topic's list, and the comparisons available are exactly the ones above plus the qualitative reading of the curve: which body is hotter (the one whose peak is further left), which emits more in total (the one whose curve is higher everywhere), which emits more at one specified wavelength (read the two heights at that wavelength).

How Topic 15.4 is tested, and the telescope question

The AP Physics 2 exam is 3 hours long: 42 multiple-choice questions in 85 minutes and 4 free-response questions in 95 minutes, each section worth 50 percent. Unit 15 carries a 12 to 15 percent weighting on the multiple-choice section, and a four-function, scientific, or graphing calculator is allowed on both.

What the topic's skill list implies:

1.C, sketch qualitatively. Draw the intensity-per-unit-wavelength curve, or a second curve for a different temperature on the same axes. Label the axes. Put the hotter peak to the left and the hotter curve above.
2.C, compare. Two bodies, two temperatures, two areas. Which is hotter, which radiates more.
3.B, apply a model to make a claim. Why must a perfect absorber emit. Why does the spectrum depend only on temperature. Why can classical physics not produce the observed distribution.
3.C, justify with evidence. Point at the graph or the printed relationship and say what it shows.

The topic also answers one of the unit's four essential questions, why do infrared telescopes need to be cooled, and the reasoning is a good test of whether the topic has landed. The CED poses the question and leaves the answer to you, but every ingredient is in 15.4.A.

By 15.4.A.1, matter spontaneously converts some of its internal thermal energy into electromagnetic energy, so the telescope's own structure is radiating whether you want it to or not. By Wien's law, a structure sitting at around room temperature, near $300$ K, has $\lambda_{max} = \frac{2.90 \times 10^{-3}}{300} = 9.7 \times 10^{-6}$ m, which is about $9.7$ micrometres, squarely in the infrared. So an uncooled infrared telescope glows brightly in exactly the band it is trying to observe, and its own emission drowns the faint signal from the sky. Cooling the instrument pushes its peak to longer wavelengths and, by the $T^4$ dependence, cuts its total emission steeply: dropping from $300$ K to $75$ K reduces the emitted power by a factor of $4^4 = 256$ .

A checklist before you leave the topic. Can you state the blackbody definition and derive why it must emit? Can you sketch the curve with correct axis labels and add a hotter one correctly? Can you use Wien's law in both directions without losing the metres-to-nanometres conversion? Can you get a factor of change in power from a factor of change in peak wavelength, without touching a constant? Can you say what the CED does and does not commit to about Planck's law?

Where it sits: this is the first of the three experiments in the unit that force light to be quantized, and it hands off to Topic 15.5, which makes the same demand from a completely different measurement. It also reaches back to Topic 9.1, where temperature is tied to the microscopic motion this radiation is drawing its energy from.

Wien's law in both directions

(a) A blackbody's emission peaks at a wavelength of $500 \text{ nm}$ . Find its absolute temperature. (b) The same body is cooled until its temperature is $2900 \text{ K}$ . Find the new peak wavelength. (c) Which way did the peak move, and is that consistent with 15.4.A.3.ii?

(a) Rearrange the printed Wien relation $\lambda_{max} = \dfrac{b}{T}$ for temperature: $T = \dfrac{b}{\lambda_{max}}$ .
Convert the wavelength to metres first, because Wien's constant is printed in metre kelvin: $500 \text{ nm} = 500 \times 10^{-9} \text{ m} = 5.00 \times 10^{-7} \text{ m}$ .
Substitute: $T = \dfrac{2.90 \times 10^{-3} \text{ m} \cdot \text{K}}{5.00 \times 10^{-7} \text{ m}} = 5.80 \times 10^3 \text{ K}$ , that is $5800 \text{ K}$ . The metres cancel and kelvin is left, which is the check that the conversion was done.
(b) Now go the other way with the same equation: $\lambda_{max} = \dfrac{2.90 \times 10^{-3} \text{ m} \cdot \text{K}}{2900 \text{ K}} = 1.00 \times 10^{-6} \text{ m}$ , which is $1000 \text{ nm}$ .
(c) The temperature fell by a factor of $5800/2900 = 2.00$ , and the peak wavelength rose by a factor of $1000/500 = 2.00$ . Cooler body, longer peak wavelength, exactly the inverse proportionality 15.4.A.3.ii describes.
Worth noting where those two wavelengths sit in the ordering of 14.4.A.3.i, which runs from long to short as radio, microwave, infrared, visible, ultraviolet, X-ray, gamma. The $500 \text{ nm}$ peak is in the visible band; the $1000 \text{ nm}$ peak has moved past the red end into the infrared. The body has stopped looking bright and started merely feeling warm, and it never stopped radiating.

(a) $T = 5800 \text{ K}$ . (b) $\lambda_{max} = 1.00 \times 10^{-6} \text{ m}$ , that is $1000 \text{ nm}$ . (c) The peak moved to a longer wavelength when the body cooled, which is the inverse relationship in 15.4.A.3.ii.

Total power from a hot sphere

A spherical blackbody of radius $0.10 \text{ m}$ is held at $1500 \text{ K}$ . (a) Find the rate at which it emits energy. (b) The temperature is then raised to $3000 \text{ K}$ . Find the new rate, and state the factor of change. (c) How much energy does the hotter sphere emit in one minute?

(a) The printed law is $P = A \sigma T^4$ , and $A$ is the surface area. For a sphere, the Geometry and Trigonometry table on the same equation sheet gives $S = 4\pi r^2$ , so $A = 4\pi (0.10 \text{ m})^2 = 4\pi (0.010) = 0.1257 \text{ m}^2$ .
Take the fourth power on its own, so the exponent is easy to audit: $T^4 = (1500)^4 = 5.0625 \times 10^{12} \text{ K}^4$ .
Now multiply through: $\sigma T^4 = (5.67 \times 10^{-8})(5.0625 \times 10^{12}) = 2.870 \times 10^{5} \text{ W/m}^2$ , and $P = (0.1257)(2.870 \times 10^{5}) = 3.61 \times 10^{4} \text{ W}$ , about $36 \text{ kW}$ .
(b) Do this one as a ratio rather than repeating the arithmetic. The area has not changed, so $\dfrac{P_2}{P_1} = \left(\dfrac{T_2}{T_1}\right)^4 = \left(\dfrac{3000}{1500}\right)^4 = 2^4 = 16$ .
So $P_2 = 16 \times 3.61 \times 10^{4} = 5.77 \times 10^{5} \text{ W}$ . Doubling the temperature of a $36 \text{ kW}$ radiator turns it into a $577 \text{ kW}$ radiator without changing its size at all.
(c) Power is energy per unit time, so energy is power times time. One minute is $60 \text{ s}$ : $E = (5.77 \times 10^{5} \text{ W})(60 \text{ s}) = 3.46 \times 10^{7} \text{ J}$ .
A check on the setup rather than the arithmetic: the units of $\sigma$ are W/(m $^2$ K $^4$ ), so multiplying by m $^2$ and by K $^4$ has to leave watts. It does. If an answer to this kind of question comes out in joules, a time has been multiplied in by mistake somewhere.

(a) $P = 3.61 \times 10^{4} \text{ W}$ . (b) $P = 5.77 \times 10^{5} \text{ W}$ , sixteen times as much. (c) $3.46 \times 10^{7} \text{ J}$ in one minute.

From a shift in the peak to a factor of change in power

A blackbody of fixed surface area has its peak emission at $900 \text{ nm}$ . It is then heated until the peak sits at $300 \text{ nm}$ . Without using a calculator for the main result, find the factor by which the emitted power changes. Then verify with numbers, taking the surface area as $1.0 \text{ m}^2$ .

Step one, wavelength to temperature. Wien's law gives $T \propto 1/\lambda_{max}$ , so dividing the peak wavelength by 3 multiplies the absolute temperature by 3. No constants required: $\dfrac{T_2}{T_1} = \dfrac{\lambda_{max,1}}{\lambda_{max,2}} = \dfrac{900}{300} = 3$ .
Step two, temperature to power. Stefan-Boltzmann gives $P \propto T^4$ at fixed area, so $\dfrac{P_2}{P_1} = \left(\dfrac{T_2}{T_1}\right)^4 = 3^4 = 81$ .
So the power increases by a factor of 81. Note that both constants, $b$ and $\sigma$ , cancelled out of the ratio, which is the whole point of working this way for a skill 2.D question.
The two steps can be combined once and reused: substituting $T = b/\lambda_{max}$ into $P = A\sigma T^4$ gives $P = A\sigma b^4 / \lambda_{max}^4$ , so at fixed area $P \propto 1/\lambda_{max}^4$ . Directly, $(900/300)^4 = 81$ .
Now the numerical verification. $T_1 = \dfrac{2.90 \times 10^{-3}}{900 \times 10^{-9}} = 3222 \text{ K}$ and $T_2 = \dfrac{2.90 \times 10^{-3}}{300 \times 10^{-9}} = 9667 \text{ K}$ . The ratio is $3.00$ , as predicted.
$P_1 = (1.0)(5.67 \times 10^{-8})(3222)^4 = (5.67 \times 10^{-8})(1.078 \times 10^{14}) = 6.11 \times 10^{6} \text{ W}$ .
$P_2 = (1.0)(5.67 \times 10^{-8})(9667)^4 = (5.67 \times 10^{-8})(8.732 \times 10^{15}) = 4.95 \times 10^{8} \text{ W}$ . The ratio $\dfrac{4.95 \times 10^{8}}{6.11 \times 10^{6}} = 81.0$ , confirming the shortcut.
The moral for the exam: the ratio route took two lines and no constants; the numerical route took six lines and two constants raised to the fourth power. When a question gives you a factor of change rather than absolute values, it is asking for the first route.

The power increases by a factor of $81$ . Dividing the peak wavelength by 3 triples the absolute temperature by Wien's law, and $P \propto T^4$ then gives $3^4 = 81$ . The numerical check gives $6.11 \times 10^{6} \text{ W}$ rising to $4.95 \times 10^{8} \text{ W}$ , a ratio of $81.0$ .

Frequently asked questions

What is a blackbody in AP Physics 2?

Essential knowledge 15.4.A.2 defines a blackbody as an idealized model of matter that absorbs all radiation that falls on the body, and adds that if the body is in equilibrium at a constant temperature then it must in turn emit energy. The emission follows from conservation of energy: if everything arriving is absorbed and the temperature is not changing, energy must leave at the same rate, and radiation is the only channel available. The word idealized matters, since no real surface absorbs everything, and no emissivity factor appears in the printed Stefan-Boltzmann equation.

What is Wien's law and how do you use it?

Wien's law is lambda_max = b / T, printed in the Modern Physics group of the AP Physics 2 equation sheet, with Wien's constant b = 2.90 times 10 to the minus 3 metre kelvin from the Table of Information. Essential knowledge 15.4.A.3.ii states that the peak wavelength emitted by a blackbody, meaning the wavelength at which it emits the greatest amount of radiation per unit wavelength, decreases with increasing temperature. Two cautions: T must be an absolute temperature in kelvin, and because b carries units of metre kelvin the answer comes out in metres, so convert to nanometres afterwards if needed.

What is the Stefan-Boltzmann law on the AP Physics 2 sheet?

P = A sigma T to the fourth, printed in the Modern Physics group, with the Stefan-Boltzmann constant sigma = 5.67 times 10 to the minus 8 watts per square metre per kelvin to the fourth. Essential knowledge 15.4.A.3.iii states that the rate at which energy is emitted by a blackbody is proportional to the surface area of the body and to the temperature of the body raised to the fourth power. A is the total surface area, so 4 pi r squared for a sphere, and T is absolute temperature. Since P is a power, multiply by a time to get an energy.

What happens to a blackbody's spectrum if its temperature doubles?

Two things at once. The peak wavelength halves, because Wien's law makes lambda_max inversely proportional to absolute temperature. And the total emitted power rises by a factor of two to the fourth, which is sixteen, because the Stefan-Boltzmann law makes power proportional to T to the fourth. On a graph of intensity per unit wavelength against wavelength, that means the peak moves left and the whole curve rises. The CED's own sample multiple-choice question 11 runs this backwards, giving a halved peak wavelength and asking for the new power, with sixteen times the original as the answer.

Why do infrared telescopes need to be cooled?

Because the telescope itself radiates in the band it is trying to observe. Essential knowledge 15.4.A.1 says matter will spontaneously convert some of its internal thermal energy into electromagnetic energy, so a warm instrument is always glowing. Wien's law puts the peak of a structure at around 300 kelvin at 2.90 times 10 to the minus 3 divided by 300, about 9.7 micrometres, which is in the infrared. An uncooled infrared telescope therefore swamps the faint sky signal with its own emission. Cooling moves that peak to longer wavelengths and, through the fourth-power dependence, cuts the total emission steeply.

Is the ultraviolet catastrophe on the AP Physics 2 exam?

The phrase appears once in the AP Physics 2 CED, in one of the four essential questions on the Unit 15 opener, and nowhere in the required course content. No essential knowledge statement defines it. What the required content does say, in 15.4.A.3.i, is that the distribution of the intensity of a blackbody's spectrum as a function of temperature cannot be modeled using only classical physics concepts, and that a blackbody's spectrum is described by Planck's law, which assumes that the energy of light is quantized. Use that wording for an exam answer. Planck's law itself is not printed in the CED or on the equation sheet.

Does AP Physics 2 Topic 15.4 have a boundary statement?

No. Topic 15.4 prints no boundary statement, and neither does Topic 15.1 or Topic 15.7. Five of the eight topics in Unit 15 do have one: 15.2, 15.3, 15.5, 15.6 and 15.8. The topic's suggested skills are 1.C, 2.C, 3.B and 3.C, and 1.C, create qualitative sketches of graphs, is listed for no other topic in the unit, which is a strong hint that sketching the intensity-against-wavelength curve is examinable.