3 questions, one for each idea where we can. Answer them, then see which ideas to fix.
Question 1 of 3
A block of ice at −10°C is taken out of a freezer and placed on a table in a room at 20°C. What, if anything, does the ice emit as a result of its temperature?
Answer and reasoning
ANothing at all, since the ice is much too cold for it to glow A student who thinks only glowing objects emit radiation picks this. The ice cannot be seen to glow because it emits almost no visible light, but it emits infrared radiation, as every object does because of its temperature.
BCold radiation, which lowers the temperature of nearby objects A student who pictures cold as something that flows out of cold objects picks this. Only energy is emitted and absorbed. Nearby objects cool because they emit more radiation toward the ice than they absorb from it.
CMostly infrared radiation, converted from some of its internal energyCorrect All matter spontaneously converts some of its internal energy into electromagnetic radiation, so the ice emits radiation. At −10°C (263 K) almost all of it is infrared, which is why it cannot be seen. The ice absorbs more radiation from the warmer room than it emits, so it warms up.
DRadiation of a single wavelength, set by its temperature A student who thinks an object emits only the wavelength at the peak of its spectrum picks this. The ice's temperature sets where its emission is greatest, but the radiation it emits is spread continuously over a wide range of wavelengths.
A small object that absorbs all the radiation falling on it hangs by a thin insulating thread inside a closed, evacuated container whose walls are held at 300 K. After a long time, the object's temperature is steady at 300 K. Which claim about the object is correct?
Answer and reasoning
AIt emits radiation at the same rate as it absorbs radiationCorrect The object is a blackbody: it absorbs all the radiation from the walls that falls on it. Its temperature, and so its internal energy, is constant, so it must emit energy at the same rate at which it absorbs it. With a vacuum around it and an insulating thread, radiation is the only significant way energy is transferred.
BIt neither emits nor absorbs radiation, being as warm as the walls A student who thinks an object emits only toward colder surroundings picks this. Equal temperatures mean the net transfer is zero: the object and the walls both emit, and the object absorbs from the walls exactly as much as it emits.
CIt absorbs radiation but emits nothing, since it absorbs all that arrives A student who thinks a blackbody only absorbs picks this. If the object absorbed energy and emitted none, its internal energy and temperature would keep rising. A steady temperature shows it emits at the rate it absorbs.
DIt emits radiation, but only at the one wavelength of its peak A student who thinks a blackbody emits only its peak wavelength picks this. A blackbody emits a continuous spectrum; at 300 K it is spread over a wide range of infrared wavelengths, with the most intense emission near the peak.
Measured blackbody spectra could not be reproduced by models that used only classical physics. What did Planck assume in the model that successfully matched the measurements?
Answer and reasoning
ALight's energy varies continuously, as set by the amplitude of its waves A student who holds the classical wave picture picks this. That is exactly the assumption of the classical models that failed; Planck's model works because it treats light's energy as quantized.
BEach material emits its own spectrum, set by the atoms that make it up A student who thinks a hot object's spectrum depends on its material picks this. A blackbody's spectrum depends only on its temperature, and Planck's law contains no property of the material.
CLight's energy is quantized, emitted and absorbed in discrete amountsCorrect Planck's law, which matches the measured spectra, rests on the assumption that the energy of light is quantized: light is emitted and absorbed only in discrete amounts, not in any amount at all.
DThe short-wavelength measurements were in error and can be ignored A student who assumes that data disagreeing with a model must be wrong picks this. The measurements were reliable; it was classical physics that failed, and Planck's quantum assumption was needed to match them.
In preparation: 0 of 3 sections compiled and reviewed. The rest show key terms and common mistakes from our question bank until they are.
15.4.A.1 Thermal radiation Fix
Thermal radiation
Electromagnetic radiation that matter emits because of its temperature: the matter spontaneously converts some of its internal (thermal) energy into electromagnetic energy. Every object emits it; objects near room temperature emit mostly infrared, which the eye cannot see.
Students often think The electromagnetic radiation an object emits because of its temperature is the visible light of its glow, so an object too cool to glow emits none and a glowing object emits only visible light. In fact No. All matter converts some of its internal energy into electromagnetic radiation. Objects too cool to glow emit mainly infrared, which the eye cannot see; glowing objects emit visible light and much infrared as well.
Students often think Cold objects radiate 'cold', which spreads out from them and lowers the temperature of objects nearby. In fact No. Only energy is emitted and absorbed. Objects near a cold object cool because they emit more radiation toward it than they absorb from it.
15.4.A.2 Blackbody Fix
Blackbody
An idealized model of matter that absorbs all the radiation that falls on it, reflecting none. The name describes what it absorbs, not what it emits: a hot blackbody, such as a star modeled as one, can be very bright.
Blackbody at constant temperature
A blackbody in equilibrium with its surroundings at a constant temperature, exchanging energy only by radiation, absorbs energy from the radiation falling on it, so it must emit energy at the same rate; otherwise its internal energy, and its temperature, would change.
Students often think Only the hotter of two objects emits radiation, so an object emits none toward surroundings that are at the same or a higher temperature. In fact No. Every object emits radiation at a rate set by its temperature and surface, and absorbs radiation from its surroundings. At equal temperatures the two rates are equal, so the net transfer is zero, but neither rate is zero.
Students often think A blackbody only absorbs radiation and does not emit any, which is why it is called black. In fact No. A blackbody absorbs all the radiation that falls on it, and it also emits radiation because of its temperature. A blackbody in equilibrium with its surroundings, exchanging energy only by radiation, emits energy at the same rate as it absorbs it.
15.4.A.3 Continuous spectrum Fix
Continuous spectrum
Radiation that includes every wavelength over a range, with no gaps, rather than a set of separate wavelengths. A blackbody's spectrum is continuous, and its shape depends only on the body's temperature, not on the material.
Intensity per unit wavelength
The power per unit area emitted in a small wavelength interval, divided by the width of that interval. A graph of intensity per unit wavelength against wavelength models a blackbody's spectrum; its SI unit is W/m³ (W/m² per meter of wavelength), though graphs often use arbitrary units.
Planck's law and quantized energy
Planck's law describes how a blackbody's intensity per unit wavelength depends on wavelength and temperature. It matches the measured spectra because it assumes that the energy of light is quantized: light is emitted and absorbed only in discrete amounts.
Failure of classical physics for blackbodies
Models that used only classical physics predicted an intensity per unit wavelength that keeps rising as the wavelength gets shorter, whereas measured spectra rise to a peak and fall toward zero at short wavelengths. The spectrum cannot be modeled with classical physics alone.
Peak wavelength, λmax
The wavelength at which a blackbody emits the greatest intensity per unit wavelength: the wavelength at the top of its spectrum graph. The body also emits at many other wavelengths. SI unit: meter (m); often given in nm.
Wien's law
λmax = b/T, where T is the absolute temperature and b = 2.90 × 10⁻³ m·K. The peak wavelength is inversely proportional to the absolute temperature: doubling T halves λmax, so hotter bodies look bluer.
Power emitted by a blackbody
The rate at which a blackbody emits energy as radiation, P, in watts (W). It depends on both the body's surface area and its temperature.
Stefan-Boltzmann law
P = AσT⁴, where A is the surface area of the body (4πr² for a sphere of radius r), T its absolute temperature and σ = 5.67 × 10⁻⁸ W/(m²·K⁴). Doubling T multiplies P by 16; doubling the radius of a sphere multiplies P by 4.
Absolute temperature in radiation laws
Wien's law and the Stefan-Boltzmann law use temperature in kelvins, T(K) = T(°C) + 273. Using Celsius temperatures gives wrong values and wrong ratios.
Students often think A blackbody emits only one wavelength, its peak wavelength λmax; the rest of the spectrum graph does not represent radiation it emits. In fact No. A blackbody's spectrum is continuous: it emits over a wide range of wavelengths. The peak wavelength is only the one at which the intensity per unit wavelength is greatest.
Students often think The spectrum an object emits because of its temperature is set by the material it is made of, each material giving its own characteristic wavelengths. In fact Not for a blackbody. A blackbody emits a continuous spectrum whose shape depends only on its temperature: blackbodies of different materials at the same temperature emit identical spectra.
8 more questions. Every wrong answer here is a real mistake students make, and you see why it is wrong as soon as you answer.
Question 1 of 8
A lump of carbon and a lump of iron are both at 1500 K, and both are modeled as blackbodies. For each, a graph is drawn of the intensity per unit wavelength of the radiation it emits as a function of wavelength. How do the two graphs compare?
Answer and reasoning
AThey are continuous curves peaking at wavelengths set by each material A student who thinks each material gives off its own characteristic wavelengths picks this. That is true of the line spectra of gases, but a blackbody's spectrum depends only on its temperature, so both peak at the same wavelength.
BThey are the same continuous curve, since the temperatures are equalCorrect A blackbody emits a continuous spectrum whose shape depends only on its temperature. Both lumps are at 1500 K, so the graphs of intensity per unit wavelength are identical, whatever the materials.
CThe carbon's curve is lower, since black materials absorb, not emit A student who thinks black objects only absorb picks this. In the blackbody model both lumps absorb all the radiation falling on them, and both emit the spectrum set by their common temperature.
DEach is one spike at the single wavelength fixed by 1500 K A student who thinks a blackbody emits only its peak wavelength picks this. Each lump emits a continuous spectrum over a wide range of wavelengths; 1500 K fixes where the curve peaks, not a single wavelength.
A tungsten filament at 3000 K glows yellowish white. It is modeled as a blackbody, and the graph shows the intensity per unit wavelength of the radiation it emits as a function of wavelength λ, with the band of visible wavelengths shaded. Which description of the emitted radiation does the graph support?
Answer and reasoning
AJust the one wavelength at the graph's peak, about 1000 nm A student who thinks a blackbody emits only its peak wavelength picks this. The graph shows emission at every wavelength where the curve is above the axis; the peak is where the intensity per unit wavelength is greatest.
BVisible wavelengths alone, being the light that it glows with A student who equates thermal radiation with visible glow picks this. The shaded visible band holds only a small part of the area under the curve; most of the energy is emitted as infrared, which the eye cannot see.
CA set of separate wavelengths, each characteristic of tungsten A student who thinks each material emits its own characteristic wavelengths picks this. The graph is a smooth, continuous curve, and in the blackbody model its shape is set by the temperature alone, not by the material.
DA continuous spectrum, most of whose energy lies in the infraredCorrect The curve is above the axis at every wavelength from the ultraviolet to beyond 3000 nm, so the spectrum is continuous. Its peak is near 1000 nm, in the infrared, and most of the area under the curve lies at wavelengths longer than 700 nm, beyond the visible band.
The graph shows the intensity per unit wavelength emitted by a blackbody as a function of wavelength λ. The absolute temperature of the blackbody is then doubled. At what wavelength does the spectrum of the heated blackbody peak?
Answer and reasoning
A1.6 × 10³ nm A student who thinks hotter bodies emit at longer wavelengths picks this, doubling 800 nm. λmax = b/T decreases as T increases, so the peak moves to a shorter wavelength.
B8.0 × 10² nm A student who thinks heating only makes the curve taller picks this. The heated body does emit more at every wavelength, but its peak also moves: λmax = b/T halves when T doubles.
C5.0 × 10¹ nm A student who uses the fourth power from the Stefan-Boltzmann law picks this: 800 nm ÷ 2⁴ = 50 nm. It is the emitted power that depends on T⁴; the peak wavelength is inversely proportional to T itself.
D4.0 × 10² nmCorrect From the graph, the spectrum peaks at 800 nm. By Wien's law, λmax = b/T, the peak wavelength is inversely proportional to the absolute temperature, so doubling T halves it: 800 nm ÷ 2 = 4.0 × 10² nm.
Working From the graph λmax = 800 nm. Wien's law λmax = b/T, so λ₂/λ₁ = T₁/T₂ = 1/2: λ₂ = 800 nm/2 = 4.0 × 10² nm. (×2 gives 1.6 × 10³ nm; unchanged 8.0 × 10² nm; ÷2⁴ gives 50 nm.)
Stars P and Q are modeled as blackbodies. Star P looks reddish, star Q looks bluish white, and P emits more power in total than Q does. Which claim about the stars' surface temperatures, with its reasoning, is correct?
Answer and reasoning
AQ's is higher, as Q emits blue light alone while P emits red light alone A student who thinks a blackbody emits only one wavelength picks this. Each star emits a continuous spectrum across the visible range and beyond; its color shows where its emission is strongest, not the only light it emits.
BQ's is higher, as its spectrum peaks at a shorter wavelength than P'sCorrect A bluish-white star emits most strongly at shorter wavelengths than a reddish one, so its spectrum peaks at a shorter wavelength. By Wien's law, λmax = b/T, a shorter peak wavelength means a higher temperature. P's greater total power comes from its much larger surface area.
CP's is higher, as redder light comes from hotter objects than bluer light A student who links red with hot picks this. λmax = b/T: hotter blackbodies peak at shorter wavelengths, so the reddish star is the cooler one.
DP's is higher, since P emits more power and power is set by temperature A student who thinks a blackbody's power depends only on its temperature picks this. P = AσT⁴ also depends on the surface area: a large, cooler star can emit more power than a small, hotter one.
Working Color gives the peak wavelength: bluish white → shorter λmax than reddish. λmax = b/T, so TQ > TP. P's larger power is consistent with a much larger surface area (P = AσT⁴).
A small metal sphere of diameter 0.10 m is held at 727°C. Modeling the sphere as a blackbody, what power does it emit? Use σ = 5.67 × 10⁻⁸ W/(m²·K⁴).
Answer and reasoning
A1.8 × 10³ WCorrect The radius is 0.050 m, so the surface area is A = 4πr² = 4π(0.050 m)² = 0.0314 m². The absolute temperature is 727 + 273 = 1000 K. P = AσT⁴ = (0.0314 m²)(5.67 × 10⁻⁸ W/(m²·K⁴))(1000 K)⁴ = 1.8 × 10³ W.
B5.0 × 10² W A student who substitutes the Celsius temperature picks this: (0.0314)(5.67 × 10⁻⁸)(727)⁴ = 5.0 × 10² W. The Stefan-Boltzmann law needs the absolute temperature, 1000 K.
C4.5 × 10² W A student who uses the cross-sectional area πr² = 0.00785 m² picks this. The sphere radiates from its whole surface, A = 4πr², four times as large.
D7.1 × 10³ W A student who uses the diameter, 0.10 m, as the radius picks this: 4π(0.10 m)² = 0.126 m², four times the true area. The radius is half the diameter, 0.050 m.
Working r = 0.10 m/2 = 0.050 m; A = 4πr² = 4π(0.050 m)² = 3.14 × 10⁻² m². T = 727 + 273 = 1000 K. P = AσT⁴ = (3.14 × 10⁻² m²)(5.67 × 10⁻⁸ W/(m²·K⁴))(1000 K)⁴ = 1.78 × 10³ W ≈ 1.8 × 10³ W. (727 K: 5.0 × 10² W; πr²: 4.5 × 10² W; r = 0.10 m: 7.1 × 10³ W.)
A blackbody sphere of radius R at absolute temperature T₁ emits power P. A second blackbody sphere, of radius 2R, emits the same power P. What is the absolute temperature T₂ of the second sphere?
Answer and reasoning
AT₂ = T₁ A student who thinks a blackbody's power depends only on its temperature picks this. The larger sphere has four times the surface area, so at temperature T₁ it would emit four times the power; to emit only P it must be cooler.
BT₂ = T₁/⁴√2 A student who doubles the area when the radius doubles picks this: 2T₂⁴ = T₁⁴. The area 4πR² is proportional to R², so it is multiplied by 4, giving T₂⁴ = T₁⁴/4 and T₂ = T₁/√2.
CT₂ = T₁/4 A student who takes the power to be proportional to T rather than T⁴ picks this: 4T₂ = T₁. With P = AσT⁴, four times the area needs T⁴ to be divided by 4, so T is divided by ⁴√4 = √2.
DT₂ = T₁/√2Correct P = AσT⁴ with A = 4πR². The second sphere has 4 times the area, so to emit the same power its T⁴ must be 4 times smaller: T₂⁴ = T₁⁴/4, so T₂ = T₁/⁴√4 = T₁/√2.
Working P = 4πR²σT₁⁴ = 4π(2R)²σT₂⁴ ⇒ T₂⁴ = T₁⁴/4 ⇒ T₂ = T₁/√2. (Area ignored: T₂ = T₁. Area ∝ R: 2T₂⁴ = T₁⁴, T₂ = T₁/⁴√2. P ∝ T: 4T₂ = T₁, T₂ = T₁/4.)
A blackbody at absolute temperature T₀ has peak wavelength λ₀. It is then heated to absolute temperature T. Planck's constant is h and the speed of light is c. Which expression gives the energy of a photon whose wavelength equals the blackbody's new peak wavelength?
Answer and reasoning
AhcT/(λ₀T₀)Correct Wien's law, λmax = b/T, means λmax T is the same at every temperature, so λnew T = λ₀T₀ and the new peak wavelength is λ₀T₀/T. A photon of that wavelength has energy E = hc/λ = hcT/(λ₀T₀), which is larger than hc/λ₀ when T > T₀, since the hotter blackbody peaks at a shorter wavelength.
Bhc(T₀/T)/λ₀ A student who thinks the peak wavelength increases with temperature takes λmax ∝ T, so the new peak is λ₀T/T₀ and E = hcT₀/(λ₀T), and picks this. Wien's law is λmax = b/T: a hotter blackbody peaks at a shorter wavelength, λ₀T₀/T, so the photon energy at the peak increases with T.
ChcT⁴/(λ₀T₀⁴) A student who thinks the peak wavelength depends on the fourth power of T, as the emitted power does, takes λmax ∝ 1/T⁴, so the new peak is λ₀T₀⁴/T⁴, and picks this. The fourth power belongs to P = AσT⁴; Wien's law is λmax = b/T, so the new peak is λ₀T₀/T.
Dhc/λ₀ A student who thinks heating a blackbody makes it emit more strongly but leaves its peak wavelength unchanged keeps the peak at λ₀ and picks this. Heating also moves the peak to a shorter wavelength, λ₀T₀/T, so the photon energy at the peak increases to hcT/(λ₀T₀).
Working λmax T = b is constant, so λnew = λ₀T₀/T and E = hc/λnew = hcT/(λ₀T₀). Errors: λ ∝ T → hcT₀/(λ₀T); λ ∝ 1/T⁴ → hcT⁴/(λ₀T₀⁴); peak unchanged → hc/λ₀.
The spectrum of a hot star, modeled as a blackbody, has its peak at a wavelength of 116 nm. The radius of the star is 3.0 × 10⁹ m. What power does the star emit? Use b = 2.90 × 10⁻³ m·K and σ = 5.67 × 10⁻⁸ W/(m²·K⁴).
Answer and reasoning
A6.3 × 10²⁹ W A student who uses the cross-sectional area πR² for the radiating area picks this, one quarter of the correct power. The star emits from its whole spherical surface, A = 4πR².
B2.5 × 10³⁰ WCorrect Wien's law gives the temperature: T = b/λmax = (2.90 × 10⁻³ m·K)/(1.16 × 10⁻⁷ m) = 2.50 × 10⁴ K. The star radiates from its whole surface, A = 4πR² = 1.13 × 10²⁰ m², so P = AσT⁴ = (1.13 × 10²⁰ m²)(5.67 × 10⁻⁸ W/(m²·K⁴))(2.50 × 10⁴ K)⁴ = 2.5 × 10³⁰ W.
C2.2 × 10¹⁰ W A student who thinks the emitted power depends only on temperature, not on size, computes σT⁴ = (5.67 × 10⁻⁸)(2.50 × 10⁴)⁴ and picks this. σT⁴ is the power per square meter of surface; it must be multiplied by the surface area 4πR².
D1.6 × 10¹⁷ W A student who takes the power to be proportional to T rather than T⁴ picks this: AσT = (1.13 × 10²⁰)(5.67 × 10⁻⁸)(2.50 × 10⁴). The Stefan-Boltzmann law has the absolute temperature raised to the fourth power.
Working Wien's law: T = b/λmax = (2.90 × 10⁻³ m·K)/(116 × 10⁻⁹ m) = 2.50 × 10⁴ K. Surface area A = 4πR² = 4π(3.0 × 10⁹ m)² = 1.13 × 10²⁰ m². P = AσT⁴ = (1.13 × 10²⁰ m²)(5.67 × 10⁻⁸ W/(m²·K⁴))(2.50 × 10⁴ K)⁴ = 2.5 × 10³⁰ W. (πR²: 6.3 × 10²⁹ W; σT⁴ with no area: 2.2 × 10¹⁰; AσT: 1.6 × 10¹⁷ W.)
Compiled from the AP Physics 2 Course and Exam Description (effective Fall 2024, 2026 reissue) and our question bank · Specialist review in progress. How these pages are made · Free, no account
