AP Physics 2 Quiz: Blackbody Radiation

What this quiz covers

This quiz focuses on Blackbody Radiation, giving you a quick way to practice the rules, question types, and explanations that matter most for AP Physics 2.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. The peak wavelength increases and the total intensity decreases.
2. The peak wavelength decreases and the total intensity increases. (correct answer)
3. The peak wavelength stays constant and the total intensity stays constant.
4. The object emits only higher-energy photons and no lower-energy photons.

Explanation: This question tests understanding of blackbody radiation. As a blackbody's temperature increases, two key changes occur in its emission spectrum: the peak wavelength shifts to shorter wavelengths (following Wien's displacement law, where λ_peak is inversely proportional to temperature), and the total intensity increases dramatically (following the Stefan-Boltzmann law, where total power is proportional to $T^4$). When the metal sphere is heated from 600 K to 900 K, its peak wavelength decreases by a factor of 600/900 = 2/3, while its total intensity increases by a factor of (900/600)^4 = 5.06. Choice D incorrectly suggests that only high-energy photons are emitted, reflecting the misconception that hot objects stop emitting low-energy radiation entirely. Remember that higher temperature always means shorter peak wavelength and greater total intensity across all wavelengths.

Question 2

1. It decreases, shifting the peak toward shorter wavelengths. (correct answer)
2. It increases, shifting the peak toward longer wavelengths.
3. It remains the same because wavelength is independent of temperature.
4. It becomes undefined because hotter objects emit only one wavelength.

Explanation: This question tests understanding of blackbody radiation. When a blackbody's temperature increases, Wien's displacement law tells us that the peak wavelength is inversely proportional to temperature (λ_peak × T = constant). As the ceramic plate heats from 300 K to 600 K, its temperature doubles, so the peak wavelength must halve, shifting from longer wavelengths toward shorter wavelengths. Simultaneously, the Stefan-Boltzmann law ensures that the total emitted power (proportional to the area under the curve) increases as T^4, growing by a factor of 16. Choice C incorrectly claims wavelength is independent of temperature, reflecting the misconception that emission properties are fixed material characteristics. The key principle is that higher temperature always shifts the peak toward shorter wavelengths (higher frequencies, higher photon energies).

Question 3

1. The spectrum shifts to longer wavelengths and intensity decreases at all wavelengths.
2. The spectrum is unchanged because emission is independent of temperature.
3. The spectrum shifts to shorter wavelengths and intensity increases at all wavelengths. (correct answer)
4. The spectrum becomes a single wavelength because hotter objects emit monochromatically.

Explanation: This question tests understanding of blackbody radiation. When a blackbody's temperature increases, the entire emission spectrum undergoes two simultaneous changes: the peak shifts to shorter wavelengths (following Wien's displacement law), and the intensity increases at every single wavelength across the spectrum. This means the curve not only shifts left on a wavelength plot but also grows taller everywhere, resulting in a much greater total emitted power. Choice B incorrectly claims the spectrum is unchanged with temperature, reflecting the misconception that thermal emission is a fixed property of materials. The key principle is that temperature affects both the spectral distribution (peak position) and the absolute intensity at all wavelengths.

Question 4

1. The total emitted power per unit area increases and the peak wavelength decreases. (correct answer)
2. The total emitted power per unit area decreases and the peak wavelength increases.
3. The total emitted power per unit area is unchanged and the peak wavelength is unchanged.
4. The object emits only ultraviolet photons and no infrared photons.

Explanation: This question tests understanding of blackbody radiation. When the filament temperature doubles from 1000 K to 2000 K, two fundamental changes occur: the peak wavelength decreases by half (from Wien's law, λ_peak ∝ 1/T), and the total emitted power per unit area increases by a factor of 16 (from Stefan-Boltzmann law, P ∝ $T^4$). The spectrum maintains its continuous distribution but becomes taller at all wavelengths, with the peak shifting toward the blue/UV end of the spectrum. Choice D incorrectly suggests the object emits only UV photons, reflecting the misconception that hot objects stop emitting infrared radiation. The crucial insight is that blackbodies always emit a continuous spectrum; higher temperatures shift the peak shorter and increase intensity everywhere.

Question 5

1. The peak wavelength stays the same, and the total emitted intensity decreases.
2. The object emits only shorter wavelengths and emits no radiation at longer wavelengths.
3. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)
4. The peak wavelength shifts to longer wavelengths, and the total emitted intensity decreases.

Explanation: This question tests understanding of blackbody radiation. As a blackbody's temperature rises, its emission spectrum undergoes predictable changes: the wavelength of peak emission decreases inversely with temperature (Wien's displacement law), and the total radiated power increases as the fourth power of temperature (Stefan-Boltzmann law). The cavity radiator exemplifies this: doubling temperature from 500 K to 1000 K halves the peak wavelength from 5.8 μm to 2.9 μm, while dramatically increasing the curve's height. Choice D incorrectly suggests that longer wavelengths disappear, but blackbodies emit at all wavelengths—only the relative intensities change with temperature. Remember: temperature increase always shifts the peak toward shorter wavelengths while boosting intensity everywhere.

Question 6

1. The peak wavelength shifts to longer wavelengths, and the total emitted intensity increases.
2. The peak wavelength stays constant, and the total emitted intensity increases.
3. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)
4. The spectrum is independent of temperature, so both peak wavelength and intensity are unchanged.

Explanation: This question tests understanding of blackbody radiation. When a blackbody's temperature increases, two fundamental changes occur in its emission spectrum: the peak wavelength becomes shorter (Wien's law states λ_peak = b/T where b is Wien's constant), and the intensity increases at every wavelength, resulting in greater total emitted power (Stefan-Boltzmann law). The kiln data demonstrates this perfectly: at 800 K the peak is at 3.6 μm, while at 1200 K it shifts to 2.4 μm (shorter wavelength), with increased intensity across the spectrum. Choice B incorrectly claims the peak shifts to longer wavelengths, reversing the actual temperature dependence. The key principle: hotter objects glow with shorter-wavelength light and emit more total energy.

Question 7

1. The peak wavelength shifts to longer wavelengths, and the total emitted intensity decreases.
2. The peak wavelength stays the same, and the total emitted intensity increases.
3. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)
4. The spectrum is independent of temperature, so both peak wavelength and intensity are unchanged.

Explanation: This question tests understanding of blackbody radiation. When a blackbody's temperature doubles, its emission spectrum transforms according to two principles: the peak wavelength halves (Wien's law: λ_peak = 2898 μm·K / T), and the total emitted power increases by a factor of 16 (Stefan-Boltzmann: P ∝ $T^4$). The rock data perfectly illustrates this: doubling from 300 K to 600 K shifts the peak from 9.7 μm to 4.8 μm (half the wavelength) while greatly increasing intensity. Choice A incorrectly predicts longer wavelengths and decreased intensity, completely reversing both temperature dependencies. Remember: higher temperature always means shorter peak wavelength and dramatically increased emission.

Question 8

1. The peak wavelength stays constant, and the total emitted intensity stays constant.
2. The peak wavelength shifts to longer wavelengths, and the total emitted intensity increases.
3. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)
4. The source emits only photons with energies above the original peak energy.

Explanation: This question tests understanding of blackbody radiation. Temperature changes in blackbodies produce predictable spectral shifts: the peak wavelength decreases inversely with temperature (Wien's displacement law), while the intensity at every wavelength increases, raising total emitted power substantially (Stefan-Boltzmann law). The laboratory source demonstrates this: heating from 1000 K to 1500 K shifts the peak from 2.9 μm to 1.9 μm (shorter wavelength) and increases the spectrum's intensity. Choice D incorrectly suggests that only high-energy photons are emitted above a threshold, but blackbodies emit a continuous spectrum at all wavelengths regardless of temperature. The universal rule: increasing temperature shifts peaks to shorter wavelengths while boosting intensity everywhere.

Question 9

1. It increases by a factor of 2. (correct answer)
2. It stays the same because only intensity changes with temperature.
3. It decreases by a factor of 2.
4. It increases by a factor of 4.

Explanation: This question tests understanding of blackbody radiation. Wien's displacement law states λ_peak = b/T, so if the peak wavelength decreases from 4 μm to 2 μm (halves), the temperature must double. This inverse relationship between peak wavelength and temperature is fundamental to blackbody radiation. The spectrum also becomes more intense at all wavelengths as temperature increases. Choice A incorrectly suggests temperature decreases when wavelength decreases, reversing the actual relationship. The key strategy is to apply Wien's law: when peak wavelength changes by a factor, temperature changes by the reciprocal factor.

Question 10

1. The peak wavelength stays constant, and the total emitted intensity stays constant.
2. The spectrum is independent of temperature, so the peak wavelength changes but intensity does not.
3. The peak wavelength shifts to longer wavelengths, and the total emitted intensity increases.
4. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)

Explanation: This question tests understanding of blackbody radiation. When temperature increases, a blackbody's emission spectrum exhibits two characteristic behaviors: the peak wavelength shifts to shorter values (following Wien's law where λ_peak × T = constant), and the intensity increases at all wavelengths, raising the total emitted power. The steel plate data confirms this pattern: increasing from 600 K to 900 K moves the peak from 4.8 μm to 3.2 μm (shorter wavelength) while increasing overall intensity. Choice A incorrectly reverses the wavelength shift direction, suggesting peaks move to longer wavelengths with heating. The fundamental rule: higher temperature means both shorter peak wavelength and greater emission intensity.

Question 11

1. The peak wavelength shifts to longer wavelengths, and the total emitted intensity increases.
2. The spectrum is independent of temperature, so the peak wavelength stays fixed while intensity changes randomly.
3. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)
4. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity decreases.

Explanation: This question tests understanding of blackbody radiation. Doubling a blackbody's temperature produces two distinct effects on its emission spectrum: the peak wavelength halves (Wien's law gives λ_peak = constant/T), and the total radiated power increases by a factor of 16 (Stefan-Boltzmann law: P ∝ $T^4$). The oven data confirms this: doubling from 700 K to 1400 K shifts the peak from 4.1 μm to 2.1 μm (half the original) while dramatically increasing intensity. Choice C incorrectly pairs the correct wavelength shift with decreased intensity, contradicting the fundamental principle that hotter objects radiate more power. Key concept: temperature increase always means both shorter peak wavelength and greater emission intensity.

Question 12

1. The peak wavelength shifts to longer wavelengths, and the total emitted intensity decreases.
2. The peak wavelength stays the same, and the total emitted intensity stays the same.
3. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)
4. The object emits only higher-energy photons and stops emitting lower-energy photons.

Explanation: This question tests understanding of blackbody radiation. As temperature increases, the blackbody spectrum changes in two key ways: the peak wavelength shifts to shorter wavelengths (following Wien's displacement law, where λ_peak ∝ 1/T), and the total emitted intensity increases dramatically (following the Stefan-Boltzmann law, where total power ∝ $T^4$). The data confirms this: heating from 3000 K to 4000 K shifts the peak from 970 nm to 725 nm (shorter wavelength) and increases overall intensity. Choice D incorrectly suggests that lower-energy photons stop being emitted, but blackbodies emit at all wavelengths—the distribution just shifts. Remember: higher temperature always means shorter peak wavelength and greater total intensity.

Question 13

1. The object emits only shorter wavelengths and emits no longer wavelengths.
2. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)
3. The peak wavelength shifts to longer wavelengths, and the total emitted intensity decreases.
4. The peak wavelength stays the same, and the total emitted intensity stays the same.

Explanation: This question tests understanding of blackbody radiation. When temperature doubles, blackbody emission follows precise mathematical relationships: the peak wavelength halves according to Wien's law (λ_peak × T = 2898 μm·K), and the total emitted intensity increases sixteen-fold per Stefan-Boltzmann law (I ∝ $T^4$). The sphere's behavior matches perfectly: doubling from 400 K to 800 K shifts the peak from 7.2 μm to 3.6 μm (exactly half) while greatly increasing overall intensity. Choice C incorrectly claims that longer wavelengths vanish, but blackbodies emit continuous spectra at all wavelengths—only the distribution shape changes with temperature. Remember: higher temperature shifts the spectrum toward shorter wavelengths while maintaining emission at all frequencies.

Question 14

1. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)
2. The peak wavelength shifts to longer wavelengths, and the total emitted intensity increases.
3. The peak wavelength stays constant, and the total emitted intensity increases.
4. The spectrum is independent of temperature, so peak wavelength and intensity do not change.

Explanation: This question tests understanding of blackbody radiation. Increasing temperature from 2000 K to 5000 K (2.5× increase) produces dramatic spectral changes: the peak wavelength decreases by the same factor per Wien's law (from 1450 nm to 580 nm, exactly 2.5× shorter), while the total emitted power increases by (2.5)^4 ≈ 39 times following Stefan-Boltzmann law. These two effects—shorter peak wavelength and vastly increased intensity—always occur together when blackbody temperature rises. Choice B incorrectly reverses the wavelength shift direction, contradicting the fundamental inverse relationship between temperature and peak wavelength. The key principle: hotter blackbodies emit shorter-wavelength light with much greater total power.

Question 15

1. The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases. (correct answer)
2. The peak wavelength shifts to longer wavelengths, and the total emitted intensity increases.
3. The peak wavelength stays constant, and the total emitted intensity decreases.
4. The filament emits only higher-frequency light and stops emitting infrared radiation.

Explanation: This question tests understanding of blackbody radiation. As temperature increases, blackbody emission follows two fundamental laws: Wien's displacement law dictates that peak wavelength decreases with rising temperature (λ_peak ∝ 1/T), while Stefan-Boltzmann law ensures total emitted power increases dramatically (P ∝ $T^4$). The filament data demonstrates both effects: heating from 2500 K to 3000 K shifts the peak from 1160 nm to 970 nm (shorter wavelength) and raises the entire spectrum. Choice D incorrectly claims that infrared emission stops, but blackbodies emit at all wavelengths—the spectrum simply shifts toward shorter wavelengths while maintaining emission across all frequencies. Key insight: temperature increase always produces shorter peak wavelengths and higher total intensity.

Question 16

1. It increases because hotter objects radiate more strongly at all wavelengths.
2. It stays the same because the material determines $\lambda_{\text{peak}}$.
3. It decreases because the spectrum shifts toward shorter wavelengths. (correct answer)
4. It becomes zero because all photons become infinitely energetic.

Explanation: This question tests understanding of blackbody radiation. Wien's displacement law states that the peak wavelength is inversely proportional to temperature: λ_peak = b/T, where b is Wien's constant. When temperature doubles from 300 K to 600 K, the peak wavelength halves, meaning it decreases and shifts toward shorter wavelengths. The spectrum also becomes taller (higher intensity) at all wavelengths due to the Stefan-Boltzmann law. Choice C incorrectly assumes that material properties determine the peak wavelength, but for blackbodies, only temperature matters. The transferable strategy is that higher temperature always shifts the spectrum toward shorter wavelengths (higher frequencies/energies).

Question 17

1. It emits the same photon energies, but a greater number of photons at each energy.
2. Its emission spectrum is independent of temperature because emissivity is constant.
3. It emits only higher-energy photons, and lower-energy photons are no longer emitted.
4. It emits higher-energy photons on average and also continues emitting lower-energy photons. (correct answer)

Explanation: This question tests understanding of blackbody radiation. As temperature increases, the peak of the blackbody spectrum shifts to shorter wavelengths (higher energies), and the intensity increases at all wavelengths according to Planck's law. The spectrum doesn't cut off at low energies; instead, the entire curve shifts and scales up, meaning the object emits more photons at every wavelength while the peak moves to higher energies. Choice B incorrectly suggests that lower-energy photons stop being emitted, which violates the continuous nature of blackbody spectra. The key strategy is to remember that temperature affects both the peak position and the overall intensity, with higher temperatures producing more radiation at all wavelengths while shifting the peak toward shorter wavelengths.

Question 18

1. It increases because shorter peak wavelength corresponds to higher photon energy. (correct answer)
2. It stays constant because photon energy depends only on the material.
3. It becomes infinite because intensity increases at all wavelengths.
4. It decreases because the spectrum shifts to shorter wavelength.

Explanation: This question tests understanding of blackbody radiation. Since E = hc/λ and the peak wavelength is inversely proportional to temperature (Wien's law), the peak photon energy is directly proportional to temperature. When temperature doubles from 1000 K to 2000 K, the peak wavelength halves, which means the peak photon energy doubles. The spectrum shifts toward shorter wavelengths (higher frequencies), corresponding to more energetic photons at the peak. Choice C incorrectly assumes photon energy depends on material properties rather than temperature for blackbodies. The key principle is that higher temperature shifts the spectrum toward shorter wavelengths, which means higher photon energies at the peak.

Question 19

1. $\lambda_{\text{peak}} \propto T^2$
2. $\lambda_{\text{peak}} \propto T$
3. $\lambda_{\text{peak}}$ is constant for a given object regardless of $T$.
4. $\lambda_{\text{peak}} \propto \dfrac{1}{T}$ (correct answer)

Explanation: This question tests understanding of blackbody radiation. The observation that peak wavelength halves (from 2.0 μm to 1.0 μm) when temperature doubles demonstrates Wien's displacement law: λ_peak = b/T. This inverse relationship means λ_peak ∝ 1/T, making the peak wavelength inversely proportional to absolute temperature. As temperature increases, the peak shifts to shorter wavelengths while total emitted intensity increases according to the Stefan-Boltzmann law. Choice A incorrectly suggests direct proportionality, which would mean higher temperatures produce longer wavelengths—opposite to what we observe. The key insight is that temperature and peak wavelength are inversely related: doubling temperature halves the peak wavelength.

Question 20

1. The hotter blackbody emits only higher-frequency radiation and none at lower frequencies.
2. The spectrum depends only on surface area, so both spectra have the same shape and peak.
3. The hotter blackbody emits more total power and has a peak at shorter wavelength. (correct answer)
4. The hotter blackbody emits less total power because its peak wavelength is shorter.

Explanation: This question tests understanding of blackbody radiation. When temperature doubles from 300 K to 600 K, two key changes occur: the peak wavelength halves (Wien's law: λ_peak ∝ 1/T) and the total emitted power increases by a factor of 16 (Stefan-Boltzmann law: P ∝ T⁴). The hotter blackbody emits more radiation at all wavelengths, with its peak shifted to shorter wavelengths. The spectrum maintains its characteristic shape but is compressed horizontally and stretched vertically. Choice D incorrectly claims the hotter object stops emitting at lower frequencies, but blackbody spectra always include all wavelengths. The principle to remember is that higher temperature means both a shift to shorter peak wavelength and increased emission at all wavelengths.