Blackbody Radiation
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AP Physics 2 › Blackbody Radiation
A blackbody’s peak wavelength changes from $10,\mu\text{m}$ at 290 K to about $5,\mu\text{m}$ at 580 K, and intensity increases. As temperature increases, which relationship is supported for peak wavelength?
$\lambda_{\text{peak}}$ increases because hotter objects emit only higher-energy photons.
$\lambda_{\text{peak}}$ is directly proportional to $T$.
$\lambda_{\text{peak}}$ is inversely proportional to $T$.
$\lambda_{\text{peak}}$ is independent of $T$.
Explanation
This question tests understanding of blackbody radiation. The data shows peak wavelength halving when temperature doubles (10 μm at 290 K to 5 μm at 580 K), confirming Wien's displacement law: λ_peak is inversely proportional to T. This inverse relationship means λ_peak × T = constant for all blackbodies. Choice A incorrectly suggests direct proportionality, reflecting the misconception that higher temperature means longer wavelengths. For blackbody radiation, peak wavelength always varies as 1/T, shifting to shorter wavelengths as temperature increases.
An ideal blackbody’s temperature increases from 400 K to 800 K; the peak shifts to shorter wavelength and the overall intensity rises. As temperature increases, which statement about emitted power per unit area is correct?
It increases by a factor of 2 because the peak wavelength halves.
It increases by a factor of 16 because total emission scales as $T^4$.
It increases by a factor of 4 because intensity is proportional to $T^2$.
It increases by a factor of 8 because photon energy is proportional to $T^3$.
Explanation
This question tests understanding of blackbody radiation. The Stefan-Boltzmann law states that total emitted power per unit area is proportional to T⁴: P/A = σT⁴. When temperature doubles from 400 K to 800 K, the emitted power increases by a factor of 2⁴ = 16. This dramatic increase occurs because blackbodies emit more radiation at all wavelengths as temperature rises, with the peak also shifting to shorter wavelengths. Choice B incorrectly suggests T² dependence, confusing the Stefan-Boltzmann law with other physical relationships. The strategy is to remember that total radiated power scales as the fourth power of temperature, explaining why hot objects glow so much more brightly.
A matte metal sphere is heated from 600 K to 900 K. Its emitted spectrum shifts so the peak wavelength decreases and the total intensity increases. As temperature increases, which statement correctly describes the spectrum?
The peak wavelength increases and the total intensity decreases.
The object emits only higher-energy photons and no lower-energy photons.
The peak wavelength decreases and the total intensity increases.
The peak wavelength stays constant and the total intensity stays constant.
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.
A star’s surface temperature increases slightly while remaining a near-blackbody. Which statement correctly describes how the color associated with peak emission changes?
It shifts toward longer wavelengths, so the peak appears redder.
It shifts toward shorter wavelengths, so the peak appears bluer.
It is unchanged because emitted color depends only on the star’s composition.
It becomes monochromatic because hotter objects emit at only one wavelength.
Explanation
This problem addresses blackbody radiation. As a star's surface temperature increases, Wien's displacement law dictates that the peak wavelength shifts to shorter values, moving from the red end toward the blue end of the visible spectrum. This is why hotter stars appear bluer while cooler stars appear redder. Choice A incorrectly suggests the peak shifts toward longer wavelengths, showing the misconception that temperature and wavelength are directly related. To understand stellar colors, remember that higher temperature always means shorter peak wavelength, producing bluer light.
A blackbody is heated so its peak wavelength decreases and the total emitted intensity increases. As temperature increases, which statement correctly compares spectra at two temperatures?
The hotter spectrum peaks at longer wavelength but has greater total intensity.
The hotter spectrum peaks at shorter wavelength but has the same total intensity.
The hotter spectrum peaks at shorter wavelength and has greater total intensity.
The spectra are identical because blackbody radiation does not depend on temperature.
Explanation
This question tests understanding of blackbody radiation. When temperature increases, two simultaneous changes occur: the peak wavelength decreases (Wien's law) and the total intensity increases dramatically (Stefan-Boltzmann law). The hotter spectrum shows both a leftward shift of the peak and greater area under the curve. Choice B incorrectly suggests constant total intensity, reflecting the misconception that only the peak position changes with temperature. Higher temperature always produces both shorter peak wavelength and greater total radiated power.
A blackbody source is heated so its spectrum’s peak shifts left (to shorter wavelength) and the area under the curve increases. As temperature increases, which statement correctly describes the spectrum?
The peak shifts to longer wavelength and the total emitted power increases.
The peak shifts to shorter wavelength and the total emitted power decreases.
The peak shifts to shorter wavelength and the total emitted power increases.
The peak position and total emitted power do not change with temperature.
Explanation
This question tests understanding of blackbody radiation. When a blackbody is heated, two key changes occur: the peak wavelength decreases (shifts left toward shorter wavelengths) and the total emitted power increases (larger area under the curve). These changes follow Wien's displacement law and Stefan-Boltzmann law respectively. Choice C incorrectly pairs the wavelength shift with decreasing power, reflecting confusion about how temperature affects emission intensity. Remember that increasing temperature always shifts the peak to shorter wavelengths AND increases total radiated power.
Two identical cavities emit blackbody radiation at 500 K and 1000 K. The hotter cavity’s spectrum has a higher peak and a peak at shorter wavelength. Which statement is correct about total emitted power?
The 500 K cavity emits more total power because it has more long-wavelength radiation.
Both cavities emit the same total power because blackbody radiation depends only on surface area.
The 1000 K cavity emits only shorter-wavelength photons, so total power cannot be compared.
The 1000 K cavity emits more total power because the spectrum’s overall intensity is greater.
Explanation
This question tests understanding of blackbody radiation. The Stefan-Boltzmann law states that total emitted power is proportional to T⁴, so doubling temperature increases power by a factor of 16. The 1000 K cavity emits far more total power than the 500 K cavity, with higher intensity across all wavelengths. Choice B incorrectly assumes temperature independence, reflecting the misconception that only surface area matters for thermal emission. Remember that higher temperature always means greater total radiated power, scaling as the fourth power of absolute temperature.
A ceramic object is heated so its spectrum’s peak wavelength moves from 2.0 µm to 1.0 µm while the emitted intensity increases. As temperature increases, which relation between temperature and peak wavelength is supported?
$\lambda_{\text{peak}} \propto \dfrac{1}{T}$
$\lambda_{\text{peak}} \propto T$
$\lambda_{\text{peak}} \propto T^2$
$\lambda_{\text{peak}}$ is constant for a given object regardless of $T$.
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.
A blackbody is warmed from 300 K to 600 K. The peak of its intensity spectrum shifts to shorter wavelengths and the curve becomes taller. As temperature increases, what happens to the peak wavelength $\lambda_{\text{peak}}$?
It becomes zero because all photons become infinitely energetic.
It increases because hotter objects radiate more strongly at all wavelengths.
It decreases because the spectrum shifts toward shorter wavelengths.
It stays the same because the material determines $\lambda_{\text{peak}}$.
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).
A hot object’s emission peak moves from 700 nm to 500 nm as it is heated, and its intensity increases. As temperature increases, which statement about photon emission is most accurate?
It emits photons of higher average energy, but it still emits some lower-energy photons.
It emits radiation at the same wavelengths because wavelength depends only on size.
It emits the same photon energies as before, just with higher speed.
It emits only photons with wavelength shorter than 500 nm after heating.
Explanation
This question tests understanding of blackbody radiation. When the emission peak shifts from 700 nm to 500 nm, the temperature has increased by a factor of 700/500 = 1.4 according to Wien's law. The blackbody continues to emit photons across all wavelengths, but with higher average energy as the spectrum shifts toward shorter wavelengths. The intensity increases at all wavelengths, with more dramatic increases at shorter wavelengths. Choice B incorrectly suggests emission is limited to wavelengths shorter than 500 nm, but blackbody spectra always include all wavelengths. The key insight is that higher temperature shifts the spectrum toward shorter wavelengths (higher energies) while maintaining continuous emission across all wavelengths.