Blackbody Radiation
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AP Physics 2 › Blackbody Radiation
A blackbody is heated, and its emission curve shifts and grows. As temperature increases, which statement correctly describes the peak intensity and peak wavelength together?
Peak intensity increases, and peak wavelength shifts to longer values.
Peak intensity decreases, and peak wavelength shifts to shorter values.
Peak intensity is unchanged, and peak wavelength is unchanged.
Peak intensity increases, and peak wavelength shifts to shorter values.
Explanation
This question tests blackbody radiation. As temperature increases, two key changes occur: the peak wavelength shifts to shorter values (Wien's law) and the peak intensity increases. The intensity increase happens because the total radiated power grows as T⁴, and this increased energy is distributed across the spectrum with the highest concentration at the peak. Choice B incorrectly states the peak shifts to longer wavelengths, revealing confusion about temperature's effect on the spectrum. When analyzing blackbody curves, remember that heating always shifts the peak left (shorter λ) and up (higher intensity).
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 stays constant and the total intensity stays constant.
The peak wavelength decreases and the total intensity increases.
The peak wavelength increases and the total intensity decreases.
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.
A ceramic plate behaves approximately as a blackbody. When heated from 300 K to 600 K, the peak of its emitted spectrum moves to shorter wavelength while the curve’s area increases. As temperature increases, what happens to the peak wavelength?
It increases, shifting the peak toward longer wavelengths.
It decreases, shifting the peak toward shorter wavelengths.
It becomes undefined because hotter objects emit only one wavelength.
It remains the same because wavelength is independent of temperature.
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).
A filament approximates a blackbody at 1200 K, then at 2400 K. As temperature increases, which statement correctly describes photons emitted at long wavelengths?
Long-wavelength photons are no longer emitted because hotter objects emit only short wavelengths.
Long-wavelength photons are still emitted, and the spectrum’s peak shifts to longer wavelengths.
Long-wavelength photons are still emitted, and the spectrum’s peak stays at the same wavelength.
Long-wavelength photons are still emitted, but the spectrum’s peak shifts to shorter wavelengths.
Explanation
This question addresses blackbody radiation. When a blackbody's temperature increases, its emission spectrum shifts such that the peak moves to shorter wavelengths, but the blackbody continues to emit at all wavelengths including long ones. The intensity at long wavelengths actually increases with temperature, though not as dramatically as at short wavelengths. Choice A incorrectly states that long-wavelength emission stops, reflecting the misconception that hot objects emit only short-wavelength radiation. The key principle is that blackbodies emit a continuous spectrum at all temperatures; higher temperature shifts the peak but enhances emission everywhere.
A ceramic sphere approximates a blackbody at 300 K and then at 600 K. Which statement correctly describes how the peak wavelength changes?
It shifts to longer wavelengths because hotter objects radiate more strongly at redder light.
It stays at the same wavelength because blackbody peak depends only on material type.
It disappears because a hotter blackbody emits only at a single wavelength.
It shifts to shorter wavelengths because increasing temperature increases typical photon energy.
Explanation
This question examines blackbody radiation. Wien's displacement law states that the peak wavelength is inversely proportional to temperature (λ_peak ∝ 1/T), so doubling the temperature from 300 K to 600 K halves the peak wavelength. This shift toward shorter wavelengths occurs because higher temperatures produce more energetic photons on average, and photon energy is inversely related to wavelength. Choice A incorrectly suggests the peak shifts to longer wavelengths, reflecting the misconception that "hotter means redder" when actually hotter means bluer. To analyze blackbody spectra, always apply the rule that increasing temperature shifts the peak toward shorter wavelengths (bluer light).
A blackbody radiator is measured at temperature $T$ and then at $2T$. Which statement correctly describes the peak wavelength $\lambda_{\text{peak}}$?
$\lambda_{\text{peak}}$ halves because peak wavelength is inversely proportional to temperature.
$\lambda_{\text{peak}}$ becomes zero because higher temperature eliminates any peak.
$\lambda_{\text{peak}}$ is unchanged because peak wavelength is independent of temperature.
$\lambda_{\text{peak}}$ doubles because peak wavelength is proportional to temperature.
Explanation
This problem tests blackbody radiation. Wien's displacement law states that λ_peak × T = constant, meaning peak wavelength is inversely proportional to temperature. When temperature doubles from T to 2T, the peak wavelength halves, shifting the spectrum toward shorter wavelengths (higher energies). This relationship arises because higher temperatures produce more energetic photons on average. Choice A incorrectly claims wavelength doubles, showing the misconception that wavelength and temperature are directly proportional. To solve blackbody problems, remember that peak wavelength and temperature are inversely related: λ_peak ∝ 1/T.
A blackbody radiator’s temperature increases, and the measured spectrum shows a shorter peak wavelength and greater intensity at every wavelength. As temperature increases, which statement correctly describes the spectrum?
The spectrum shifts to shorter wavelengths and intensity increases at all wavelengths.
The spectrum shifts to longer wavelengths and intensity decreases at all wavelengths.
The spectrum becomes a single wavelength because hotter objects emit monochromatically.
The spectrum is unchanged because emission is independent of temperature.
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.
A blackbody filament is warmed from 1000 K to 2000 K. The emitted spectrum becomes taller and its peak shifts to a shorter wavelength. As temperature increases, which statement correctly describes the emitted radiation?
The total emitted power per unit area decreases and the peak wavelength increases.
The total emitted power per unit area is unchanged and the peak wavelength is unchanged.
The total emitted power per unit area increases and the peak wavelength decreases.
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.
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 shorter wavelengths, so the peak appears bluer.
It shifts toward longer wavelengths, so the peak appears redder.
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 small cavity radiator is a near-ideal blackbody. When its temperature rises from 500 K to 1000 K, the peak wavelength moves from about 5.8 μm to about 2.9 μm and the curve grows taller. As temperature increases, which statement best describes the emitted spectrum?
The peak wavelength shifts to shorter wavelengths, and the total emitted intensity increases.
The peak wavelength stays the same, and the total emitted intensity decreases.
The peak wavelength shifts to longer wavelengths, and the total emitted intensity decreases.
The object emits only shorter wavelengths and emits no radiation at longer wavelengths.
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.