Math Tools 5.3 Working with the Stefan-Boltzmann and Wien's Laws The Stefan-Boltzmann law can be used to estimate the flux and luminosity of Earth. Earth's average temperature is ( 288 mathrm{~K} ), so the flux from its surface is: [ egin{aligned} mathcal{F} & =sigma T^{4} \ & =left(5.67 imes 10^{-8} mathrm{~W} / mathrm{m}^{2} mathrm{~K}^{4} ight) imes(288 mathrm{~K})^{4} \ & =390 mathrm{~W} / mathrm{m}^{2} end{aligned} ] The luminosity is the flux multiplied by the surface area ( (A) ) of Earth. Surface area is given by ( 4 pi R^{2} ), and the radius of Earth is ( 6,378,000 ) meters, or ( 6.378 imes 10^{6} ) meters. So the luminosity is: [ egin{aligned} L & =mathcal{F} imes A \ & =mathcal{F} imes 4 pi R^{2} \ & =left(390 mathrm{~W} / mathrm{m}^{2} ight) imesleft[4 pileft(6.378 imes 10^{6} mathrm{~m} ight)^{2} ight] \ & approx 2 imes 10^{17} mathrm{~W} end{aligned} ] This means that Earth emits the equivalent of 2,000,000,000,000,000 (2 million billion) 100-W lightbulbs. This is an enormous amount of energy, but not anywhere close to the amount emitted by the Sun. Wien's law also proves useful to astronomers as they study the universe. If they measure the spectrum of an object emitting thermal radiation and find where the peak in the spectrum is, they can use Wien's law to calculate the temperature of the object. For example, astronomers cannot drop a thermometer onto the Sun to directly measure its surface temperature, but they can observe the spectrum of the light coming from the Sun and estimate its surface temperature. The peak in the Sun's spectrum occurs at a wavelength of about ( 500 mathrm{~nm} ). Wien's law can be written as: [ T=frac{2,900,000 mathrm{~nm} mathrm{~K}}{lambda_{ ext {peak }}} ] Plugging the observed peak of the spectrum of the Sun ( left(lambda_{ ext {peak }}=500 mathrm{~nm} ight) ) into this equation gives: [ T=frac{2,900,000 mathrm{~nm} mathrm{~K}}{500 mathrm{~nm}}=5800 mathrm{~K} ] This is how you can know the surface temperature of the Sun. Suppose you want to calculate the peak wavelength at which Earth radiates. Plugging Earth's average temperature of ( 288 mathrm{~K} ) into Wien's law gives: [ lambda_{ ext {peak }}=frac{2,900,000 mathrm{~nm} mathrm{~K}}{288 mathrm{~K}}=10,100 mathrm{~nm}=10.1 mu mathrm{m} ] Earth's radiation peaks in the infrared region of the spectrum.
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Second, we know that we can use Wien's law to calculate the peak wavelength of the radiation. Wien's law is given by: \[ T=\frac{2,900,000 \mathrm{~nm} \mathrm{~K}}{\lambda_{\text {peak }}} \] Show more…
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Blackbody Radiation and Continuous Spectra A recurring theme throughout the study of quantum physics has been the omnipresence of Planck's constant h. This universal constant of the microscopic world first made its appearance in 1900 in the study of the radiation of so-called blackbodies. A blackbody is a substance that absorbs radiation of all wavelengths and radiates in a continuous spectrum at all wavelengths. It is given the name blackbody because an object that absorbs light at all wavelengths appears black to the human eye. By the end of the 19th century, several properties of blackbody radiation had been established. First, the total intensity I (the average rate of radiation of energy per unit surface area) emitted from a blackbody was shown to be proportional to the fourth power of its temperature: I = σT^4. This is called the Stefan-Boltzmann law for a blackbody. The constant of proportionality σ is known as the Stefan-Boltzmann constant and was determined to be σ = 5.67×10^−8 W/(m^2⋅K^4). It had also been discovered that the wavelength at which the radiation intensity was maximum varied inversely with temperature. This result, known as the Wien displacement law, is written λmT = 2.90×10^−3 m⋅K, where λm is the wavelength with the greatest radiated intensity. One aspect of blackbody radiation that remained unexplained was the full wavelength dependence of the intensity of the radiation, I(λ). In 1900, largely through trial and error, Max Planck formulated the following equation that successfully explained the wavelength dependence of the intensity: I(λ) = 2πhc^2 / (λ^5(e^(hc/λkBT)−1)), where h is Planck's constant, c is the speed of light in vacuum, and kB is Boltzmann's constant. Planck justified his law by claiming that different modes of electromagnetic oscillations within the cavity could only emit radiation in increments of energy equal to Planck's constant h multiplied by the frequency f. At first, Planck did not believe in that idea himself, but the revolutionary concept of quantization (or "clumping") of energy paved the way for the "quantum revolution" in physics. Part A Consider a blackbody that radiates with an intensity I1 at a room temperature of 300K. At what intensity I2 will this blackbody radiate when it is at a temperature of 400K? Express your answer in terms of I1. I2 = Part B At what wavelength λm would the intensity of blackbody radiation be at a maximum when the blackbody is at 2900K? Express your answer in meters to two significant figures. λm = Part C An astronomer is trying to estimate the surface temperature of a star with a radius of 5.0×10^8m by modeling it as an ideal blackbody. The astronomer has measured the intensity of radiation due to the star at a distance of 2.5×10^13m and found it to be equal to 0.055W/m2. Given this information, what is the temperature of the surface of the star? Express your answer in kelvins to two significant digits. T =
Timothy J.
The radiation emitted by a blackbody at temperature $T$ has a frequency distribution given by the Planck spectrum: $$ \epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right) $$ where $\epsilon_{T}(f)$ is the energy density of the radiation per unit increment of frequency, $v$ (for example, in watts per square meter per hertz), $h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}$ is Planck's constant, $k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{~m}^{2} \mathrm{~kg} \mathrm{~s}^{-2} \mathrm{~K}^{-1}$ is the Boltzmann constant, and $c$ is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits $f \rightarrow 0$ and $f \rightarrow \infty$ with a single peak in between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value. a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature. b) Evaluate the peak frequency at temperature $T=6.00 \cdot 10^{3} \mathrm{~K}$, approximately the temperature of the photosphere (surface) of the Sun. c) Evaluate the peak frequency at temperature $T=2.735 \mathrm{~K}$, the temperature of the cosmic background microwave radiation. d) Evaluate the peak frequency at temperature $T=300 . \mathrm{K}$, which is approximately the surface temperature of Earth.
The radiation emitted by a blackbody at temperature $T$ has a frequency distribution given by the Planck spectrum: $$\epsilon_{T}(f)=\frac{2 \pi h}{c^{2}}\left(\frac{f^{3}}{e^{h f / k_{\mathrm{B}} T}-1}\right)$$ where $\epsilon_{T}(f)$ is the energy density of the radiation per unit increment of frequency, $v$ (for example, in watts per square meter per hertz), $h=6.626 \cdot 10^{-34} \mathrm{~J} \mathrm{~s}$ is Planck's constant, $k_{\mathrm{B}}=1.38 \cdot 10^{-23} \mathrm{JK}^{-1}$ is the Boltzmann constant, and $c$ is the speed of light in vacuum. (We'll derive this distribution in Chapter 36 as a consequence of the quantum hypothesis of light, but here it can reveal something about radiation. Remarkably, the most accurately and precisely measured example of this energy distribution in nature is the cosmic microwave background radiation.) This distribution goes to zero in the limits $f \rightarrow 0$ and $f \rightarrow \infty$ with a single peak between those limits. As the temperature is increased, the energy density at each frequency value increases, and the peak shifts to a higher frequency value. a) Find the frequency corresponding to the peak of the Planck spectrum, as a function of temperature. b) Evaluate the peak frequency at $T=6.00 \cdot 10^{3} \mathrm{~K}$, approximately the temperature of the photosphere (surface) of the Sun. c) Evaluate the peak frequency at $T=2.735 \mathrm{~K}$, the temperature of the cosmic background microwave radiation. d) Evaluate the peak frequency at $T=300 . \mathrm{K},$ which is approximately the surface temperature of Earth.
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