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Approximations to Planck's law for the spectral emissive power are the Wien and Rayleigh-Jeans spectral distributions, which are useful for the extreme low and high limits of the product \(\lambda T\), respectively. (a) Show that the Planck distribution will have the form $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{\lambda^{5}} \exp \left(-\frac{C_{2}}{\lambda T}\right) $$ when \(C_{2} / \lambda T\) sor 1 and determine the error (compared to the exact distribution) for the condition \(\lambda T=\) \(2898 \mu \mathrm{m} \cdot \mathrm{K}\). This form is known as Wien's law. (b) Show that the Planck distribution will have the form $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{C_{2}} \frac{T}{\lambda^{4}} $$ when \(C_{2} / \lambda T \& 1\) and determine the error (compared to the exact distribution) for the condition \(\lambda T=100,000 \mu \mathrm{m} \cdot \mathrm{K}\). This form is known as the Rayleigh-Jeans law.

Short Answer

Expert verified
The Wien's approximation of the Planck distribution is given by \(E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{\lambda^{5}} \exp\left(-\frac{C_{2}}{\lambda T}\right)\) and is valid when \(C_2 / \lambda T \gg 1\). The error for the condition \(\lambda T=2898 \mu\mathrm{m} \cdot\mathrm{K}\) can be calculated as the absolute difference between the exact Planck distribution and Wien's approximation. The Rayleigh-Jeans approximation of the Planck distribution is given by \(E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{C_{2}} \frac{T}{\lambda^{4}}\) and is valid when \(C_2 / \lambda T \ll 1\). The error for the condition \(\lambda T=100{,}000\,\mu\mathrm{m} \cdot\mathrm{K}\) can be calculated as the absolute difference between the exact Planck distribution and the Rayleigh-Jeans approximation.

Step by step solution

01

Part (a): Deriving Wien's Law

To derive Wien's Law, we are given that the Planck distribution will have the following form when \(C_2 / \lambda T \gg 1\): $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{\lambda^{5}} \exp\left(-\frac{C_{2}}{\lambda T}\right) $$ This approximation holds true when \(C_2 / \lambda T \gg 1\). The Planck distribution's exact formula is given by $$ E_{\lambda, b}(\lambda, T) = \frac{2 \pi h c^2}{\lambda^5}\frac{1}{\mathrm{e}^{\frac{hc}{\lambda k_\mathrm{B} T}}-1} $$ where \(h\) is the Planck constant, \(c\) is the speed of light, and \(k_\mathrm{B}\) is the Boltzmann constant. We can rewrite this formula with the constants \(C_1 = 2 \pi h c^2\) and \(C_2 = hc/k_\mathrm{B}\) as: $$ E_{\lambda, b}(\lambda, T) = \frac{C_1}{\lambda^5}\frac{1}{\mathrm{e}^{\frac{C_2}{\lambda T}}-1} $$ To obtain the Wien's approximation, first note that when \(\frac{C_2}{\lambda T} \gg 1\), \(\mathrm{e}^{\frac{C_2}{\lambda T}} \gg 1\). Thus, we can approximate the denominator \(\mathrm{e}^{\frac{C_2}{\lambda T}} - 1 \approx \mathrm{e}^{\frac{C_2}{\lambda T}}\). Then, we have: $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{\lambda^{5}} \exp{\left(-\frac{C_{2}}{\lambda T}\right)} $$ which is the Wien's law.
02

Part (a): Error Calculation for Wien's Law

For the given condition, \(\lambda T = 2898 \mu\mathrm{m} \cdot\mathrm{K}\), the exact Planck distribution is: $$ E_{\lambda, b}^{\mathrm{exact}}(\lambda, T) = \frac{C_1}{\lambda^5}\frac{1}{\mathrm{e}^{\frac{C_2}{\lambda T}}-1} $$ And the Wien's approximation is: $$ E_{\lambda, b}^{\mathrm{Wien}}(\lambda, T) = \frac{C_{1}}{\lambda^{5}} \exp\left(-\frac{C_{2}}{\lambda T}\right) $$ The error is given as the absolute difference between the exact Planck distribution and Wien's approximation: $$ \mathrm{Error} = |E_{\lambda, b}^{\mathrm{exact}}(\lambda, T) - E_{\lambda, b}^{\mathrm{Wien}}(\lambda, T)| $$
03

Part (b): Deriving the Rayleigh-Jeans Law

To derive the Rayleigh-Jeans law, we are given that the Planck distribution will have the following form when \(C_2 / \lambda T \ll 1\): $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{C_{2}} \frac{T}{\lambda^{4}} $$ Using the same technique, we'll rewrite the exact Planck distribution formula with the given constants \(C_1\) and \(C_2\). When \(\frac{C_2}{\lambda T} \ll 1\), we can use the approximation \(\mathrm{e}^{\frac{C_2}{\lambda T}} \approx 1 + \frac{C_2}{\lambda T}\). Thus, the denominator becomes \(\mathrm{e}^{\frac{C_2}{\lambda T}} - 1 \approx \frac{C_2}{\lambda T}\). Then, we have: $$ E_{\lambda, b}(\lambda, T) \approx \frac{C_{1}}{C_{2}} \frac{T}{\lambda^{4}} $$ which is the Rayleigh-Jeans law.
04

Part (b): Error Calculation for Rayleigh-Jeans Law

For the given condition, \(\lambda T = 100{,}000\,\mu\mathrm{m} \cdot\mathrm{K}\), the exact Planck distribution is: $$ E_{\lambda, b}^{\mathrm{exact}}(\lambda, T) = \frac{C_1}{\lambda^5}\frac{1}{\mathrm{e}^{\frac{C_2}{\lambda T}}-1} $$ And the Rayleigh-Jeans approximation is: $$ E_{\lambda, b}^{\mathrm{Rayleigh-Jeans}}(\lambda, T) = \frac{C_{1}}{C_{2}} \frac{T}{\lambda^{4}} $$ The error is given as the absolute difference between the exact Planck distribution and the Rayleigh-Jeans approximation: $$ \mathrm{Error} = |E_{\lambda, b}^{\mathrm{exact}}(\lambda, T) - E_{\lambda, b}^{\mathrm{Rayleigh-Jeans}}(\lambda, T)| $$

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wien's Law
Understanding the behavior of radiation emitted by a black body is crucial in the field of physics, particularly in thermodynamics and quantum mechanics. Wien's Law offers an important approximation for the spectral emissive power at the high-frequency end of the spectrum. This law indicates how the wavelength at which a black body emits radiation most strongly is inversely proportional to its temperature.

Using Planck's law as a basis, Wien's Law simplifies the complex relationship for cases where the product of wavelength \times temperature, denoted \text(\(\textlambda T\)), is small. This law states that the spectral emissive power is approximately proportional to \text(\(\frac{C_{1}}{\textlambda^{5}} \exp\textleft(-\frac{C_{2}}{\textlambda T}\textright)\)), revealing the emission characteristics for hotter objects that emit primarily at shorter wavelengths. For students, it's helpful to note that this approximation holds when the exponential term in Planck's law is much greater than one, thus simplifying the denominator of the distribution function. An error analysis, when comparing to the exact Planck distribution, can demonstrate the practicality of this approximation.
Rayleigh-Jeans Law
In contrast to Wien's Law, the Rayleigh-Jeans Law applies to the opposite end of the spectrum, specifically dealing with low frequencies or long wavelengths. As the product of wavelength \times temperature increases, Wien's Law becomes less accurate, and Rayleigh-Jeans Law becomes a more appropriate model.

This law simplifies Planck's radiation formula under the condition that the product \text(\(\textlambda T\)) is large, providing an approximation given by \text(\(\frac{C_{1}}{C_{2}} \frac{T}{\textlambda^{4}}\)). The Rayleigh-Jeans Law is derived by assuming that the exponential term in Planck's distribution can be approximated by its first term in a Taylor series expansion. Students should be aware that this approximation becomes increasingly inaccurate at high frequencies due to the 'ultraviolet catastrophe,' a problem historically fixed by Planck's quantization approach. Understanding when to use the Rayleigh-Jeans approximation and recognizing its limitations is fundamental for accurately modeling the behavior of electromagnetic radiation at long wavelengths.
Spectral Emissive Power
Spectral emissive power is a fundamental concept when describing the thermal radiation emitted per unit area of a body across a given wavelength. This quantity dictates how much energy is emitted at a particular wavelength and is the central element of both Wien's and Rayleigh-Jeans laws. It is imperative to comprehend that the spectral emissive power is wavelength-dependent, which implies that a body at a certain temperature will not emit the same amount of energy across all wavelengths.

In the context of the black body radiation described by Planck's law, the spectral emissive power helps us understand how energy distribution varies with temperature and wavelength. For educational purposes, this concept underlines the idea that a hotter object does not just emit more energy overall, but also peaks at shorter wavelengths, as described by Wien's Displacement Law, while a cooler object emits more at longer wavelengths. Students could benefit from working through problem sets that involve calculating the spectral emissive power under different scenarios to solidify their understanding of this concept.
Black Body Radiation
The phenomenon of black body radiation provides the framework for the study of thermally emitted radiation from objects. A black body is an idealized physical body that perfectly absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence, and emits radiation at a characteristic rate dependent on temperature. This concept is critical because it sets the stage for Planck's law and consequently for Wien's and Rayleigh-Jeans laws.

Planck's law, which describes the radiation emitted by a black body in thermal equilibrium, represents the combined contributions of many modes of electromagnetic radiation. It considers the quantum nature of light and was pivotal in the development of quantum mechanics. For students grappling with this concept, understanding black body radiation as an idealization helps in studying how real materials emit and absorb radiation. Simplifications such as Wien's Law for high-frequency radiation and Rayleigh-Jeans Law for low-frequency radiation highlight specific aspects of this emission spectrum and are useful approximations under the right conditions. Practical exercises that include calculating the characteristics of a black body spectrum can provide students with insights into how energy is distributed across different wavelengths and why this matters in various scientific applications.

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Most popular questions from this chapter

A manufacturing process involves heating long copper rods, which are coated with a thin film, in a large furnace whose walls are maintained at an elevated temperature \(T_{w}\). The furnace contains quiescent nitrogen gas at 1-atm pressure and a temperature of \(T_{x}=T_{w}\). The film is a diffuse surface with a spectral emissivity of \(\varepsilon_{\lambda}=0.9\) for \(\lambda \leq 2 \mu \mathrm{m}\) and \(\varepsilon_{\lambda}=0.4\) for \(\lambda>2 \mu \mathrm{m}\). (a) Consider conditions for which a rod of diameter \(D\) and initial temperature \(T_{i}\) is inserted in the furnace, such that its axis is horizontal. Assuming validity of the lumped capacitance approximation, derive an equation that could be used to determine the rate of change of the rod temperature at the time of insertion. Express your result in terms of appropriate variables. (b) If \(T_{w}=T_{\mathrm{o}}=1500 \mathrm{~K}, T_{i}=300 \mathrm{~K}\), and \(D=10 \mathrm{~mm}\), what is the initial rate of change of the rod temperature? Confirm the validity of the lumped capacitance approximation. (c) Compute and plot the variation of the rod temperature with time during the heating process.

An opaque, horizontal plate has a thickness of \(L=21 \mathrm{~mm}\) and thermal conductivity \(k=25 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Water flows adjacent to the bottom of the plate and is at a temperature of \(T_{x, w}=25^{\circ} \mathrm{C}\). Air flows above the plate at \(T_{x, a}=260^{\circ} \mathrm{C}\) with \(h_{a}=40 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The top of the plate is diffuse and is irradiated with \(G=1450 \mathrm{~W} / \mathrm{m}^{2}\), of which \(435 \mathrm{~W} / \mathrm{m}^{2}\) is reflected. The steady-state top and bottom plate temperatures are \(T_{t}=43^{\circ} \mathrm{C}\) and \(T_{b}=35^{\circ} \mathrm{C}\), respectively. Determine the transmissivity, reflectivity, absorptivity, and emissivity of the plate. Is the plate gray? What is the radiosity associated with the top of the plate? What is the convection heat transfer coefficient associated with the water flow?

A proposed method for generating electricity from solar irradiation is to concentrate the irradiation into a cavity that is placed within a large container of a salt with a high melting temperature. If all heat losses are neglected, part of the solar irradiation entering the cavity is used to melt the salt while the remainder is used to power a Rankine cycle. (The salt is melted during the day and is resolidified at night in order to generate electricity around the clock.) Consider conditions for which the solar power entering the cavity is \(q_{\mathrm{sal}}=7.50 \mathrm{MW}\) and the time rate of change of energy stored in the salt is \(\dot{E}_{\mathrm{st}}=3.45 \mathrm{MW}\). For a cavity opening of diameter \(D_{s}=1 \mathrm{~m}\), determine the heat transfer to the Rankine cycle, \(q_{R}\). The temperature of the salt is maintained at its melting point, \(T_{\text {salt }}=T_{\text {m }}=1000^{\circ} \mathrm{C}\). Neglect heat loss by convection and irradiation from the surroundings.

Two small surfaces, \(A\) and \(B\), are placed inside an isothermal enclosure at a uniform temperature. The enclosure provides an irradiation of \(6300 \mathrm{~W} / \mathrm{m}^{2}\) to each of the surfaces, and surfaces A and B absorb incident radiation at rates of 5600 and \(630 \mathrm{~W} / \mathrm{m}^{2}\), respectively. Consider conditions after a long time has elapsed. (a) What are the net heat fluxes for each surface? What are their temperatures? (b) Determine the absorptivity of each surface. (c) What are the emissive powers of each surface? (d) Determine the emissivity of each surface.

Two plates, one with a black painted surface and the other with a special coating (chemically oxidized copper) are in earth orbit and are exposed to solar radiation. The solar rays make an angle of \(30^{\circ}\) with the normal to the plate. Estimate the equilibrium temperature of each plate assuming they are diffuse and that the solar flux is \(1368 \mathrm{~W} / \mathrm{m}^{2}\). The spectral absorptivity of the black painted surface can be approximated by \(\alpha_{\lambda}=0.95\) for \(0 \leq \lambda \leq \infty\) and that of the special coating by \(\alpha_{\lambda}=0.95\) for \(0 \leq \lambda<3 \mu \mathrm{m}\) and \(\alpha_{\lambda}=0.05\) for \(\lambda \geq 3 \mu \mathrm{m}\).

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