/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 Estimate the wavelength correspo... [FREE SOLUTION] | 91Ó°ÊÓ

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Estimate the wavelength corresponding to maximum emission from each of the following surfaces: the sun, a tungsten filament at \(2500 \mathrm{~K}\), a heated metal at \(1500 \mathrm{~K}\), human skin at \(305 \mathrm{~K}\), and a cryogenically cooled metal surface at \(60 \mathrm{~K}\). Estimate the fraction of the solar emission that is in the following spectral regions: the ultraviolet, the visible, and the infrared.

Short Answer

Expert verified
The maximum emission wavelength for the given surfaces are: 1. Sun: \( 5 * 10^{-7} m \) (approx.) 2. Tungsten filament: \( 1.16 * 10^{-6} m \) 3. Heated metal: \( 1.93 * 10^{-6} m \) 4. Human skin: \( 9.5 * 10^{-6} m \) 5. Cryogenically cooled metal surface: \( 4.83 * 10^{-5} m \) Qualitatively, we know that most of the solar radiation lies in the visible and infrared regions, as the Earth's atmosphere absorbs much of the ultraviolet radiation. The exact fractions cannot be calculated based on Wien's Law alone.

Step by step solution

01

Recall Wien's Displacement Law

Recall that Wien's Displacement Law states that the wavelength of maximum emission (λ_max) is inversely proportional to the temperature (T) of the black body: \( \lambda_{max} = \frac{b}{T} \) where b is Wien's constant, approximately equal to \( 2.898 * 10^{-3} mK \).
02

Calculate the wavelength for each surface

We will now find the wavelength corresponding to the maximum emission for each surface using the given temperatures and Wien's Displacement Law: 1. Sun: T = 5778 K (approximately) 2. Tungsten filament: T = 2500 K 3. Heated metal: T = 1500 K 4. Human skin: T = 305 K 5. Cryogenically cooled metal surface: T = 60 K Calculate \( \lambda_{max} \) using \( \lambda_{max} = \frac{b}{T} \) for each surface: 1. Sun: \( \lambda_{max} = \frac{2.898 * 10^{-3}}{5778} \) = \(5 * 10^{-7} m \) (approximately) 2. Tungsten filament: \( \lambda_{max} = \frac{2.898 * 10^{-3}}{2500} \) = \( 1.16 * 10^{-6} m \) 3. Heated metal: \( \lambda_{max} = \frac{2.898 * 10^{-3}}{1500} \) = \( 1.93 * 10^{-6} m \) 4. Human skin: \( \lambda_{max} = \frac{2.898 * 10^{-3}}{305} \) = \( 9.5 * 10^{-6} m \) 5. Cryogenically cooled metal surface: \( \lambda_{max} = \frac{2.898 * 10^{-3}}{60} \) = \( 4.83 * 10^{-5} m \)
03

Estimate the fraction of solar emission in different spectral regions

To estimate the fraction of solar emission in the ultraviolet, visible, and infrared spectral regions, consider the following wavelength ranges: 1. Ultraviolet: 100 nm - 400 nm 2. Visible: 400 nm - 700 nm 3. Infrared: 700 nm - 1 mm The sun's maximum emission wavelength is about 500 nm (from Step 2), which is in the visible range. It is not possible to calculate the exact fractions just based on Wien's Law. However, qualitatively, we do know that: - Sun emits a significant portion of its radiation in the visible range, which is why sunlight appears white. - Most of the solar radiation lies in the visible and infrared regions since the Earth's atmosphere absorbs much of the ultraviolet radiation. In conclusion, we estimated the wavelength corresponding to maximum emission for different surfaces using Wien's Displacement Law and qualitatively analyzed the distribution of solar emission in various spectral regions.

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Key Concepts

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

Wavelength of Maximum Emission
Understanding the concept of the wavelength of maximum emission is vital when studying thermal radiation from objects. According to Wien's Displacement Law, every object emits radiation across a spectrum of wavelengths, but there's a particular wavelength where the emission is at its peak. This is known as the wavelength of maximum emission, denoted as \( \lambda_{max} \).

The wavelength of maximum emission can be calculated using the formula \( \lambda_{max} = \frac{b}{T} \), where b is Wien's constant, roughly 2.898 \times 10^{-3} m\cdot K, and T is the absolute temperature in kelvins. For instance, we can determine that the sun, with a temperature of approximately 5778 K, has a maximum emission wavelength of 5 \times 10^{-7} m, which is in the visible spectrum. As the temperature of an object decreases, the peak emission moves to longer wavelengths. This explains why objects at room temperature emit primarily in the infrared range, and not in visible light.

Learning how to apply Wien's Law provides a fundamental understanding of thermal physics and helps in predicting the behavior of different bodies at various temperatures, a concept especially useful in fields such as astronomy and materials science.
Blackbody Radiation
Blackbody radiation refers to the idealized spectral emission of a body that absorbs all incident electromagnetic radiation, regardless of wavelength or angle of incidence. This theoretical object, called a blackbody, re-emits radiation in a characteristic spectrum that depends solely on its temperature, not on its shape or composition.

In the context of our exercise, different materials such as a tungsten filament, heated metal, human skin, and cryogenically cooled metal surface, act as approximate blackbodies. Each has a unique emission spectrum that peaks at a wavelength calculable through Wien's law. It's fascinating to note that as the temperature of a blackbody increases, not only does its total emitted radiation increase exponentially (as per the Stefan-Boltzmann Law), but also the peak wavelength of emission shifts to shorter wavelengths, moving from the infrared towards the ultraviolet spectrum as the body gets hotter.

To visualize the practicality of blackbody radiation concepts, consider a filament inside a lightbulb, which emits light due to being heated to high temperatures; or the Earth, which emits infrared radiation due to its lower temperature compared to the sun.
Solar Emission Spectrum
The solar emission spectrum is a representation of the range of wavelengths emitted by the sun. While the sun emits radiation across a wide range of wavelengths, including ultraviolet (UV), visible, and infrared (IR), it is the temperature of the sun's surface, approximately 5778 K, that dictates the peak of its emission spectrum. As noted, the sun's maximum emission falls within the visible range, particularly around 500 nm.

A qualitative estimation of the sun's emission fractions in different spectral regions, as highlighted in the exercise, reveals that a significant portion of solar radiation is emitted in the visible range. This visibility forms the basis of our day-night cycle and vision. The reason the sun appears white to our eyes is because it emits a broad range of wavelengths in the visible spectrum. Notably, the Earth's atmosphere plays a critical role in filtering the sun's radiation—absorbing much of the UV while allowing most of the visible and IR to pass through.

Understanding the solar emission spectrum is fundamental in disciplines such as environmental science, astrophysics, and climate studies because it helps explain the heat balance of the Earth and the various biochemical and physical processes that are driven by solar radiation.

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Most popular questions from this chapter

A diffuse, opaque surface at \(700 \mathrm{~K}\) has spectral emissivities of \(\varepsilon_{\lambda}=0\) for \(0 \leq \lambda \leq 3 \mu \mathrm{m}, \varepsilon_{\lambda}=0.5\) for \(3 \mu \mathrm{m}<\lambda \leq 10 \mu \mathrm{m}\), and \(\varepsilon_{\lambda}=0.9\) for \(10 \mu \mathrm{m}<\lambda<\infty\). A radiant flux of \(1000 \mathrm{~W} / \mathrm{m}^{2}\), which is uniformly distributed between 1 and \(6 \mu \mathrm{m}\), is incident on the surface at an angle of \(30^{\circ}\) relative to the surface normal. Calculate the total radiant power from a \(10^{-4} \mathrm{~m}^{2}\) area of the surface that reaches a radiation detector positioned along the normal to the area. The aperture of the detector is \(10^{-5} \mathrm{~m}^{2}\), and its distance from the surface is \(1 \mathrm{~m}\).

A wet towel hangs on a clothes line under conditions for which one surface receives solar irradiation of \(G_{S}=900 \mathrm{~W} / \mathrm{m}^{2}\) and both surfaces are exposed to atmospheric (sky) and ground radiation of \(G_{\text {tam }}=\) \(200 \mathrm{~W} / \mathrm{m}^{2}\) and \(G_{\mathrm{g}}=250 \mathrm{~W} / \mathrm{m}^{2}\), respectively. Under moderately windy conditions, airflow at a temperature of \(27^{\circ} \mathrm{C}\) and a relative humidity of \(60 \%\) maintains a convection heat transfer coefficient of \(20 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) at both surfaces. The wet towel has an emissivity of \(0.96\) and a solar absorptivity of \(0.65\). As a first approximation the properties of the atmospheric air may be evaluated at a temperature of \(300 \mathrm{~K}\). Determine the temperature \(T_{s}\) of the towel. What is the corresponding evaporation rate for a towel that is \(0.75 \mathrm{~m}\) wide by \(1.50 \mathrm{~m}\) long?

A small disk \(5 \mathrm{~mm}\) in diameter is positioned at the center of an isothermal, hemispherical enclosure. The disk is diffuse and gray with an emissivity of \(0.7\) and is maintained at \(900 \mathrm{~K}\). The hemispherical enclosure, maintained at \(300 \mathrm{~K}\), has a radius of \(100 \mathrm{~mm}\) and an emissivity of \(0.85\). Calculate the radiant power leaving an aperture of diameter \(2 \mathrm{~mm}\) located on the enclosure as shown.

A radiation thermometer is a radiometer calibrated to indicate the temperature of a blackbody. A steel billet having a diffuse, gray surface of emissivity \(0.8\) is heated in a furnace whose walls are at \(1500 \mathrm{~K}\). Estimate the temperature of the billet when the radiation thermometer viewing the billet through a small hole in the furnace indicates \(1160 \mathrm{~K}\).

Photovoltaic materials convert sunlight directly to electric power. Some of the photons that are incident upon the material displace electrons that are in turn collected to create an electric current. The overall efficiency of a photovoltaic panel, \(\eta\), is the ratio of electrical energy produced to the energy content of the incident radiation. The efficiency depends primarily on two properties of the photovoltaic material, (i) the band gap, which identifies the energy states of photons having the potential to be converted to electric current, and (ii) the interband gap conversion efficiency, \(\eta_{\mathrm{bg}}\), which is the fraction of the total energy of photons within the band gap that is converted to electricity. Therefore, \(\eta=\eta_{\mathrm{bg}} F_{\mathrm{bg}}\) where \(F_{\mathrm{bg}}\) is the fraction of the photon energy incident on the surface within the band gap. Photons that are either outside the material's band gap or within the band gap but not converted to electrical energy are either reflected from the panel or absorbed and converted to thermal energy. Consider a photovoltaic material with a band gap of \(1.1 \leq B \leq 1.8 \mathrm{eV}\), where \(B\) is the energy state of a photon. The wavelength is related to the energy state of a photon by the relationship \(\lambda=1240 \mathrm{eV} \cdot \mathrm{nm} / B\). The incident solar irradiation approximates that of a blackbody at \(5800 \mathrm{~K}\) and \(G_{S}=1000 \mathrm{~W} / \mathrm{m}^{2}\). (a) Determine the wavelength range of solar irradiation corresponding to the band gap. (b) Determine the overall efficiency of the photovoltaic material if the interband gap efficiency is \(\eta_{\text {bog }}=0.50\) (c) If half of the incident photons that are not converted to electricity are absorbed and converted to thermal energy, determine the heat absorption per unit surface area of the panel.

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