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Daylight and incandescent light may be approximated as a blackbody at the effective surface temperatures of \(5800 \mathrm{~K}\) and \(2800 \mathrm{~K}\), respectively. Determine the wavelength at maximum emission of radiation for each of the lighting sources.

Short Answer

Expert verified
Answer: The wavelength at maximum emission for daylight is 500 nm, and for incandescent light, it is 1040 nm.

Step by step solution

01

Understanding Wien's Displacement Law

Wien's Displacement Law relates the wavelength at maximum emission to the temperature of the blackbody, which is given by: \(\lambda_{max} = \dfrac{b}{T}\) Where \(\lambda_{max}\) is the wavelength at maximum emission, \(b\) is Wien's displacement constant (\(b \approx 2.9 \times 10^{-3} \mathrm{~m K}\)), and \(T\) is the temperature of the blackbody in kelvins. We will apply this formula to both lighting sources to find the wavelengths at maximum emission.
02

Calculate wavelength for daylight source

Given the temperature of the daylight source as \(T = 5800 \mathrm{~K}\). Then using Wien's displacement law: \(\lambda_{max_{daylight}} = \dfrac{b}{T_{daylight}}\) \(\lambda_{max_{daylight}} = \dfrac{2.9 \times 10^{-3} \mathrm{~m K}}{5800 \mathrm{~K}}\) \(\lambda_{max_{daylight}} = 5\times10^{-7} \mathrm{~m} = 500 \mathrm{~nm}\) The wavelength at maximum emission for daylight source is \(500 \mathrm{~nm}\).
03

Calculate wavelength for incandescent light source

Given the temperature of the incandescent light source as \(T = 2800 \mathrm{~K}\). Then using Wien's displacement law: \(\lambda_{max_{incandescent}} = \dfrac{b}{T_{incandescent}}\) \(\lambda_{max_{incandescent}} = \dfrac{2.9 \times 10^{-3} \mathrm{~m K}}{2800 \mathrm{~K}}\) \(\lambda_{max_{incandescent}} = 1.04 \times 10^{-6} \mathrm{~m} = 1040 \mathrm{~nm}\) The wavelength at maximum emission for the incandescent light source is \(1040 \mathrm{~nm}\).
04

Conclusion

In conclusion, the wavelength at maximum emission for daylight is \(500 \mathrm{~nm}\), and for incandescent light, it is \(1040 \mathrm{~nm}\).

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Key Concepts

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

Blackbody Radiation
Blackbody radiation is a concept in physics that describes how objects emit radiation or light based on their temperature. Imagine a perfect object that absorbs all incoming light and reflects none. This theoretical object is a 'blackbody'. In reality, no perfect blackbody exists, but many real-world objects approximate this behavior.

When a blackbody is heated, it emits light, visible or invisible, depending on its temperature. This emission covers a range of wavelengths. The intensity of the emitted radiation changes with wavelength, having a specific distribution. This distribution is called the blackbody radiation spectrum. The spectrum peaks at particular wavelengths, which shift as temperature changes.

Key characteristics of blackbody radiation include:
  • The radiation emitted is continuous and covers a broad range of wavelengths.
  • The peak wavelength is inversely proportional to the temperature of the blackbody. This means as temperature increases, the peak wavelength decreases.
  • Blackbody radiation plays a vital role in many fields, from understanding stellar objects to designing energy-efficient lighting.
Understanding blackbody radiation is crucial for developing an intuition about how temperature affects light emission.
Wavelength at Maximum Emission
The wavelength at maximum emission is a critical concept in understanding blackbody radiation. It defines the wavelength at which the intensity of light is highest for a given temperature. Wien's Displacement Law gives us a clear mathematical framework to determine this wavelength based on temperature.

Wien's Displacement Law: This law states that the product of the absolute temperature of a blackbody and the wavelength at maximum emission is constant. The formula is given by:
\[\lambda_{max} = \dfrac{b}{T}\]
Here, \(\lambda_{max}\) denotes the wavelength at which the emission is maximum, \(b\) is Wien's displacement constant \( \approx 2.9 \times 10^{-3} \mathrm{~m \cdot K} \), and \(T\) is the temperature in Kelvin.

This relationship tells us:
  • As the temperature of the blackbody increases, the peak wavelength shifts toward shorter wavelengths (e.g., from infrared to visible range).
  • For cooler objects, the peak is in the infrared range, explaining why some objects glow less visibly as they cool down.
Applying this understanding allows us to calculate light characteristics for various temperature-based sources, just like we see in the sunlight and incandescent lighting examples.
Daylight and Incandescent Light
Daylight and incandescent light are two common sources of artificial and natural lighting in our everyday environment. Both can be understood as emitters of blackbody radiation, albeit at different temperatures.

Daylight: Daylight can be considered equivalent to a blackbody emitting at a temperature of about \(5800 \mathrm{~K}\). The wavelength at maximum emission for sunlight is calculated using Wien's law and found to be approximately \(500 \mathrm{~nm}\), which falls in the visible spectrum, contributing to the bright and vibrant light we perceive.

Incandescent Light: Traditional incandescent bulbs operate at a lower temperature, around \(2800 \mathrm{~K}\). When applying Wien’s displacement law here, the maximum emission wavelength is about \(1040 \mathrm{~nm}\), which lies in the infrared region. This calculation explains why incandescent lights are warm and not as energy-efficient as other light types, as much of the energy is lost as heat.

These differences highlight why daylight appears more balanced in color versus the warm tint of incandescent lighting.
  • Daylight covers a broad spectrum, shining most intensely at visible wavelengths.
  • The warmth of incandescent light is due to its peak emission in the infrared.
Understanding these principles aids not only in designing various lighting but also in appreciating the natural and artificial light interactions.

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Most popular questions from this chapter

A cordless telephone is designed to operate at a frequency of \(8.5 \times 10^{8} \mathrm{~Hz}\). Determine the wavelength of these telephone waves.

A surface is exposed to solar radiation. The direct and diffuse components of solar radiation are 350 and \(250 \mathrm{~W} / \mathrm{m}^{2}\), and the direct radiation makes a \(35^{\circ}\) angle with the normal of the surface. The solar absorptivity and the emissivity of the surface are \(0.24\) and \(0.41\), respectively. If the surface is observed to be at \(315 \mathrm{~K}\) and the effective sky temperature is \(256 \mathrm{~K}\), the net rate of radiation heat transfer to the surface is (a) \(-129 \mathrm{~W} / \mathrm{m}^{2}\) (b) \(-44 \mathrm{~W} / \mathrm{m}^{2}\) (c) \(0 \mathrm{~W} / \mathrm{m}^{2}\) (d) \(129 \mathrm{~W} / \mathrm{m}^{2}\) (e) \(537 \mathrm{~W} / \mathrm{m}^{2}\)

A long metal sheet that can be approximated as a blackbody is being conveyed through a water bath to be cooled. In order to prevent thermal burn on people handling the sheet, it must exit the water bath at a temperature below \(45^{\circ} \mathrm{C}\). A radiometer is placed normal to and at a distance of \(0.5 \mathrm{~m}\) from the sheet surface to monitor the exit temperature. The radiometer receives radiation from a target area of \(1 \mathrm{~cm}^{2}\) of the metal sheet surface. When the radiometer detects that the metal sheet temperature is not below \(45^{\circ} \mathrm{C}\), an alarm will go off to warn that the sheet is not safe to touch. Determine the irradiation on the radiometer that the warning alarm should be triggered.

Consider a 4-cm-diameter and 6-cm-long cylindrical rod at \(1000 \mathrm{~K}\). If the emissivity of the rod surface is \(0.75\), the total amount of radiation emitted by all surfaces of the rod in \(20 \mathrm{~min}\) is (a) \(43 \mathrm{~kJ}\) (b) \(385 \mathrm{~kJ}\) (c) \(434 \mathrm{~kJ}\) (d) \(513 \mathrm{~kJ}\) (e) \(684 \mathrm{~kJ}\)

Consider an opaque plate that is well insulated on the edges and it is heated at the bottom with an electric heater. The plate has an emissivity of \(0.67\), and is situated in an ambient surrounding temperature of \(7^{\circ} \mathrm{C}\) where the natural convection heat transfer coefficient is \(7 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To maintain a surface temperature of \(80^{\circ} \mathrm{C}\), the electric heater supplies \(1000 \mathrm{~W} / \mathrm{m}^{2}\) of uniform heat flux to the plate. Determine the radiosity of the plate under these conditions.

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