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Chapter 12: Problem 75

A black body is at temperature of \(2880 \mathrm{~K}\). The energy of radiation emitted by this object between wavelength \(4990 \AA\) and \(5000 \AA\) is \(U_{1}\), between \(9900 \AA\) and \(10000 \AA\) is \(U_{2}\) and between \(14990 \AA\) and \(15000 \AA\) is \(U_{3}\), The Wein's constant is \(b=2.88 \times 10^{-3} \mathrm{~m} \mathrm{~K}\) then (a) \(U_{2}>U_{1}\) (b) \(U_{2}>U_{3}\) (c) \(U_{1}=U_{3}

Short Answer

Expert verified
(c) \(U_{1}=U_{3}<U_{2}\)

Step by step solution

01

Understand Wien's Displacement Law

Wien's Displacement Law states that the wavelength at which the emission of a black body spectrum is maximized, is inversely proportional to its temperature. The formula is \( \lambda_{max} T = b \), where \( \lambda_{max} \) is the peak wavelength, \( T \) is the temperature, and \( b \) is Wien's constant.
02

Calculate Peak Wavelength (\( \lambda_{max} \))

Given the temperature \( T = 2880 \, K \) and using the Wien's constant \( b = 2.88 \times 10^{-3} \, m \, K \), calculate \( \lambda_{max} \) using \( \lambda_{max} = \frac{b}{T} \). Substituting in, \( \lambda_{max} = \frac{2.88 \times 10^{-3}}{2880} \, m \). Simplifying gives \( \lambda_{max} \approx 1000 \times 10^{-9} = 1000 \, nm = 10000 \, \AA \).
03

Determine Energy Range Relations

Since \( \lambda_{max} \approx 10000 \, \AA \), the radiation intensity is highest at \( 10000 \, \AA \). Hence, at wavelength ranges close to this peak, emitted energy is higher. The given wavelength ranges are \( 4990 \, \AA - 5000 \, \AA \), \( 9900 \, \AA - 10000 \, \AA \), and \( 14990 \, \AA - 15000 \, \AA \). \( U_2 \) corresponds directly to \( \lambda_{max} \), making it the highest energy. \( U_1 \) and \( U_3 \) are symmetric and both further from \( \lambda_{max} \), likely resulting in similar energy values.

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Key Concepts

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

Black Body Radiation
Black body radiation refers to the theoretical concept of an idealized object that absorbs all incident electromagnetic radiation, regardless of frequency or angle. Such an object is called a "black body." The interesting part about black bodies is that they re-emit this absorbed energy in a characteristic spectrum.
  • Black bodies are perfect emitters, meaning they emit the maximum possible radiation at a given temperature for each wavelength.
  • The pattern of emission solely depends on the temperature, not on the material or surface characteristics.
The radiation emitted by a black body forms a spectrum that is continuous and depends on temperature. Understanding this spectrum is crucial for grasping concepts like Wien's Displacement Law, as it helps predict peak wavelength and intensity of radiation.
Peak Wavelength
The peak wavelength is the wavelength at which a black body emits its maximum radiation. According to Wien's Displacement Law, peak wavelength is inversely related to temperature.
For a black body at a higher temperature, the peak wavelength will shift to shorter lengths (e.g., from red towards blue), explaining why hotter objects often appear more blueish.
  • The formula for Wien's Law is: \( \lambda_{max} = \frac{b}{T} \)where \( \lambda_{max} \) is the peak wavelength, \( T \) is the temperature in Kelvin, and \( b \) is Wien's constant.
  • This inverse relationship means that an increase in temperature results in a shorter peak wavelength.
In our exercise, the calculated peak wavelength was approximately \( 10000 \text{ 脜} \) for a temperature of \( 2880 \text{ K} \), indicating that most radiation occurs closer to this wavelength.
Radiation Intensity
Radiation intensity refers to the amount of energy emitted by a black body per unit wavelength. The intensity is the highest at the peak wavelength, which for the black body at \( 2880 \text{ K} \), occurs at approximately \( 10000 \text{ 脜} \).
  • The intensity varies with the temperature of the black body and the wavelength of the emitted radiation.
  • For wavelengths close to \( \lambda_{max} \), the radiation intensity will be high.
In our example, the comparisons among \( U_1 \), \( U_2 \), and \( U_3 \) were made based on their proximity to the peak wavelength. Since \( U_2 \) was centered around the peak, it had the highest radiation intensity, compared to \( U_1 \) and \( U_3 \), which were further from this peak.
Temperature
Temperature is a crucial factor in determining the radiation properties of a black body. When considering black body radiation, temperature directly influences both the peak wavelength and intensity of the emitted radiation.
  • Higher temperatures result in shorter peak wavelengths, meaning the emission shifts into higher energy or "bluer" test parts of the spectrum.
  • With increasing temperature, a black body emits more overall radiation, meaning all intensities increase, which often involves a noticeable change in visible color.
In the original exercise, the temperature was given as \( 2880 \text{ K} \), which allowed calculation of the peak wavelength using Wien's Law. This temperature also explained the distribution and intensity of radiation across the specified wavelength ranges. Understanding how temperature affects these factors can help predict and explain different phenomena like the color of stars.

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Most popular questions from this chapter

A wall has two layers \(A\) and \(B\), each made of diffeent materials, The thickness of both the layers is the same. The thermal conductivity of \(A, K_{A}=3 K_{B^{*}}\) The temperature across the wall is \(20^{\circ} \mathrm{C}\). In thermal equilibrium (a) the temperature difference across \(A=150^{\circ} \mathrm{C}\) (b) rate of heat transfer across \(A\) is more than across \(B\) (c) rate of heat transfer across both is same (d) temperature difference across \(A\) is \(5^{\circ} \mathrm{C}\) :

\(\triangle 0.1 \mathrm{~kg}\) stecl ball falls from a height of \(10 \mathrm{~m}\) and bounecs to a height \(7 \mathrm{~m}\). (a) Why docs il nol bounce back to its original height? (b) If all the dissipated cnergy were absorbed by the ball as heat, how much will its temperatuc rise? (specific heat of steel \(=0.11 \mathrm{~K} \mathrm{cal} / \mathrm{k} \mathrm{g}^{\circ} \mathrm{C}, 1 \mathrm{cal}=4.2 \mathrm{~J}\) )

\(100 \mathrm{~g}\) of steam at \(100^{\circ} \mathrm{C}\). is passed into \(200 \mathrm{~g}\) of water and \(20 \mathrm{~g}\) of ice at \(0^{\circ} \mathrm{C}\) in a calorimeler whosc water cquivalent is \(50 \mathrm{~g} \cdot\left(L_{\text {stam }}=540 \mathrm{cal} / \mathrm{g}\right.\) and \(\left.L_{\mathrm{ioc}}=80 \mathrm{cal} / \mathrm{g} .\right) .\) The observed result is (a) the temperature of the system becomes \(169^{\circ} \mathrm{C}\). (b) half of the ice is meled and the temperature of the system remains \(0^{\circ} \mathrm{C}\). (c) the temperature remains \(100^{\circ} \mathrm{C}\) and \(53 \mathrm{~g}\) of stcam condensos. (d) the temperatue remains \(100^{\circ} \mathrm{C}\). and the entire steam condenses.

Statement-1: Heat is transferred most rapidly during radiation. Statement-2 : Transfer of heat take place with the speed of light during radiation.

\(\Lambda 10 \mathrm{ohm}\) resistance is kept in a liquid of mass \(100 \mathrm{gm}\) contained in a calorimeler of water cquivalent \(20 \mathrm{gm}\). Calculate the specific heat if a current of 4 amp. is passed for 7 minutes which raises the temperature through \(50^{\circ} \mathrm{C}\).
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