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Chapter 12: Problem 84

When the wavelength corresponding to maximum emission of radiation is halved, the radiant power of the body becomes \(\begin{array}{lll}\text { (a) Double } & \text { (b) Sixteen times } & \text { (c) } \text { Half }\end{array}\) (d) Four times

Short Answer

Expert verified
The radiant power becomes sixteen times larger.

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 is maximum is inversely proportional to the absolute temperature of the body. The law is given by the formula \( \lambda_{max} T = b \), where \( \lambda_{max} \) is the peak wavelength, \( T \) is the absolute temperature, and \( b \) is Wien's displacement constant.
02

Relate Temperature with Wavelength Change

Given the wavelength corresponding to maximum emission is halved, we use Wien's Law: \( \lambda_{max, new} = \frac{1}{2} \lambda_{max} \). So, keeping the product constant, the temperature doubles, i.e., \( T_{new} = 2T \).
03

Use Stefan-Boltzmann Law for Radiant Power

According to the Stefan-Boltzmann Law, the total radiant power (\( P \)) radiated per unit area of a black body is proportional to the fourth power of its absolute temperature. It is given by \( P = \sigma T^4 \), where \( \sigma \) is the Stefan-Boltzmann constant.
04

Calculate the Change in Radiant Power

Since the temperature doubles when the wavelength is halved (\( T_{new} = 2T \)), substitute into Stefan-Boltzmann Law: \( P_{new} = \sigma (2T)^4 = 16 \sigma T^4 = 16P \). Thus, the radiant power becomes sixteen times the original power.

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Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is crucial in understanding how much energy a black body radiates. It tells us that the total energy emitted per unit surface area of a black body is proportional to the fourth power of its absolute temperature. This means small changes in temperature can cause big changes in the emitted energy.

This law is mathematically expressed as:\[P = \sigma T^4\]Where:
  • \(P\) is the radiant power per unit area.
  • \(\sigma\) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \text{Wm}^{-2}\text{K}^{-4}\).
  • \(T\) is the absolute temperature in Kelvin.
By knowing the temperature, we can easily calculate how much radiant energy is being emitted. When applied, this law helps us understand phenomena like why the Sun emits so much more energy compared to a cooler star.
Black Body Radiation
Black body radiation is a concept describing an idealized physical body that absorbs all incident electromagnetic radiation, regardless of frequency or angle of incidence. Such a body emits radiation in a manner that depends solely on its temperature.

The emitted radiation, known as black body radiation, has a characteristic spectrum that only depends on the temperature of the body. It is important because:
  • This spectrum peaks at a wavelength determined by the temperature, described by Wien's Displacement Law.
  • It helps in determining the temperature of stars and other astronomical objects based on spectral analysis.
  • It serves as a baseline in radiation studies, helping us understand real-world objects that approximate black bodies.
Understanding this concept is vital in physics as it lays the groundwork for quantum mechanics and explains why objects change color with temperature.
Radiant Power Calculation
Radiant power calculation involves determining the total power emitted by an object as a function of its temperature and emissive properties. The calculation is based on principles such as the Stefan-Boltzmann Law and Wien's Displacement Law.

When the peak wavelength of emitted radiation is halved, as per Wien's Displacement Law, the temperature doubles. To calculate the change in radiant power:
  • Use Wien's Law: \(\lambda_{max, new} = \frac{1}{2} \lambda_{max}\) thus \(T_{new} = 2T\).
  • Insert the new temperature into the Stefan-Boltzmann formula: \(P_{new} = \sigma (2T)^4\).
  • Simplify: \(P_{new} = 16\sigma T^4\).
This results in the new power being sixteen times the original power. Such calculations are useful in practical scenarios, from assessing stellar luminosity to energy outputs in industrial applications.

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Most popular questions from this chapter

Figure shows a lagged copper bar \(A B\) whose ends are pressed against metal tanks at \(100^{\circ} \mathrm{C}\) and \(0^{\circ} \mathrm{C}\) but are separated from them by layers of dirt. The length of the bar is \(10 \mathrm{~cm}\) and the dirt laycr are \(0.1 \mathrm{~mm}\) thick. The conductive of dirt is \(0.001\) times that of copper. the temperature dilference of coppor bar is (a) \(5(\mathrm{j} \mathrm{C}\) (b) \(20^{\circ} \mathrm{C}\) (c) \(33.4^{\circ} \mathrm{C}\) (d) \(60^{\circ} \mathrm{C}\)

\(\triangle 10 \mathrm{~kW}\) drilling machine is used to drill a bore in a small aluminium block of mass \(8.0 \mathrm{~kg}\). 1low much is the rise in temperature of the block in \(2.5\) minute? \(\Lambda\) ssuming \(50 \%\) of power is used up in heating the machine itself or lost the surroundings specific heat of aluminium \(=0.91 \mathrm{~J} / \mathrm{g}-{ }^{\circ} \mathrm{C}\)

\(10 \mathrm{gm}\) of water at \(70^{\circ} \mathrm{C}\) is mixed with \(5 \mathrm{gm}\) of water at \(30^{\circ} \mathrm{C}\). Find the temperature of the mixture in equilibrium.

\(\Lambda\) pond of water at \(0^{\circ} \mathrm{C}\) is covered with layer of ice \(4 \mathrm{~cm}\) thick if air temperature is \(-10^{\circ} \mathrm{C}\) (constant), how long it takes for ice thickness to increase to \(8 \mathrm{~cm} ? K_{\text {ia }}=2 \mathrm{~W} / \mathrm{m}^{\circ} \mathrm{C}, L_{f}=80 \mathrm{cal} / \mathrm{gm}\), \(\rho_{\text {ice }}=900 \mathrm{~kg} / \mathrm{m}^{3}\)

The ends of a melal rod are kepl at temperalures \(\theta_{1}\) and \(\theta_{2}\) with \(\theta_{2}>\theta_{1}\). The rate of flow of heat along the rod is direculy proportional to (a) Lhe length of tho rod (b) the diameler of the rod (c) the cross-scetional area of the rod (d) the temperature difference \(\left(\theta_{2}-\theta_{1}\right)\) bctween the ends of the rod
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