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A black body has maximum wavelength \(\lambda_{m}\) at \(2000 \mathrm{~K}\). Its corresponding wavelength at \(3000 \mathrm{~K}\) will be \(\quad\) [Kerala CET 2005] (a) \(\frac{3}{2} \lambda_{m}\) (b) \(\frac{2}{3} \lambda_{m}\) (c) \(\frac{16}{81} \lambda_{m}\) (d) \(\frac{81}{16} \lambda_{m}\)

Short Answer

Expert verified
The maximum wavelength at 3000 K is \( \frac{2}{3} \lambda_{m} \) (option b).

Step by step solution

01

Understanding 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 the temperature of the body. Mathematically, it is expressed as \( \lambda_{m} = \frac{b}{T} \), where \( b \) is Wien's constant.
02

Applying Wien's Displacement Law for Both Temperatures

We have two temperatures: 2000 K and 3000 K. From Wien's law, \( \lambda_{m} \propto \frac{1}{T} \). Thus, for two different temperatures, \( \frac{\lambda_{m_1}}{\lambda_{m_2}} = \frac{T_2}{T_1} \).
03

Substituting the Given Values

Let \( \lambda_{1} \) be the maximum wavelength at 2000 K and \( \lambda_{2} \) be at 3000 K. Using the proportion, \( \frac{\lambda_{1}}{\lambda_{2}} = \frac{3000}{2000} = \frac{3}{2} \).
04

Solving for the New Wavelength

Rearrange the equation to find \( \lambda_{2} \): \( \lambda_{2} = \frac{2}{3} \lambda_{1} \). Thus, the maximum wavelength at the new temperature 3000 K is \( \frac{2}{3} \lambda_{m} \).

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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 is an important concept in physics, especially thermal physics. Imagine an ideal "perfect" black body: it absorbs all incident light, regardless of wavelength or angle. When this black body is heated, it emits radiation that corresponds to its temperature. This emitted radiation covers a wide spectrum of wavelengths, yet not all wavelengths are emitted equally.
  • As the temperature of the black body increases, the emitted radiation becomes more intense.
  • The distribution of the emitted wavelengths shifts, concentrating more intensity towards shorter wavelengths.
This phenomenon is critical to understanding how temperature affects the radiation of the body, and it's what Wien's Displacement Law helps to describe.
Maximum Wavelength
The maximum wavelength, in the context of black body radiation, refers to the specific wavelength at which the emission of radiation is strongest. Wien's Displacement Law gives us the mathematical relationship to understand this. It states that the maximum wavelength \(\lambda_m\) is inversely proportional to the absolute temperature \(T\) of the black body. This relationship can be expressed with the equation \( \lambda_{m} = \frac{b}{T} \), where \(b\) is Wien's constant.
  • When the temperature increases, the maximum wavelength decreases, indicating a shift towards shorter wavelengths such as blue light.
  • This shifting of maximum wavelength is why hot objects, like stars, glow differently depending on their temperature.
This concept is crucial for astronomers, who use it to determine the temperatures of stars by observing the color of their emitted light.
Thermal Physics
Thermal physics is a branch of physics that deals with heat, temperature, and the energetic properties of systems. It encompasses concepts like thermodynamics, statistical mechanics, and black body radiation. In the realm of thermal physics, understanding radiation phenomena is crucial. Black body radiation exemplifies how objects transfer energy in the form of electromagnetic waves as a function of their temperature.
  • Thermal physics helps explain how energy is distributed in a system and how this affects observable quantities like temperature and pressure.
  • It provides insights into the behavior of gases, liquids, and solids at different temperatures.
Wien's Displacement Law fits neatly in thermal physics as it ties the macroscopic observable effect—color change due to temperature change—with the concept of energy distribution in a hot object. This law is indispensable in fields like astrophysics and manufacturing, where temperature management is key.

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Most popular questions from this chapter

In an industrial process \(10 \mathrm{~kg}\) of water per hour is to be heated from \(20^{\circ} \mathrm{C}\) to \(80^{\circ} \mathrm{C}\). To do this steam at \(150^{\circ} \mathrm{C}\) is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at \(90^{\circ} \mathrm{C}\). How many kg of steam is required per hour? (Specific heat of steam \(=1\) calorie per \(\mathrm{g}^{\circ} \mathrm{C}\), Latent heat of vaporisation \(=540 \mathrm{cal} / \mathrm{g}\) ) (a) \(1 \mathrm{~g}\) (b) \(1 \mathrm{~kg}\) (c) \(10 \mathrm{~g}\) (d) \(10 \mathrm{~kg}\)

An oxygen cylinder of volume \(30 \mathrm{~L}\) has an initial gauge pressure of \(15 \mathrm{~atm}\) and a temperature of \(27^{\circ} \mathrm{C}\). After some oxygen is withdrawn from the cylinder the gauge pressure drops to 11 atm and its temperature drops to \(17^{\circ} \mathrm{C} .\) The mass of oxygen taken out of the cylinder \(\left(R=8.31 . \mathrm{J} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\right)\). molecular mas of \(\mathrm{O}_{2}=32 \mathrm{u}\) ) is [NCERT] (a) \(0.14 \mathrm{~g}\) (b) \(0.02 \mathrm{~g}\) (c) \(0.14 \mathrm{~kg}\) (d) \(0.014 \mathrm{~kg}\)

Two rods of same length and material transfer a given amount of heat in \(12 \mathrm{~s}\), when they are joined end to end (i.e., in series). But when they are joined in parallel, they will transfer same heat under same conditions in (a) 245 (b) \(3 \underline{5}\) (c) \(48 \mathrm{~s}\) (d) \(1.5 \mathrm{~s}\)

Water of volume \(2 \mathrm{~L}\) in a container is heated with a coil of \(1 \mathrm{~kW}\) at \(27^{\circ} \mathrm{C}\). The lid of the container is open and energy dissipates at rate of \(160 \mathrm{~J} / \mathrm{s}\). In how much time temperature will rise from \(27^{\circ} \mathrm{C}\) to \(77^{\circ} \mathrm{C}\) ? [Given specific heat of water is \(4.2 \mathrm{~kJ} / \mathrm{kg}\) ] (a) \(8 \min 20 \mathrm{~s}\) (b) \(6 \min 2 \mathrm{~s}\) (c) 7 min (d) \(14 \mathrm{~min}\)

The average energy and the rms speed of molecules in a sample of oxygen gas at \(400 \mathrm{~K}\) are \(7.21 \times 10^{-21} \mathrm{~J}\) and \(524 \mathrm{~ms}^{-1}\) respectively. The corresponding values at \(800 \mathrm{~K}\) are nearly (a) \(14.42 \times 10^{-21} 3,1048 \mathrm{~ms}^{-1}\) (b) \(10.18 \times 10^{-21} J, 741 \mathrm{~ms}^{-1}\) (c) \(7.21 \times 10^{-21} \mathrm{~J}, 1048 \mathrm{~ms}^{-1}\) (d) \(14.42 \times 10^{-21} \mathrm{~J}, 741 \mathrm{~ms}^{-1}\)

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