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For an enclosure maintained at \(2000 \mathrm{~K}\), the maximum radiation occurs at wavelength \(\lambda_{\mathrm{m}^{\prime}}\). If the temperature is raised to \(3000 \mathrm{~K}\), the peak will shift to (a) \(0.5 \lambda_{\mathrm{m}}\) (b) \(\lambda_{m}\) (c) \(\frac{2}{3} \lambda_{m}\) (d) \(\frac{3}{2} \lambda_{m}\)

Short Answer

Expert verified
The peak wavelength shifts to \( \frac{2}{3} \lambda_m \), which corresponds to option (c).

Step by step solution

01

Identify the Law to Use

This problem involves calculating the shift in peak wavelength due to a change in temperature. We will use Wien's Displacement Law, which states that \( \lambda_m T = b \), where \( \lambda_m \) is the peak wavelength, \( T \) is the absolute temperature, and \( b \) is a constant (Wien's constant).
02

Set Up the Equation for Both Temperatures

First, set up the equation based on Wien’s Law for the initial and final temperatures. For the initial temperature: \( \lambda_m \cdot 2000 = b \). For the new temperature: \( \lambda_{m'} \cdot 3000 = b \). These two equations will allow us to find the ratio of the wavelengths at different temperatures.
03

Relate the Equations

Since \( b \) is the same constant in both equations, equate them: \( \lambda_m \cdot 2000 = \lambda_{m'} \cdot 3000 \). This will allow us to solve for \( \lambda_{m'} \) in terms of \( \lambda_m \).
04

Solve for New Wavelength

Rearrange the equation to find \( \lambda_{m'} \):\[ \lambda_{m'} = \frac{2000}{3000} \lambda_m \]Simplify this fraction to find \( \lambda_{m'} = \frac{2}{3} \lambda_m \).
05

Conclude with the Correct Option

Now compare the result \( \lambda_{m'} = \frac{2}{3} \lambda_m \) with the given options to determine the correct answer. The correct choice is (c) \( \frac{2}{3} \lambda_m \).

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Key Concepts

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

Blackbody Radiation
Blackbody radiation refers to the phenomenon where an object absorbs all the electromagnetic radiation that falls upon it, without reflecting any. These objects, known as black bodies, emit radiation in a continuous spectrum due to their temperature. This radiation is called blackbody radiation, and its study is fundamental in understanding how objects behave thermally when exposed to electromagnetic energy.
One of the key characteristics of blackbody radiation is that the distribution of emitted radiation frequency depends only on the temperature of the black body. The radiation emitted covers a range of wavelengths, with a peak wavelength where the emission is most intense. As the temperature of the black body increases, the peak wavelength shifts to shorter wavelengths. This behavior is crucial for studying how stars emit light and energy, among other applications in thermal physics. Wien's Displacement Law helps in quantifying this shift in wavelength as the temperature changes.
Thermal Physics
Thermal physics is the study of heat, temperature, and their effects on matter. It includes several core concepts, such as temperature dependence, specific heat, and the laws of thermodynamics. Wien's Displacement Law, which is a part of thermal physics, allows us to understand how the peak wavelength of emitted radiation from a black body shifts as the temperature changes.
In thermal physics, temperature is measured in absolute terms using Kelvin. The behavior of substances at different temperatures can help us understand not just theoretical physics, but also practical science. The application of these concepts can be seen in various fields such as meteorology, material science, and chemistry.
  • Thermal equilibrium is a key aspect where no net flow of thermal energy occurs between objects.
  • Specific heat is another important concept, describing the amount of heat required to change the temperature of a substance by a certain amount.
Temperature Dependence of Wavelength
The concept of temperature dependence of wavelength is central to Wien's Displacement Law. It states that the wavelength at which a black body's emission is strongest is inversely proportional to its temperature. As the temperature increases, the peak wavelength decreases. This dependency provides insight into how objects emit energy differently at different temperatures.
Wien's Displacement Law can be mathematically expressed as \( \lambda_m T = b \), where \( \lambda_m \) is the peak wavelength, \( T \) is the absolute temperature, and \( b \) is Wien's constant. This relationship signifies that the product of the peak wavelength and absolute temperature is a constant for all black bodies. Hence, when the temperature of a black body increases, the peak wavelength must decrease to maintain the constant value.
This principle is widely used in astrophysics and cosmology, particularly in the study of stars and other celestial objects, where temperature variations determine the colors and types of light emitted. This understanding helps scientists estimate the temperatures of distant stars by analyzing the spectrum of light they emit.

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Most popular questions from this chapter

Three rods cach of same length and cross section are joined in scrics. The thermal conductivity of the materials are \(k, 2 k\) and \(3 k\) respectively. If one end is kepl al \(200^{\circ} \mathrm{C}\) and the other at \(100^{\circ} \mathrm{C}\). What would be the temperature of the junctions in the steady state? \Lambdassume that no heat is lost duc to radiation from the sides of the rods.

A body cools in a surrounding of constant temperature \(30^{\circ} \mathrm{C}\). It's heat capacity is \(2 \mathrm{~J} / \mathrm{C}\). Initial temperature of the body is \(40^{\circ} \mathrm{C}\). Assume Newton's law of cooling is valid. The body cools to \(36^{\circ} \mathrm{C}\) in 10 minutes. In further 10 minutes it will cool from \(36 \mathrm{C}\) to (a) \(34.8 \mathrm{C}\) (b) \(32.1 \%\) (c) \(32.8^{\circ} \mathrm{C}\) (d) \(33.6^{\circ} \mathrm{C}\)

In an experiment on the specilic heat of a metal a \(0.20 \mathrm{~kg}\) block of the metal at \(150^{\circ} \mathrm{C}\) is dropped in a copper calorimeter (of water equivalent \(0.025 \mathrm{~kg}\) ) containing \(150 \mathrm{~cm}^{3}\) of water at \(27^{\circ} \mathrm{C}\). The final temperature is \(40^{\circ} \mathrm{C}\). Calculalc the spocific heat ol the metal.

The temperature of the two outer surface of a composite slab consisting of two materials having cocflicient of thermal conductivities \(K\) and \(2 K\), thickness \(x\) and \(4 x\) respectively are \(T_{3}\) and \(T_{1}\left(T_{2}>T_{1}\right) .\) The rate of heat transfer through the slab, in a steady state is \(F\left[\frac{A\left(T_{2}-T_{1}\right) K}{x}\right]\), the value of \(F\) is (a) 1 (b) \(1 / 2\) (c) \(2 / 3\) (d) \(1 / 3\)

By which of the following methods could a cup of hot tea lose heat, if placed on metallic table in a classroom? (a) Conduction (b) Convection (c) Radiation (d) Evaporaton of liquid

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