Problem 137 - Master Resource Book in JEE Main Physics

Question

Two bodies \(A\) and \(B\) have thermal emissivities of \(0.01\) and \(0.81\) respectively. The outer surface areas of the two bodies are the same. The two bodies emit total radiant power at the same rate. The wavelength \(\lambda_{B}\) corresponding to maximum spectral radiancy in the radiation from \(B\) is shifted from the wavelength corresponding to maximum spectral radiancy in the radiation from \(A\), by \(1.00 \mu \mathrm{m}\). If the temperature of \(A\) is \(5802 \mathrm{~K}\) (a) The temperature of \(B\) is \(1934 \mathrm{~K}\) (b) \(\lambda_{B}=1.5 \mu \mathrm{m}\) (c) The temperature of \(B\) is \(11604 \mathrm{~K}\) (d) The temperature of \(B\) is \(2901 \mathrm{~K}\)

Step-by-Step Solution

Answer

The temperature of B is 1934 K, and the wavelength of B is 1.5 µm.

Step 1: Understand the Problem Context

Given two bodies \(A\) and \(B\) with the same outer surface area, but different emissivities \(e_A = 0.01\) and \(e_B = 0.81\). They emit radiant power at the same rate. The temperature of \(A\) is \(5802\, \text{K}\), and the wavelength for its maximum spectral radiancy is shifted by \(1.00 \, \mu \text{m}\) from that of \(B\). The task is to find the correct conditions for body \(B\), considering its temperature and the wavelength of maximum spectral radiancy.

Step 2: Use the Stefan-Boltzmann Law

The Stefan-Boltzmann law states that the total power emitted by a black body is given by:\[ P = e \sigma A T^4 \]where \(P\) is the power, \(e\) the emissivity, \(\sigma\) the Stefan-Boltzmann constant, \(A\) the area, and \(T\) the temperature.For \(A\) and \(B\), since \(P_A = P_B\) and \(A_A = A_B\), we have:\[ e_A T_A^4 = e_B T_B^4 \]Substitute the known values:\[ 0.01 \times 5802^4 = 0.81 \times T_B^4 \]

Step 3: Solve for Temperature of B

Continuing from the equation:\[ 0.01 \times 5802^4 = 0.81 \times T_B^4 \]Solve for \(T_B\):\[ T_B^4 = \frac{0.01}{0.81} \times 5802^4 \]\[ T_B = \left( \frac{0.01}{0.81} \right)^{1/4} \times 5802 \]\[ T_B \approx 1934 \, \text{K} \]

Step 4: Apply Wien's Displacement Law

Wien's Displacement Law relates the temperature \(T\) of a black body to the wavelength of maximum emission \(\lambda_{max}\):\[ \lambda_{max} T = b \]where \(b\) is Wien's constant, approximately \(2.898 \times 10^{-3} \, \text{m \cdot K}\).For \(A\), \(\lambda_A T_A = b\).For \(B\), \(\lambda_B T_B = b\).We know \(\lambda_A - \lambda_B = 1.00 \, \mu\text{m}\).Calculate \(\lambda_A\) using \(\lambda_A = \frac{b}{T_A}\) and substitute to find \(\lambda_B\).

Step 5: Solve for Wavelength of B

Using \(\lambda_A = \frac{b}{T_A}\), we find:\[ \lambda_A = \frac{2.898 \times 10^{-3}}{5802} \approx 5.00 \times 10^{-7} \, \text{m} = 0.500 \, \mu\text{m}\]Given \(\lambda_A - \lambda_B = 1.00 \, \mu\text{m}\), solve for \(\lambda_B\):\[ \lambda_B = \lambda_A + 1.00 \, \mu\text{m} = 0.500 \, \mu\text{m} + 1.00 \, \mu\text{m} = 1.50 \, \mu\text{m} \]

Step 6: Conclusion Based on Solution Steps

According to the calculations, the solutions match the options given:1. The temperature of \(B\) is \(1934 \, \text{K}\) (matching option (a)).2. The wavelength \(\lambda_B\) for \(B\) is \(1.5 \, \mu\text{m}\) (matching option (b)).Thus, the temperature of \(B\) and the shifted wavelength match options (a) and (b).

Key Concepts

Stefan-Boltzmann law

The Stefan-Boltzmann law is a principle in physics that explains how the total energy radiated per unit surface area of a black body is proportional to the fourth power of its temperature. This can be mathematically expressed as \[ P = e \sigma A T^4 \], where:

• \( P \) is the power radiated per unit area.
• \( e \) denotes the emissivity of the material, a measure of how efficiently a surface radiates energy.
• \( \sigma \) is the Stefan-Boltzmann constant, a fundamental physical constant.
• \( A \) is the surface area.
• \( T \) is the absolute temperature of the body in Kelvin.

This law was used in the problem to compare two bodies, \( A \) and \( B \), which have the same surface area but different emissivities and temperatures. By equating the power \( P \) for both bodies, given their identical power output, we deduce a relationship between their emissivities and temperatures: \( e_A T_A^4 = e_B T_B^4 \). Here, the known values for emissivities and temperature of body \( A \) allow us to solve for the unknown temperature of body \( B \).

Wien's displacement law

Wien's displacement law relates the temperature of a black body to the wavelength at which it emits radiation most strongly. The law is expressed by the formula \[ \lambda_{max} T = b \], where:

• \( \lambda_{max} \) is the peak emission wavelength.
• \( T \) is the absolute temperature of the black body.
• \( b \), Wien's displacement constant, equals approximately \( 2.898 \times 10^{-3} \, \text{m} \cdot \text{K} \).

In the given problem, this law helps us find the peak wavelength for body \( B \) given that the peak wavelength for \( A \) is shifted by \( 1.00 \, \mu \text{m} \). Once we know the temperature of body \( A \) from the previous calculations, we can find its peak wavelength. We used this wavelength difference to solve for the peak wavelength for radiation from body \( B \), completing the relationship provided by Wien's law.

temperature calculation

Temperature calculation in this context involves using known temperature values and mathematical relationships to determine an unknown temperature. Given the emissivity values and surface area being constant, we first calculate the temperature of body \( B \) using the Stefan-Boltzmann law:

• We start with the equation \( e_A T_A^4 = e_B T_B^4 \).
• Insert the known values of emissivity for both bodies as well as the temperature of body \( A \) (\( 5802 \, \text{K} \)).
• Solving for \( T_B \), the temperature of body \( B \), requires mathematically rearranging the equation to \[ T_B = \left( \frac{e_A}{e_B} \right)^{1/4} \times T_A \].

Using this approach, we find that \( T_B \) is approximately \( 1934 \, \text{K} \). This temperature allows us to further utilize Wien’s displacement law to solve for related values, ensuring that all calculations align with the laws of thermodynamics. Such calculations highlight the interplay between emissivity, temperature, and wavelength in thermal physics.