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Energy of radiation emitted by a black b...

Energy of radiation emitted by a black body at temperature 3000 K is `u_(1)` for wavelength between 8000 `Å` and `9000Å`, `u_(2)` for wavelength between 9000 `Å` and `10000Å` and `u_(3)` forwavelength between `10000Åand 11000Å`.Which of the following is true? [Wien’s constant `b=2.88xx10^(-3)mK]`

A

`u_(1)gtu_(2)gtu_(3)`

B

`u_(3)gtu_(2)gtu_(1)`

C

`u_(2)gtu_(2)andu_(2)gtu_(3)`

D

`u_(2)gtu_(1)=u_(2)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to analyze the energy of radiation emitted by a black body at a given temperature using Wien's displacement law. Let's break down the steps: ### Step-by-Step Solution: 1. **Understand Wien's Displacement Law**: Wien's displacement law states that the wavelength (λm) at which the intensity of radiation is maximum is inversely proportional to the temperature (T) of the black body. The formula is given by: \[ \lambda_m \cdot T = b \] where \( b \) is Wien's constant. 2. **Calculate the Maximum Wavelength**: Given: - \( T = 3000 \, K \) - \( b = 2.88 \times 10^{-3} \, m \cdot K \) We can rearrange the formula to find \( \lambda_m \): \[ \lambda_m = \frac{b}{T} = \frac{2.88 \times 10^{-3}}{3000} \] Converting \( b \) to meters: \[ \lambda_m = \frac{2.88 \times 10^{-3}}{3000} = 0.00096 \, m = 9600 \, Å \] 3. **Identify the Range of Wavelengths**: The problem specifies three ranges of wavelengths: - \( U_1 \): between 8000 Å and 9000 Å - \( U_2 \): between 9000 Å and 10000 Å - \( U_3 \): between 10000 Å and 11000 Å Since \( \lambda_m = 9600 \, Å \), it falls within the range of \( U_2 \). 4. **Determine the Energy Relations**: According to the properties of black body radiation: - The energy emitted by a black body is maximum at the wavelength corresponding to \( \lambda_m \). - As the wavelength increases beyond \( \lambda_m \), the energy emitted decreases. Therefore, we can conclude: - \( U_2 \) (the range containing 9600 Å) will have the maximum energy. - \( U_1 \) will have less energy than \( U_2 \). - \( U_3 \) will have even less energy than \( U_2 \). 5. **Final Conclusion**: The relationship between the energies can be summarized as: \[ U_2 > U_1 \quad \text{and} \quad U_2 > U_3 \] ### Answer: The correct statement is: **\( U_2 > U_1 \) and \( U_2 > U_3 \)**.

To solve the problem, we need to analyze the energy of radiation emitted by a black body at a given temperature using Wien's displacement law. Let's break down the steps: ### Step-by-Step Solution: 1. **Understand Wien's Displacement Law**: Wien's displacement law states that the wavelength (λm) at which the intensity of radiation is maximum is inversely proportional to the temperature (T) of the black body. The formula is given by: \[ \lambda_m \cdot T = b ...
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