The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 μm. If the peak wavelength of emitted radiation changes to 2.90 μm, then the temperature (in K) of the black body is
The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 μm. If the peak wavelength of emitted radiation changes to 2.90 μm, then the temperature (in K) of the black body is
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![**Title: Understanding Black Body Radiation and Temperature Relation**
**Content:**
The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 µm. If the peak wavelength of emitted radiation changes to 2.90 µm, then the temperature (in K) of the black body is ________.
**Explanation:**
For those studying black body radiation, it's crucial to understand the relationship between the temperature of a black body and the peak wavelength of its emitted radiation. According to **Wien's Displacement Law**, the product of the temperature (T) of a black body and the wavelength (λ) at which it emits the most radiation is a constant, given by:
\[ \lambda_{\text{peak}} T = b \]
where \( b \) is Wien's displacement constant approximately equal to \( 2.898 \times 10^{-3} \, \text{m} \cdot \text{K} \).
Given:
- Initial temperature, \( T_1 = 2000 \, \text{K} \)
- Initial peak wavelength, \( \lambda_1 = 1.45 \, \mu m \)
- New peak wavelength, \( \lambda_2 = 2.90 \, \mu m \)
We can find the new temperature \( T_2 \) using Wien’s law as follows:
\[ T_1 \lambda_1 = T_2 \lambda_2 \]
By substituting the known values:
\[ 2000 \, \text{K} \times 1.45 \, \mu \text{m} = T_2 \times 2.90 \, \mu \text{m} \]
Solving for \( T_2 \):
\[ T_2 = \frac{2000 \, \text{K} \times 1.45 \, \mu \text{m}}{2.90 \, \mu \text{m}} \]
\[ T_2 = 1000 \, \text{K} \]
Therefore, when the peak wavelength of emitted radiation changes to 2.90 µm, the temperature of the black body is 1000 K.
**Note:** Wien’s law illustrates a fundamental aspect of thermal radiation, showing that as the temperature of a black body increases, its peak wavelength shifts to shorter wavelengths, moving from the infrared](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbdf4ad1c-8d6e-4c6e-ab0d-20fae6a9f061%2Fdcc072bb-cf50-4197-8934-58ce7fdf0e8e%2Fm9dsmp_processed.png&w=3840&q=75)
Transcribed Image Text:**Title: Understanding Black Body Radiation and Temperature Relation**
**Content:**
The peak wavelength of radiation emitted by a black body at a temperature of 2000 K is 1.45 µm. If the peak wavelength of emitted radiation changes to 2.90 µm, then the temperature (in K) of the black body is ________.
**Explanation:**
For those studying black body radiation, it's crucial to understand the relationship between the temperature of a black body and the peak wavelength of its emitted radiation. According to **Wien's Displacement Law**, the product of the temperature (T) of a black body and the wavelength (λ) at which it emits the most radiation is a constant, given by:
\[ \lambda_{\text{peak}} T = b \]
where \( b \) is Wien's displacement constant approximately equal to \( 2.898 \times 10^{-3} \, \text{m} \cdot \text{K} \).
Given:
- Initial temperature, \( T_1 = 2000 \, \text{K} \)
- Initial peak wavelength, \( \lambda_1 = 1.45 \, \mu m \)
- New peak wavelength, \( \lambda_2 = 2.90 \, \mu m \)
We can find the new temperature \( T_2 \) using Wien’s law as follows:
\[ T_1 \lambda_1 = T_2 \lambda_2 \]
By substituting the known values:
\[ 2000 \, \text{K} \times 1.45 \, \mu \text{m} = T_2 \times 2.90 \, \mu \text{m} \]
Solving for \( T_2 \):
\[ T_2 = \frac{2000 \, \text{K} \times 1.45 \, \mu \text{m}}{2.90 \, \mu \text{m}} \]
\[ T_2 = 1000 \, \text{K} \]
Therefore, when the peak wavelength of emitted radiation changes to 2.90 µm, the temperature of the black body is 1000 K.
**Note:** Wien’s law illustrates a fundamental aspect of thermal radiation, showing that as the temperature of a black body increases, its peak wavelength shifts to shorter wavelengths, moving from the infrared
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