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Chapter 6: Problem 59

Interstellar space is filled with radiation of wavelength 970\mum. This radiation is considered to be a remnant of the "big bang." What is the corresponding blackbody temperature of this radiation?

Short Answer

Expert verified
The corresponding blackbody temperature of the radiation with a wavelength of 970 micrometers is approximately \(2.988 \times 10^3 \: \text{K}\).

Step by step solution

01

Understand Wien's Displacement Law

Wien's Displacement Law relates the wavelength of the peak of the blackbody radiation emitted by an object to its temperature. The formula for this law is given by: \[ \lambda_{\text{max}} = \frac{b}{T} \] where \(\lambda_{\text{max}}\) is the wavelength of the peak radiation, \(b\) is Wien's constant (approximately \(2.898 \times 10^{-3} \: \text{m} \: \text{K}\)), and \(T\) is the blackbody temperature in Kelvin.
02

Convert the wavelength to meters

The given wavelength is 970 μm (micrometers). To use this value in the formula, we must convert it to meters. 1 μm is equal to \(10^{-6}\) m. So, the conversion is: \[970 \: \text{μm} \times \frac{10^{-6} \: \text{m}}{1 \: \text{μm}} = 970 \times 10^{-6} \: \text{m} = 9.70 \times 10^{-4} \text{m}\]
03

Calculate the blackbody temperature

Now, we will insert the given wavelength and Wien's constant into the formula, and solve for the temperature: \[ \frac{b}{T} = \lambda_{\text{max}} \] \[ T = \frac{b}{\lambda_{\text{max}}} \] Plug in the values of \(b\) and \(\lambda_{\text{max}}\) into the equation: \[ T = \frac{2.898 \times 10^{-3} \: \text{m} \: \text{K}}{9.70 \times 10^{-4} \: \text{m}} \]
04

Compute the final answer

Now, we will calculate the value of the temperature: \[ T = \frac{2.898 \times 10^{-3} \: \text{m} \: \text{K}}{9.70 \times 10^{-4} \: \text{m}} = 2.988 \times 10^{3} \: \text{K} \]
05

Report the result

The corresponding blackbody temperature of this radiation is approximately \(2.988 \times 10^3 \: \text{K}\).

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