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Chapter 6: Problem 57

(a) For what temperature is the peak of blackbody radiation spectrum at \(400 \mathrm{nm} ?\) (b) If the temperature of a blackbody is \(800 \mathrm{K},\) at what wavelength does it radiate the most energy?

Short Answer

Expert verified
(a) The temperature at which the peak of the blackbody radiation spectrum is at \(400 nm\) is approximately \(7245 K\). (b) If the temperature of a blackbody is \(800 K\), it radiates the most energy at a wavelength of approximately \(3,623 nm\).

Step by step solution

01

Convert Wavelength to Meters

To be consistent with the units of Wien's constant, which are in meters and Kelvins, we need to convert the given peak wavelength from nanometers to meters. \(400 nm = 400 \times 10^{-9} m\) Now we have the peak wavelength in meters, which is \(4 \times 10^{-7}\) m.
02

Calculate the Temperature for Part (a)

Using Wien's Displacement Law, we can solve for temperature. \(λ_{max} T = b\) Given: \(λ_{max} = 4 \times 10^{-7} m\) and \(b = 2.898 \times 10^{-3} mK\). We need to find T. Rearrange the formula to find temperature: \[T = \frac{b}{λ_{max}}\] Substituting the values: \(T = \frac{2.898 \times 10^{-3} mK}{4 \times 10^{-7} m} = 7,245 K\) So, the temperature at which the peak of the blackbody radiation spectrum is at \(400 nm\) is approximately \(7245 K\).
03

Calculate the Peak Wavelength for Part (b)

Using Wien's Displacement Law, we can solve for the peak wavelength. \(λ_{max} T = b\) Given: \(T = 800 K\) and \(b = 2.898 \times 10^{-3} mK\). We need to find \(λ_{max}\). Rearrange the formula to find the peak wavelength: \[λ_{max} = \frac{b}{T}\] Substituting the values: \(λ_{max} = \frac{2.898 \times 10^{-3} mK}{800 K} = 3.623 \times 10^{-6}\) m Now convert the peak wavelength to nanometers: \(3.623 \times 10^{-6} m = 3,623 nm\) So, if the temperature of a blackbody is \(800 K\), it radiates the most energy at a wavelength of approximately \(3,623 nm\).

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