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Chapter 14: Problem 66

If the maximum intensity of radiation for a blackbody is found at $2.65 \mu \mathrm{m},$ what is the temperature of the radiating body?

Short Answer

Expert verified
Answer: The temperature of the radiating body is approximately 1094.34 K.

Step by step solution

01

Convert wavelength to meters

We are given \(\lambda_{max}\) in micrometers. Let's convert it to meters: $$\lambda_{max} = 2.65 \, \mu \mathrm{m} = 2.65 \times 10^{-6} \, \mathrm{m}$$
02

Rearrange Wien's Displacement Law formula

Let's rearrange Wien's Displacement Law formula to find the temperature \(T\): $$ T = \frac{b}{\lambda_{max}}$$
03

Plug in the values and solve for T

Now we will plug in the values for \(\lambda_{max}\) and \(b\) and solve for the temperature: $$ T = \frac{2.898 \times 10^{-3} \, \text{m K}}{2.65 \times 10^{-6} \, \mathrm{m}}$$ Divide \(2.898 \times 10^{-3}\) by \(2.65 \times 10^{-6}\) to find the temperature: $$ T \approx 1094.34 \, \text{K}$$ So, the temperature of the radiating body is approximately \(1094.34 \, \text{K}\).

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