Chapter 27, Problem 6P

To determine

(a) To determine:

Temperature of the black body.

Answer to Problem 6P

Solution:

Temperature of the black body is $16111.1\text{K}$.

Explanation of Solution

Given:

Wavelength of the radiation, $\lambda =18\text{nm}=\text{18}\times {\text{10}}^{-9}\text{m}$

Formula used:

The peak wavelength of emitted radiation is calculated by the Wien’s law.

$\lambda \times T=0.0029$

Here,

$\lambda$ is the wavelength:

$T$ is the temperature

Calculation:

The temperature of the blackbody is calculated as

$\lambda \times T=0.0029$

Plugging the value in the above equation

$\begin{array}{c}18\times {10}^{-9}\times T=0.0029\\ T=16111.1\text{K}\end{array}$

To determine

(b) To determine:

Peak wavelength of black body spectrum.

Answer to Problem 6P

Solution:

Peak wavelength of black body spectrum is $\text{1}\text{.318}\mu \text{m}$.

Explanation of Solution

Given:

Temperature of the blackbody, $T=2200\text{K}$

Formula used:

The peak wavelength of emitted radiation is calculated by the Wien’s law

$\lambda \times T=0.0029$

Here,

$\lambda$ is the wavelength:

$T$ is the temperature

Calculation:

The wavelength of the blackbody spectrum is calculated as

$\lambda \times T=0.0029$

Plugging the value in the above equation

$\begin{array}{c}\lambda \times 2200=0.0029\\ \lambda =1.318\times {10}^{-6}\text{m}\\ =\text{1}\text{.318}\mu \text{m}\end{array}$