Chapter 27, Problem 5P

To determine

(a) To determine:

Peak wavelength for radiation emitted from $0°\text{C}$ ice.

Answer to Problem 5P

Solution:

Peak wavelength for radiation emitted from $0°\text{C}$ ice is $10.6\mu \text{m}$ which is far away from infrared radiation.:

Explanation of Solution

Given:

Ice at $T=0°\text{C}=\text{273}\text{K}$

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:

Peak wavelength for radiation emitted from $0°\text{C}$ ice is calculated as

$\lambda \times T=0.0029$

Plugging the values in the above equation

$\begin{array}{c}\lambda \times 273=0.0029\\ \lambda =1.06\times {10}^{-5}\text{m}\\ =\text{10}\text{.6}\mu \text{m}\end{array}$

The wavelength of the radiation is far away from the infrared radiation.

To determine

(b) To determine:

Peak wavelength for radiation emitted from a flood lamp at $3100\text{K}$.

Answer to Problem 5P

Solution:

Peak wavelength for radiation emitted from a flood lamp at $3100\text{K}$ is $935\text{nm}$ which is near infrared radiation.:

Explanation of Solution

Given:

Flood lamp at $T=\text{3100}\text{K}$

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:

Peak wavelength for radiation emitted from a flood lamp at $3100\text{K}$ is calculated as

$\lambda \times T=0.0029$

Plugging the values in the above equation

$\begin{array}{c}\lambda \times 3100=0.0029\\ \lambda =9.35\times {10}^{-7}\text{m}\\ =935\text{nm}\end{array}$

The wavelength of the radiation is near infrared radiation.

To determine

(c) To determine:

Peak wavelength for radiation emitted from helium at $4\text{K}$.

Answer to Problem 5P

Solution:

Peak wavelength for radiation emitted from helium at $4\text{K}$ is $0.7\text{mm}$ which comes in microwaveradiation.:

Explanation of Solution

Given:

Temperature of helium $T=4\text{K}$

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:

Peak wavelength for radiation emitted from helium at $4\text{K}$ is calculated as

$\lambda \times T=0.0029$

Plugging the values in the above equation

$\begin{array}{c}\lambda \times 4=0.0029\\ \lambda =0.7\times {10}^{-3}\text{m}\\ =0.7\text{mm}\end{array}$

The wavelength of the radiation comes under microwave radiation.