Question

The blackbody emission spectrum of object A peaks in the ultraviolet region of the electromagnetic spectrum at a wavelength of 200 nm. That of object B peaks in the red region, at 650 nm. Which object is hotter, and, according to Wien’s law, how many times hotter is it? According to Stefan’s law, how many times more energy per unit area does the hotter body radiate per second?

Answer 1

Answer:

A) Object A is 3.25 times hotter.

B) Object A radiates 111.6 times more energy per unit of area.

Explanation:

Wiens's law states that there is an inverse relationship between the wavelength in which there is a peak in the emission of a black body and its temperature, mathematically,

$\lambda_{peak}= \dfrac{0.0028976}{T}$,

where $T$ is the temperature in kelvins and, $\lambda_{peak}$ is the wavelenght (in meters) where the emission is in its peak.

From here, if we solve Wien's law for the temperature we get

$T=\dfrac{0.0028976}{\lambda_{peak}}$.

Now, we can easily compute the temperatures.

For object A:

$T_{A}=\dfrac{0.0028976}{200*10^{-9}}$

$T_{A}=14488K$.

For object B:

$T_{B}=\dfrac{0.0028976}{650*10^{-9}}$

$T_{B}=4458K$

From this, we get that

$T_{A}/T_{B}=3.25$,

which means that object A is 3.25 times hotter.

Stefan's Law states that a black body emits thermal radiation with power proportional to the fourth power of its temperature.

This is

$E=\sigma T^{4}$,

where  $\sigma=5.67*10^{-8}\ Wm^{-2}K^{-4}$ is call the Stefan-Boltzmann constant.

From this, power can be easily compute:

$E_{A}=(5.67*10^{-8}*(14488)^{4})=2.5*10^{9}W\\E_{B}=(5.67*10^{-8}*(4458)^{4})=22.4*10^{6}}W$,

and we can notice that

$E_{A}/E_{B}=111.6$,

which means that object A radiates 111.6 time more energy per unit of area.