Question

Calculate the intensity of thermal radiation from a blackbody by using Raleigh-Jeans Law at a temperature of 1000 K and with a wavelength of 1000 nm (infrared), 500 nm (visible), and 100 nm (ultraviolet). (3 points) (b) Recalculate the above intensities using Plank's Law

Answer 1

Explanation:

Given that,

Temperature = 1000 K

We need to calculate the  intensity of thermal radiation from a black body

For wavelength 1000 nm

Using Raleigh-Jeans Law

$B_{k}=\dfrac{8\pi kT}{\lambda^4}$

Put the value into the formula

$B_{k}=\dfrac{8\times\pi\times1.38\times10^{-23}\times1000}{(1000\times10^{-9})^{4}}$

$B_{k}=346831.828956$

$B_{k}=34.68\times10^{4}\ W$

For wavelength 500 nm

$B_{k}=\dfrac{8\times\pi\times1.38\times10^{-23}\times1000}{(500\times10^{-9})^{4}}$

$B_{k}=5549309$

$B_{k}=55.49\times10^{5}\ W$

For wavelength 100 nm

$B_{k}=\dfrac{8\times\pi\times1.38\times10^{-23}\times1000}{(100\times10^{-9})^{4}}$

$B_{k}=3.47\times10^{9}\ W$

We need to calculate the intensity

Using Plank's Law

$E=\dfrac{2hc^2}{\lambda^5}\times\dfrac{1}{e^{\dfrac{hc}{\lambda k T}}-1}$

For wavelength 1000 nm

Put the value into the formula

$E=\dfrac{2\times6.63\times10^{-34}\times(3\times10^{8})^2}{(1000\times10^{-9})^5}\times\dfrac{1}{e^{\dfrac{6.63\times10^{-34}\times3\times10^{8}}{1000\times10^{-9}\times 1.38\times10^{-23}\times1000}}-1}$

$E=65.65\times10^{6}\ W/m^2.K^4$

For wavelength 500 nm

$E=\dfrac{2\times6.63\times10^{-34}\times(3\times10^{8})^2}{(500\times10^{-9})^5}\times\dfrac{1}{e^{\dfrac{6.63\times10^{-34}\times3\times10^{8}}{500\times10^{-9}\times 1.38\times10^{-23}\times1000}}-1}$

$E=11.56\times10^{2}\ W/m^2.K^4$

For wavelength 100 nm

$E=\dfrac{2\times6.63\times10^{-34}\times(3\times10^{8})^2}{(100\times10^{-9})^5}\times\dfrac{1}{e^{\dfrac{6.63\times10^{-34}\times3\times10^{8}}{100\times10^{-9}\times 1.38\times10^{-23}\times1000}}-1}$

$E=3.032\times10^{-44}\ W/m^2.K^4$

Hence, This is the required solution.