Question

What is Stefan-Boltzmann constant? Write its dimensional formula.

Answer 1

Hint: We need to understand the application of the Stefan-Boltzmann constant to recognise it and define the constant according to its other varying physical parameters. We can use the Stefan-Boltzmann law to start with the definition.

Complete Solution Step-by-Step:
We know that the Stefan-Boltzmann law is an important law in thermodynamics. It deals with the rate of energy loss or the power radiated by a body in a given surrounding. It relates the surrounding temperature to the heat loss rate which is one of the most practical requirements in every field.
The Stefan-Boltzmann law states that the intensity of the wavelength of the energy radiated from a body is increased with the increase in the difference between the temperature of the body and the surroundings.
According to the Stefan-Boltzmann law the rate of heat energy loss or the power radiated by any body is proportional to the emissivity of the material of the body, the surface area of the body which can emit radiation and the fourth power of the temperature difference between the body and the surroundings. It is mathematically given as –
\[P\propto eA{{T}^{4}}\]
The proportionality is made into an equation by a constant known by the Stefan-Boltzmann constant. It has a value given as –
\[\sigma =5.67\times {{10}^{-8}}W{{m}^{-2}}{{K}^{4}}\]
From the Stefan-Boltzmann law, we can deduce the constant as –
\[\begin{align}
  & P=\sigma eA{{T}^{4}} \\
 & \therefore \sigma =\dfrac{P}{eA{{T}^{4}}} \\
\end{align}\]
The ratio of the R.H.S. will be always a constant for any material. The actual value of the Stefan-Boltzmann constant is given by the equation –
\[\begin{align}
  & \sigma =\dfrac{2{{\pi }^{5}}{{k}_{B}}^{4}}{15{{h}^{3}}{{c}^{2}}} \\
 & \therefore \sigma =5.67\times {{10}^{-8}}W{{m}^{-2}}{{K}^{4}} \\
\end{align}\]
Where, \[{{k}_{B}}\] is the Boltzmann constant,
h is the Planck's constant,
c is the speed of light.
The dimensional formula for each physical quantity can be given as –
\[\begin{align}
  & [{{k}_{B}}]=J{{K}^{-1}} \\
 & \Rightarrow [{{k}_{B}}]=[M{{L}^{2}}{{T}^{-2}}{{K}^{1}}] \\
 & \therefore {{[{{k}_{b}}]}^{4}}=[{{M}^{4}}{{L}^{8}}{{T}^{-8}}{{K}^{4}}] \\
 & [h]=Js \\
 & \Rightarrow [h]=[M{{L}^{2}}{{T}^{-1}}] \\
 & \therefore {{[h]}^{3}}=[{{M}^{3}}{{L}^{6}}{{T}^{-3}}] \\
 & [c]=m{{s}^{-2}} \\
 & \Rightarrow [c]=[L{{T}^{-2}}] \\
 & \therefore {{[c]}^{2}}=[{{L}^{2}}{{T}^{-4}}] \\
\end{align}\]
The dimensional formula for the constant is –
\[\begin{align}
  & [\sigma ]=\dfrac{[{{M}^{4}}{{L}^{8}}{{T}^{-8}}{{K}^{4}}]}{[{{M}^{3}}{{L}^{6}}{{T}^{-3}}][{{L}^{2}}{{T}^{-4}}]} \\
 & \therefore [\sigma ]=[M{{T}^{-3}}{{K}^{-4}}] \\
\end{align}\]
This is the required solution.

Note:
The rate of heat energy loss or the power radiated from a body is highly dependent on the emissivity of a material. Emissivity is a dimensionless quantity which is measured from a scale of 0 to 1, where 1 represents a perfect blackbody.