Physics - Unit & Dimension Question with Solution | TestHub

The dimensional formula for the Stefan-Boltzmann constant ${\sigma }_s$ is obtained from the relation $E={\sigma }_{s}At{T}^4$, where $E$ is energy, $A$ is area, $t$ is time, and $T$ is temperature:

$[{\sigma }_s]=\frac{[\text{M}{\text{L}}^2{\text{T}}^{-2}]}{[{\text{L}}^2][\text{T}][{\text{K}}^4]}=[\text{M}{\text{T}}^{-3}{\text{K}}^{-4}]$

The dimensional formula for the Boltzmann constant $k_B$ is obtained from $E=k_{B}T$:

$[k_B]=\frac{[\text{M}{\text{L}}^2{\text{T}}^{-2}]}{[\text{K}]}=[\text{M}{\text{L}}^2{\text{T}}^{-2}{\text{K}}^{-1}]$

The dimensional formula for Wien's displacement constant $b$ is obtained from ${\lambda }_{\text{max}}T=b$:

$[b]=[\text{L}][\text{K}]=[\text{L}\text{K}]$

Now, substituting these dimensions into the given expression ${\sigma }_s{k}_B^{-1}b$:

$[{\sigma }_s{k}_B^{-1}b]=\frac{[{\sigma }_s][b]}{[k_B]}$

$[{\sigma }_s{k}_B^{-1}b]=\frac{[\text{M}{\text{T}}^{-3}{\text{K}}^{-4}][\text{L}\text{K}]}{[\text{M}{\text{L}}^2{\text{T}}^{-2}{\text{K}}^{-1}]}$

$[{\sigma }_s{k}_B^{-1}b]=\frac{[\text{M}\text{L}{\text{T}}^{-3}{\text{K}}^{-3}]}{[\text{M}{\text{L}}^2{\text{T}}^{-2}{\text{K}}^{-1}]}$

$[{\sigma }_s{k}_B^{-1}b]=[{\text{L}}^{-1}{\text{T}}^{-1}{\text{K}}^{-2}]$

Answer: $[{\text{L}}^{-1}{\text{T}}^{-1}{\text{K}}^{-2}]$