Question

Consider that \(\sigma_s\), \(k_B\), and \(b\) represent Stefan-Boltzmann constant, Boltzmann constant, and Wien's displacement law constant, respectively. The dimension of \(\sigma_s k_B^{-1} b\) is:

• A. \([L^{-1}T^{-1}K^{-4}]\)
• B. \([L^{-1}T^{-1}K^{-2}]\)
• C. \([L^{-1}K^{-2}]\)
• D. \([L^{-1}T^{-1}K^{-3}]\)

Hint:

In dimensional analysis, first write dimensions of each constant separately and then combine powers systematically.

Answer 1

The Correct Option is B

Step 1: Write the dimensions of each constant.
Stefan-Boltzmann constant $[\sigma_s] = [M T^{-3} K^{-4}]$, Boltzmann constant $[k_B] = [M L^2 T^{-2} K^{-1}]$, and Wien's constant $[b] = [L K]$.
Step 2: Set up the combination.
We need $[\sigma_s\, k_B^{-1}\, b] = [\sigma_s]\,[k_B]^{-1}\,[b]$.
Step 3: Invert the Boltzmann dimension.
$[k_B]^{-1} = [M^{-1} L^{-2} T^{2} K]$.
Step 4: Multiply everything.
$[M T^{-3} K^{-4}]\,[M^{-1} L^{-2} T^{2} K]\,[L K]$.
Step 5: Collect each base quantity.
Mass: $M^{1-1} = M^0$. Length: $L^{-2+1} = L^{-1}$. Time: $T^{-3+2} = T^{-1}$. Temperature: $K^{-4+1+1} = K^{-2}$.
Step 6: Write the result.
So the dimension is $[L^{-1} T^{-1} K^{-2}]$, which is option B.
\[ \boxed{ [L^{-1} T^{-1} K^{-2}] } \]