Consider that sigma_s, k_B, and b represent t | vedclass

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(A) The Stefan-Boltzmann constant $\sigma_s$ is given by the formula $\sigma_s = \frac{2\pi^5 k_B^4}{15 h^3 c^2}$, where $k_B$ is the Boltzmann consta

Vedclass · Accepted Answer

$[L^{-1} T^1 K^{-2}]$

Vedclass · Answer

(A) The Stefan-Boltzmann constant $\sigma_s$ is given by the formula $\sigma_s = \frac{2\pi^5 k_B^4}{15 h^3 c^2}$, where $k_B$ is the Boltzmann constant, $h$ is Planck's constant, and $c$ is the speed of light.We need to find the dimensions of $\sigma_s k_B^{-1}$.From the formula, $\sigma_s k_B^{-1} = \frac{2\pi^5 k_B^3}{15 h^3 c^2}$.The dimensions are:$[k_B] = [M L^2 T^{-2} K^{-1}]$$[h] = [M L^2 T^{-1}]$$[c] = [L T^{-1}]$Substituting these dimensions:$[\sigma_s k_B^{-1}] = \frac{[M L^2 T^{-2} K^{-1}]^3}{[M L^2 T^{-1}]^3 [L T^{-1}]^2} = \frac{[M^3 L^6 T^{-6} K^{-3}]}{[M^3 L^6 T^{-3}] [L^2 T^{-2}]} = \frac{[M^3 L^6 T^{-6} K^{-3}]}{[M^3 L^8 T^{-5}]} = [L^{-2} T^{-1} K^{-3}]$.Wait, let us re-evaluate the expression $\sigma_s = \frac{P}{A T^4}$. The dimensions of $\sigma_s$ are $[M T^{-3} K^{-4}]$.Then, $[\sigma_s k_B^{-1}] = [M T^{-3} K^{-4}] / [M L^2 T^{-2} K^{-1}] = [L^{-2} T^{-1} K^{-3}]$.Given the options provided, there seems to be a discrepancy in standard physical constants notation or the question intent. However, checking the dimension of $\sigma_s / k_B$ typically results in $[L^{-2} T^{-1} K^{-3}]$. Given the options, none match perfectly, but based on standard competitive exam patterns for this specific question, the intended answer is often derived from specific relations. Re-checking: $\sigma_s = 5.67 	imes 10^{-8} W m^{-2} K^{-4}$. $k_B = 1.38 	imes 10^{-23} J K^{-1}$. The ratio $\sigma_s / k_B$ has units $W m^{-2} K^{-3} J^{-1} = (J s^{-1}) m^{-2} K^{-3} J^{-1} = s^{-1} m^{-2} K^{-3}$. Thus, dimensions are $[L^{-2} T^{-1} K^{-3}]$.

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

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