Question

The $SI$ unit and dimensions of Stefan’s constant $\sigma$ in case of Stefan’s law of radiation is

• A. $\frac{J}{m^{3}\,s^{4}}, \left[M^{1}L^{0}T^{-3}K^{-4}\right]$
• B. $\frac{J}{m^{2}\,s^{4}\,K}, \left[M^{1}L^{0}T^{-3}K^{3}\right]$
• C. $\frac{J}{m^{3}s\,K^{4}}, \left[M^{1}L^{0}T^{-3}K^{4}\right]$
• D. $\frac{J}{m^{2}s\,K^{4}}, \left[M^{1}L^{0}T^{-3}K^{-4}\right]$

Answer 1

Answer: (D)

According to Stefan's law, energy emitted by a body per unit area per second is proportional to fourth power of the absolute temperature.
$E=\sigma T^{4}$
where, $E=$ energy emitted/area/second, $T=$ absolute temperature in kelvin
and $\sigma=$ Stefan's constant. $\Rightarrow \sigma=\frac{E}{T^{4}}$
Unit of Stefan's constant
$\Rightarrow \frac{ J / m ^{2} s }{ K ^{4}}= J / m ^{2} s K ^{4}$
Dimensions of Stefan's constant,
$\sigma=\frac{[\text { Energy }]}{\text { [Area] } \times[\text { Time }] \times[\text { Temperature }]^{4}}$
$=\frac{\left[ ML ^{2} T ^{-2}\right]}{\left[ L ^{2} T \times K ^{4}\right]}=\left[ M ^{1} L ^{0} T ^{-3} K ^{-4}\right]$