Q. The wavelength of maximum intensity of radiation emitted by a star is $289.8\ nm$. The radiation intensity for the star is : (Stefan’s constant $5.67 \times 10^{- 8}Wm^{- 2}K^{- 4}$, constant $b = 2898\mu mK)$
A) $5.67 \times 10^{8}W/m^{2}$
B) $5.67 \times 10^{12}W/m^{2}$
C) $10.67 \times 10^{7}W/m^{2}$
D) $10.67 \times 10^{14}W/m^{2}$
Explanation
(a) We know $\lambda_{\max}T = b$
$\Rightarrow T = \frac{b}{\lambda_{\max}} = \frac{2898 \times 10^{- 6}}{289.8 \times 10^{- 9}} = 10^{4}K$
According to Stefan’s law
$E = \sigma T^{4} = \left( 5.67 \times 10^{- 8} \right)\left( 10^{4} \right)^{4} = 5.67 \times 10^{8}W/m^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.