Question

The wavelength of maximum intensity of radiation emitted by a star is $289.8\, nm$. The radiation intensity of the star is
(Stefans constant $= 5.67 \times 10^{-8}\, W\, m^{-2}\, K^{-4}$, Wien's constant $b = 2898 \, \mu m \,K$)

• A. $5.67 \times10^{8}\,W\,m^{-2}$
• B. $5.67 \times10^{12}\,W\,m^{-2}$
• C. $10.67 \times 10^{7}\,W\,m^{-2}$
• D. $10.67 \times 10^{14}\,W\,m^{-2}$

Answer 1

Answer: (A)

According to Wien’s displacement law,
$\lambda_{max}\,T=b$
$ \therefore T=\frac{b}{\lambda_{max}}=\frac{2898\,\mu m\,K}{289.8\,nm}$
$=\frac{2898 \times10^{-6}\,m\,K}{289.8 \times10^{-9}\,m}$
According to Stefan Boltzmann law,
Radiation intensity, $E=\sigma T^{4}$
$E=\left(5.67 \times10^{-8}\,W\,m^{-2}\,K^{-4}\right)\left(10^{4}\,K\right)^{4}$
$=5.67 \times10^{8}\,W\,m^{-2}$