Question

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

• 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. $10.67\times {10}^{14}W{m}^{-2}$

Answer 1

The correct option is A $5.67\times {10}^{8}W{m}^{-2}$

According to Wien's displacement law,

${\lambda }_{max}T=b$

$\therefore T=\frac{b}{{\lambda }_{max}}=\frac{2898\mu mK}{289.8nm}$

$\frac{2898\times {10}^{-6}mK}{289.8\times {10}^{-9}m}={10}^{4}K$

According to Stefan Boltzmann law,

Radiation intensity, $R=\sigma T^4$

$R=(5.67\times {10}^{-8}W{m}^{-2}{K}^{-4}){\left({10}^{4}K\right)}^4$

$=5.67\times {10}^{8}W{m}^{-2}$