Question

The wave length corresponding to maximum intensity of radiation emitted by a star is $289.8$nm. The intensity of radiation for the star is :

(Stefans constant $=$ 5.6x10${}^{-8}W{m}^{-2}{K}^{-4}$, Wien's displacement constant = $2898\times {10}^{-6}mK$ )

• A. 5.67 x 10${}^{8}W{m}^{-2}$
• B. 5.67 x 10${}^{4}W{m}^{-2}$
• C. 10.67 x 10${}^{7}W{m}^{-2}$
• D. 10.67 x 10${}^{4}W{m}^{-2}$

Answer 1

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

Given, Wavelength corresponding to maximum radiation $=289.8nm$

Stefans constant$=5.6\times {10}^{-8}W{m}^{-2}{K}^{-4}$

From Wien's law,
${\lambda }_{max}T$= constant (b)
$b=2898\times {10}^{-6}$
${\lambda }_{max}=289.8$
$\therefore$ $T=\frac{b}{{\lambda }_{max}}=\frac{2898\times {10}^{-6}}{289.8\times {10}^{-9}}={10}^{4}K$

Intensity of radiation E
$E=\sigma T^4=5.6\times {10}^{-8}\times {10}^{16}=5.6\times {10}^{8}W{m}^{-2}$