Given, maximum wavelength,
${\lambda }_m=289.8nm=28.98\times {10}^{-9}m$
$=2898\times {10}^{-10}m$
Stefan’s constant, $\sigma =5.67\times {10}^{-8}W{m}^{-2}{K}^{-4}$
Wien’s constant, b = 2898 $\mu mk$ = 2898 $\times {10}^{-6}mK$
According to Wien’s displacement law, the maximum wavelength is given by
${\lambda }_m=\frac{b}{T}\Rightarrow T=\frac{b}{{\lambda }_m}$ …. (i)
Substituting given values in Eq. (i), we get
$T=\frac{2898\times {10}^{-6}}{2898\times {10}^{-10}}={10}^{4}K$ …. (ii)
According to Stefan’s law the energy radiated from a source is given by
$E=\sigma Ae{T}^4$ …. (iii)
Where, A = area of source
e = emissivity (value between 0 to 1)
The intensity of radiations emitted is equal to energy radiated from a given surface area, i.e.,
$l=\frac{E}{A}=\sigma e{T}^4$ [From Eq.(iii)]
As e is very small , so
$l=\sigma T^4$
Substituting the value of T from Eq.(ii) in Eq.(iv), we get
$l=\sigma {\left(10\right)}^4=5.67\times {10}^{-8}\times {10}^{16}$
$\left[\because \sigma =5.67\times {10}^{-8}\left(given\right)\right]$
$=5.67\times {10}^{8}W{m}^{-2}$