Answer
$1.983\times 10^{-37}$ $W$
Work Step by Step
In the previous part of this problem, we have calculated that
$T=310$ $K$
For small range of wavelength ($\Delta\lambda$), the radiated power ($P$) can be written as
$P=S(\lambda)A\Delta\lambda$, ....................$(1)$
where $S(\lambda)$ is the spectral radiancy and $A$ is total surface area of the radiator.
$S(\lambda)$ is given by
$S(\lambda)=\frac{2\pi c^2h}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kT}}-1}$
Here, $T=310$ $K$ and $\lambda=500$ $nm$ $=500\times 10^{-9}$ $m$
$\therefore \lambda kT=500\times 10^{-9}\times 1.38\times 10^{-23}\times 310$ $J.m$
or, $\lambda kT\approx 2..139\times 10^{-27}$ $J.m$
and $hc=6.63\times 10^{-34}\times 3\times 10^{8}$ $J.m$
or, $hc=1.989\times 10^{-25}$ $J.m$
$\therefore \frac{hc}{\lambda kT}\approx 92.987$
Thus,
$S(\lambda)=\frac{2\times\pi\times(3\times10^8)^2\times 6.63\times 10^{-34}}{(500\times 10^{-9})^5}\frac{1}{e^{92.987}-1}$
$S(\lambda)\approx 4.958\times 10^{-25}$ $W/m^3$
Given, $A=4.00$ $cm^2$ $=0.0004$ $m^2$
and $\Delta\lambda=1.00$ $nm$ $=1\times 10^{-9}$ $m$
$\therefore$ From eq. $1$, we get
$ P=4.958\times 10^{-25}\times 0.0004\times 1\times 10^{-9}$ $W$
or, $P=1.983\times 10^{-37}$ $W$
