Answer
$3.7\times 10^{6}$
Work Step by Step
According to the classical radiation law,
$I_c=\frac{2\pi ckT}{\lambda^4}$
and according to the Planck’s radiation law,
$I_p=\frac{2\pi c^2h}{\lambda^5}\frac{1}{e^{\frac{hc}{\lambda kT}}-1}$
Therefore, the ratio of $\frac{I_c}{I_p}$ is given by
$\frac{I_c}{I_p}=\frac{\lambda kT}{hc}(e^{\frac{hc}{\lambda kT}}-1)$
Given, $T=2000$ $K$, $\lambda=400$ $nm$ $=400\times 10^{-9}$ $m$
$\therefore \lambda kT=400\times 10^{-9}\times 1.38\times 10^{-23}\times 2000$ $J.m$
or, $\lambda kT=1.104\times 10^{-26}$ $J.m$
and $hc=6.63\times 10^{-34}\times 3\times 10^{8}$ $J.m$
or, $hc=1.989\times 10^{-25}$ $J.m$
$\therefore \frac{\lambda kT}{hc}\approx 0.0555$ and $\frac{hc}{\lambda kT}\approx 18.016$
Therefore,
$\frac{I_c}{I_p}=0.0555\times(e^{18.016}-1)\approx 3.7\times 10^{6}$
