Answer
3.7723678×1019 quanta of radiation emitted per second per square meter.
Work Step by Step
Given temperature of blackbody T= 500 $^{0}C$
absolute kelvin temperature = temperature in $^{0}C$ + 273
temperature of blackbody T (in kelvin) = 500 +273= 773 K
From Wiens displacement law
$\lambda_{max} T= 2.90\times10^{-3}m.K$
$\lambda_{max}= \frac{ 2.90\times10^{-3}m.K}{T}$
$\lambda_{max}= \frac{ 2.90\times10^{-3}m.K}{773 K}$
$\lambda_{max}=0.003751617\times10^{-3} m$
$\lambda_{max}=3.751617\times10^{-6} m= 3751.617\times10^{-9} m=3751.6 nm$
frequency corresponding to above wavelength $f=\frac{c}{\lambda}$
$f=\frac{3\times10^{8} m/s}{3.751617\times10^{-6} m}$
$f=0.799655\times10^{14}/s$
$f=7.99655\times10^{13} Hz$
Energy of photon corresponding to above frequency $E=hf$
$E=6.63\times10^{-34} J.s\times7.99655\times10^{13} Hz$
$E=53.0171\times10^{-21}J$
$E=5.30171\times10^{-20}J$
Intensity of emitted radiation = $2.0W/m^ {2}$=$2.0J/s.m^ {2}$.
$E=5.30171\times10^{-20}J$ corresponds to 1 photon
so $1 J$ energy is equal to $\frac{1}{5.30171\times10^{-20}}$ Photon
so $ 2 J$ energy is equal to $\frac{2}{5.30171\times10^{-20}}$ Photon
so $ 2 J$ energy = $0.37723678\times10^{20} $
so $ 2 J$ energy = $3.7723678\times10^{19} $
Intensity $2.0J/s.m^ {2}$ means $3.7723678\times10^{19} $ quanta of radiation emitted per second per square meter.
