Answer
$1.50\times 10^{23}atoms$
Work Step by Step
We know that
$E=\frac{hc}{\lambda}$
We plug in the known values to obtain:
$E=\frac{6.626\times 10^{-34}\times 2.9979\times 10^8}{150\times 10^{-9}}$
$E=1.32\times 10^{-18}\frac{J}{photon}$
Now $\frac{1.98\times 10^5J}{1.32\times 10^{-18}\frac{J}{photon}}=1.50\times 10^{23}photons$
Similarly $1.50\times 10^{23}photons\times \frac{atom}{photon}=1.50\times 10^{23}atoms$
