## Chemistry 9th Edition

$v=3\times 10^{10}Hz$ $E=2\times 10^{-23}\frac{J}{photon}$ $E=12\frac{J}{mol}$
We know that $v=\frac{c}{\lambda}$ We plug in the known values to obtain: $v=\frac{2.9979\times 10^8}{0.01}$ $v=3\times 10^{10}Hz$ As $E=h\nu$ We plug in th known values to obtain: $E=6.626\times 10{-34}\times (3\times 10^{10})$ $E=2\times 10^{-23}\frac{J}{photon}$ We know that $h=6.626\times 10^{-34}Js$ $\implies E=2\times 10^{23}\frac{J}{photon}\times 6.022\times 10^{23}\frac{photons}{mol}$ $E=12\frac{J}{mol}$