Question

Calculate the energy in kilocalorie per mol of the photons of an electromagnetic radiation of wavelength $7600\mathring{A}$.

Answer 1

$\lambda =7600\mathring{A}=7600\times {10}^{-8}cm$
$c=3\times {10}^{10}cm{s}^{-1}$
Frequency, $v=\frac{c}{\lambda }=\frac{3\times {10}^{10}}{7600\times {10}^{-8}}=3.947\times {10}^{14}{s}^{-1}$
Energy of one photon $=hv=6.62\times {10}^{-27}\times 3.947\times {10}^{14}$
$=2.61\times {10}^{-12}erg$
Energy of one mole of photons $=2.61\times {10}^{-12}\times 6.02\times {10}^{23}$
$=15.71\times {10}^{11}erg$
Energy of one mole of photons in kilocalorie $=\frac{15.71\times {10}^{11}}{4.185\times {10}^{10}}$ $[1kcal=4.185\times {10}^{10}erg]$
$=37.538$ $kcal$ $per$ $mol$.