Question

If the energy difference between two electronic states is 46.12 kcal/mol; What will be the frequency of light emitted when the electrons dropped from higher to lower states ($Nh=$ $9.52\times {10}^{-14}$ kcal sec mol${}^{-1}$) where $N$ is the Avogadro number.

Answer 1

Energy difference $=46.12k/mol$

Energy difference per atom $=\frac{46.12}{N}kcalatm$

$N=$ Avogadro Number.

$h\nu =\frac{46.12}{N}$

$Nh\nu =46.12$

$Nh=9.52\times {10}^{-14}Kcal/sec$

$9.52\times {10}^{-14}\times \nu =46.12$

$\nu =\frac{46.12}{9.52\times {10}^{-14}}$

$=4.844\times {10}^{14}Hz$

$=489PHz$