Question

If the energy difference between two electronic states is $46.12$ $kcal$ ${mol}^{-1}$, what will be the frequency of the light emitted when the electrons drop from higher to lower states? ($Nh=9.52\times {10}^{-14}$ $kcal$ $sec$ ${mol}^{-1}$, where, $N$ is the Avogadro's number and $h$ is the Planck's constant).

Answer 1

$\Delta E=46.12$ $kcal$ ${mol}^{-1}$
According to Bohr theory, $\Delta E=Nhv$
or $v=\frac{\Delta E}{Nh}=\frac{46.12}{9.52\times {10}^{-14}}$
$=4.84\times {10}^{14}$ $cycle$ ${sec}^{-1}$.