Question

If the energy difference between two electronic states is $47.5kcalmo{l}^{-1}$ and the frequency of the light emitted when the electrons drop from a higher state to lower states is $x\times {10}^{14}$, what is the value of $x$?
$(N.h=9.5\times {10}^{-14}kcal{s}^{-1}mo{l}^{-1},\text{where, N is the Avogadro's number and h is the Planck's constant})$

Answer 1

From Planck's Quantum theory,
$\Delta E=N\times h\times \nu$
where $\nu$ is the frequency
$\Rightarrow \nu =\frac{\Delta E}{Nh}=\frac{47.5}{9.5\times {10}^{-14}}=5\times {10}^{14}{s}^{-1}$