12.4 A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom make a transition from the upper level to the lower level?

Hint: The energy of the radiation equals the difference in energy levels.

Step 1: Relate energy with the frequency.
Separation of two energy levels in an atom,
Let $$\nu$$ be the frequency of radiation emitted when the atom transits from the upper level to the lower level.
We have the relation for energy as: E =$$h\nu$$
where, h = plank's constant =$6.62×{10}^{-34}\mathrm{Js}$$\begin{array}{r}\end{array}$

Step 2: Find the frequency of the radiation.
$$\nu=\frac{E}{h}=\frac{3.68 \times 10^{-19}}{6.62 \times 10^{-32}}=5.55 \times 10^{14} \mathrm{~Hz}$$
Hence, the frequency of the radiation is