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,
E=2.3 eV=2.3×1.6×10-19=3.68×10-19J
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 Js

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 5.6×1014 Hz.