A difference of 2.3 eV separates two energy levels in an atom.

Separation of two energy levels in an atom,

E = 2.3 eV

= 2.3 × 1.6 × 10−19

= 3.68 × 10−19 J

Let ν 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 = hv

Where,

$h=$ Planck's constant $=6.62 \times 10^{-34} \mathrm{JS}$

$\therefore v=\frac{E}{h}$

$=\frac{3.68 \times 10^{-19}}{6.62 \times 10^{-32}}=5.55 \times 10^{14} \mathrm{~Hz}$

Hence, the frequency of the radiation is 5.6 × 1014 Hz.