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