Question

A difference of $2.3eV$ 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?

Answer 1

Given,
separation of two energy levels in an atom
$E=2.3eV$
Or $E=2.3\times 1.6\times {10}^{-19}$
$=3.68\times {10}^{-19}$
Let v 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:
$E=hv$
h = Plnak's constant $=6.62\times {10}^{-34}Js$
$v=\frac{E}{h}$
Substituting the values, we get
$v=\frac{3.68\times {10}^{-19}}{6.62\times {10}^{-34}}$
$v=5.5\times {10}^{14}Hz$
$\approx 5.6\times {10}^{14}Hz$
Final Answer :$5.55\times {10}^{14}Hz$