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Physics Question on Atoms

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?

Answer

Seperation of two energy levels in an atom,
E=2.3eV
=2.3×1.6×10−19=2.3×1.6×10^{-19}
=3.68×10−19J=3.68×10^{-19}J
Let vv 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=hvE=hv
Where,
h=planck's constant=6.62×10−34Js=6.62×10^{-34}Js
∴v=Eh∴v=\frac{E}{h}
=3.68×10−196.62×10−32=5.55×1014Hz=\frac{3.68×10^{-19}}{6.62×10^{-32}}=5.55×10^{14}Hz
Hence,the frequency of the radiation is 5.6×1014Hz5.6×10^{14}Hz.