A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits form the upper level to the lower level.

Explanation

Separation of two energy level in an atom
`E=2.3 eV`
`=2.3xx1.6xx10^(-19)`
`=3.68xx10^(-19)J`
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 enegy as:
`E=hv`
Where,
h=Planck's constant `=6.62xx10^(-34)Js`
`:. v=E/h`
`=(3.68xx10^(-19))/(6.62xx10^(-32))=5.55xx10^(14)Hz`
Hence, the frequency of the radiation is `5.6xx10^(14)Hz`