Question

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

Answer 1

Here `E=2.3eV = 2.3xx1.6xx10^(-19) J`
As E=hV
`therefore v=€/(h)=(2.3xx1.6xx10^(-19))/(6.6xx10^(-34))=5.6xx10^(4) Hz`