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 ?

A difference of 2.3 eV separates two energy levels in an atom. What is

1.	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» SOLUTION :It is given here `E_(n_(1)) - E_(n_(2)) = 2.3 eV = 2.3 xx 1.6 xx 10^(-19)` J `therefore ` radiation frequecy ` V= (E_(n_(1)) - E_(n_(2)))/(h) = (2.3 xx 1.6 xx 10^(-19))/(6.63 xx 10^(-34)) = 5.6 xx 10^(14)Hz`

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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 ?

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