Calculate the energy and frequency of the radiation emitted when an electron jumps from n=3 to n=2 ina hydrogen atom.

Explanation

Step-I Calculation of the energy of the radiation. According to Balmer's formula,
`DeltaE=-R_(H)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))bar(v)=109677[(1)/(n_(1)^(2))-(1)/(n_(f))]`
`DeltaE(E)=(-2.18xx10^(-18)J)((1)/(n_(1)^(2))-(1)/(n_(2)^(2)))`
For the electron, `n_(1)=2` and `n_(2)=B`
`DeltaE(E)=(-2.18xx10^(-18))((1)/(4)-(1)/(9))J=((-2.18xx10^(-18))xx5J)/(36)`
`=-3.03xx10^(-19)J`
(Electronic energy is given a negative sign)
Step II. Calculation of frequency of the radiation
`upsilon = (DeltaE)/(h)=((3.03xx10^(-19)J))/((6.626xx10^(-34)JS))=4.57xx10^(14)Hz`