1.

Energy of a hydrogen atom with principal quantum number n is shown by E = (-13.6)/(n^(2)) eV. The energy of a photon ejected when the electron jumps from n = 3 state to n = 2 state of hydrogen is approximately.

Answer»

`1.9eV`
`1.5eV`
`0.85eV`
`3.4eV`

Solution :`DeltaE=E_(3)-E_(2)=-(13.6)/(3^(2))+(13.6)/(2^(2))`
`=13.6((1)/(4)-(1)/(9))eV=1.9eV`


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