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The energy of stable states of the hydro...

The energy of stable states of the hydrogen atom is given by `E_(n)=-2.18xx10^(-8)//n^(2)[J]` where n denotes the principal quantum number.
Calculate the energy differences between `n=2` (first excited state) and `n=1` (ground state) and between `n=7` and `n=1`.

Text Solution

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The correct Answer is:
`DeltaE_(n rarr1)=E_(n)-E_(1)=2.18xx10^(-18) (1-n^(2))`
`DeltaE_(2 rarr 1)=1.635xx10^(-18) J`
`DeltaE_(7 rarr 1)=2.135xx10^(-18) J`
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