If the bineding energy of the electron of the electron in a hydrogen atom is `13.6 eV` the energy required to remove the electron from the first excited state of `Li ^(++)` is
A
`30.6eV`
B
`13.6eV`
C
`3.4eV`
D
`122.4eV`
Text Solution
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The correct Answer is:
A
`E_(n) = (13.6)/(n^(2)) Z^(2) eV//atom` for lithium ion` Z = 3 , n = 2 ` (for first excited state) `E_(n) = (13.6)/(Z^(2)) xx 3^(2) = -30.6 eV`
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