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Ionisation potential of hydrogen atom is...

Ionisation potential of hydrogen atom is 13.6 eV. If ground state of H-atom is excited by monochromatic radiations of 12.1 eV, then number of spectral lines emitted by H-atom or de-excitation will be:

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The correct Answer is:
3
`E_(a) =(-13.6)/(n^(2))" Also "E=E_(n) -E_(0)`
`12.1 =-(13.6)/(n^(2))+13.6" "therefore n=3`
Thus, de-excitation will lead spectral lines `= Sigma Delta n =3`
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