A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number 'n' this excited atom can make a transition to the first excited state by emitting a photon of first `27.2eV`. Alternatively the atom from the same excited state can make a transition to the 2nd excited state by emitting photon of energy `10.20eV` the value of `n` and `z` are given (ionization energy of hydrogen atom is 13.6eV)

Options

A. n=6 and z=3
B. n=3 and z=6
C. n=8 and z=4
D. n=4 and z=8

Answer

A

Explanation

`Delta E_(1) =27.2 =(13.6)[(1)/(4)-(1)/(n^2)]z^(2)`
`Delta E_(2) =10.2 =(13.6) [(1)/(9)-(1)/(n^2)]z^(2)` .