A hydrogen -like atom (atomic number Z) is in a higher excited state can make a transition number n .The ecxited state by succesively emitting two ophoton to the first `10.20 eV` and `17.00 Ev` respectively .Alternatively, the atom from the same excited state can make a transition to the second excited state by seccessivelyemitting two photon of energy `4.25 eV` and ` 5.95 eV` respectivel,y.Determine the values oF N AND z

Explanation

Total energy libetated during transition of element from nth shell to first excited state (i.e. second shell ) is `10.20 +17.0 = 27.20 xx 1.602 xx 10^(-12)erg`
`," (hc)/(lambda) = R_(H) xx Z^(2) hc [(1)/(2^(2)) - (1)/(n^(2))]`
`:. 27.20 xx 1.602 xx 10^(-12) = R_(H) xx Z^(2) xx h xx c [(1)/(2^(2))-(1)/(n^(2))]` ......(i)
Similarlly taotal energy libratated during transition of electron from oth to second excited state (i.e) third shell) is `4.25 xx 5.59 = 10.20 xx 1.602 xx 10^(-12) erg`
`:. 10.20 xx 1.602 xx 10^(-12) = R_(H) xx Z^(2) xx h xx c [(1)/(3^(2)) - (1)/(n^(2))]`.......(ii)
Dividing equation (i) by equation (ii) `n = 6` On subssituting the value of n in equationm (i) or (ii) we get `Z = 3`