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 successively emitting two photons of energies 10.20 e V and 17.00 eV respectively.
Alternatively the atom from the same excited state can make a transition to the second excited state by successively emitting two photons of energies 4.25 eV and 5.95 e V respectively. Determine the values of n and Z.

Explanation

So, `DeltaE_1 = E_n - E_2 = 10.20 + 17`
`= 27. 20 eV`
`DeltaE_2 = E_n - E_3 = 425 + 5.95`
`= 10.20eV`
`rArr E_n - E_2 = DeltaE_1 - DeltaE_2`
`= 27.20 - 10.20`
`=17.20eV`
So, wavelength of the emitted photon
for `n to 3 " to " n to 2`
`rArr (hc)/lambda = 17.20 xx 1.6 xx10^(-19)`
`lambda = (6.6 xx10^(-34) xx 3 xx 10^8)/(17.20 xx 1.6 xx 10^(-10))`
`= 0.72 xx 10^(-7) = 720Å`
By Balmer series,
`1/lambda = BZ^2 (1/2^2 - 1/3^2) (n to 3 " to " n to 2)`
`1/(720 xx 10^(-10) = 1.097 xx 10^7 xx Z^2 (5/36)`
`Z^2 = (36)/(720 xx 10^(-10) xx10^7 xx 1.097) xx 1/5`
`= (0.049 xx 10^3)/10^3 xx 1/5`
` = 49/5`
`Z = 7/sqrt5 = 8/2.25 ~=3`
Now `E_2 = (-13.6 Z^2)/n^2 (n = 2)`
` = - (13.6 xx 9)/(4) = - 30.6 eV`
So, energy of nth state,
`E_(n) = - 30.6 + 27.20`
= `-34eV`
So, for nth state,
`E_n = (-13.6Z^2)/(n^2)`
`- 3.4 = (-13.6 xx9)/(n^2)`
`n^2 = (13.6 xx9)/(3.4)`
` n= 6`