A hydrogen like atom (atomic number Z) is in a higher excited state of quantum number n. The excited atom can make a transition to the first excited state by successively emitting two photons of energy 10.2 eV and 17.0 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 eV, respectively Determine the values of n and Z. (Ionization energy of H-atom = 13.6 eV)

Explanation

We can write energy in the `n^(th)` state of hydrogen-like atoms as follows :
`E=-Z^2(13.6/n^2)eV`
It is given that `E_n-E_2` =10.2+17=27.2 eV
`-13.6Z^2(1/n^2-1/2^2)=27.2`
`-Z^2(1/n^2-1/2^2)=2`….(i)
Similarly we can substitute n=3 for second excited state.
`-Z^2(1/n^2-1/3^2)=(4.25+5.95)/13.6`
`-Z^2(1/n^2-1/9)`=0.75 ....(ii)
On subtracting equation (ii) from equation (i) we get the following:
`Z^2(1/4-1/9)=2-0.75`=1.25
`rArr Z^2(5/36)`=1.25
`rArr Z^2` = 36 x 0.25
`rArr` Z=3
Now we can substitute Z=3 in equation (ii)
`-9(1/n^2-1/9)=0.75`
`rArr (-9)/n^2+1`=0.75
`rArr 9/n^2`=0.25
`rArr` n=6