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 successively emitting two photons of energies `10.20` and `17.00 eV` respectively. Alternatively, the atom from the same axcited state can make a transition to the second excited state by successively emitting two photons of energy `4.25 eV` and `5.95 eV` respectively. Determine the values of n and Z `("ionisation energy of hydrogen atom"=13.6 eV)`. Given answer `=n+Z`.

Explanation

The energy emitted during transition of electron from `n^(th)` shell to first excited state ( i.e. `2^(nd)` shell ) `10.20 + 17.0 = 27.2 eV = 27.2 xx 1.6 = 10^(-19) J`
`Delta E = R_(H) Z^(2) hc [ (1)/( 2^(2)) - ( 1)/( n ^(2)) ]`
`27.2 xx 1.6 xx 10^(-19)`
`= 1.09 xx 10^(7) xx Z^(2) xx 6.6 xx 10^(-34) xx 3 xx 10^(8) [ ( 1)/( 2^(2)) - ( 1)/( n^(2)) ]` ....(1)
Similarly total energy liberated during transition of electron from nth shell to second excited state ( i.e. `3^(rd)` shell )
`= 4.25 xx 5.95 = 10.20 eV = 10.20 xx 1.6 xx 10^(-19) J `
`:. 10.20 xx 1.6 xx 10^(-19)`
`= 1.09 xx 10^(7) xx Z^(2) xx 6.6 xx 10^(-34) xx 3 xx 10^(8) [ ( 1)/( 3^(2)) - ( 1)/( n ^(2)) ] ` ......(2)
Dividing equation (1) by equation (2)
n = 2
On substituting the value of n in equation (1) or (2)
Z = 3