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 first excited state by successively emitting two photons of energies 10.20eV and 17.0 eV respectively. Alternatively, the atom from the same excited state can make a transition to second excited state by successively emitting two phons of energy 4.25 eV and 5.95eV respectively. Determine the values of n and Z.

A hydrogen like, atom ( atomic number =Z) is in a higher excited state

1.	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 first excited state by successively emitting two photons of energies 10.20eV and 17.0 eV respectively. Alternatively, the atom from the same excited state can make a transition to second excited state by successively emitting two phons of energy 4.25 eV and 5.95eV respectively. Determine the values of n and Z.
Answer» Solution :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