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 eV and 17.00 eV respectively.Alternatively the atom from the same excited state can make a transition to the second excited state by snccessively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z (ionization energy of hydrogen atom = 13.6 eV)

A hydrogen-like atom (atomic number Z) is in a higher excited state of

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 the first excited state by successively emitting two photons of energies 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 snccessively emitting two photons of energies 4.25 eV and 5.95 eV respectively. Determine the values of n and Z (ionization energy of hydrogen atom = 13.6 eV)
Answer» n=5 z=2 n=6 z=3 Solution : `E= (-2^2)/(n^2) (13.6eV)` `impliesR.hc.z^2 (1/9 - 1/n^2) = 27.2 ….. (1)` `impliesR.hc.z^2 (1/9 - 1/n^2) = 10.2 eV…(2)` `((1))/((2))= (((n^2 - 4)/(4n^2)))/(((n^2 - 9 )/(9n^2))) = (27.2)/(10.2) = 2.66` `9(n^2 - 4)xx 4 (n^2 - 9) (2.66)` `implies9n^2 - 36 = 10 . 66 n^2 - 96` `therefore 1.66 n^2 = 60 impliesn^2 =(60)/(1.66) = 36`