A hydrogen like atom with atomic number 'Z' is in higher excited state of quantum number 'n'. This excited state can make a transition to the first excited state by successively emitting two photons energies 10 eV and 17 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by emitting two photons of energies 4.25 eV adn 5.95 eV respectively. Calculate the values of 'n' and 'Z'

A hydrogen like atom with atomic number 'Z' is in higher excited state

1.	A hydrogen like atom with atomic number 'Z' is in higher excited state of quantum number 'n'. This excited state can make a transition to the first excited state by successively emitting two photons energies 10 eV and 17 eV respectively. Alternatively, the atom from the same excited state can make a transition to the second excited state by emitting two photons of energies 4.25 eV adn 5.95 eV respectively. Calculate the values of 'n' and 'Z'
Answer» Solution :For the given EXCITED state, `n_(2) = n` 1st excited state means, `n_(1) =2` 2nd excited state means, `n_(1) =3` ENERGY difference between n = 2 and n = 3 will be `= (10 + 17) - (4.25 + 5.95) = 16.8 eV` `:. 16.8 = 13.6 ((1)/(2^(2)) - (1)/(3^(2))) Z^(2) = 13.6 xx (5)/(36) xx Z^(2)` or `Z^(2) = (16.8)/(13.6) xx (36)/(5) = 9 or Z = 3` Energy difference between `n_(2) = n and n_(1) = 2` will be `= 10 + 17 eV = 27 eV` `:. 27 = 13.6 ((1)/(2^(2)) - (1)/(n^(2))) xx 3^(2)` or `3 = 13.6 ((1)/(4) - (1)/(n^(2))) or (1)/(4) - (1)/(n^(2)) = (3)/(13.6) or (1)/(n^(2)) = (1)/(4) - (3)/(13.6) = (13.6 - 12)/(4 xx 13.6)` or `n^(2) = (4 xx 13.6)/(1.6) or n = 6`