A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number n.This excited atom can make transition to the first excited state by succesively 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 successively emitting two photons of energies `4.25 eV` and ` 5.95 eV` respectively. Determine the values of n and z

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 transition to the first excited state by succesively 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 successively emitting two photons of energies `4.25 eV` and ` 5.95 eV` respectively. Determine the values of n and z
Answer» ` Delta E = Z^(2) (13.6)[(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` The electron makes a transition from state `n_(2) ` is the first excited state `(n_(1) = 2)` releasing `10.2 = 17 = 27.2 eV` energy `27.2 = Z^(2)(13.6)[(1)/(4) - (1)/(n_(2)^(2))]`........(i) When the electron comes to the second excited state `(n_(1) = 3) ` the energy is `4.25 = 5.95 = 10.2 eV` Hence `10.2 = Z^(2)(13.6) ((1)/(9) - (1)/(n_(2)^(2)))`......(ii) ON solving equation (i) and (ii) we get `Z = 3 ` and `n_(2) = 6`

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