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 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). Give answer = n + Z.
Text Solution
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(9)
Total energy liberated during tranition of electron from nth shell to first excited state, (i.e., 2nd shell)
= 10.20 + 17.0 = 27.20 eV
= 27.20 × 1.602 × 10 –12 erg
∴ 
= R H × Z 2 × hc 
∴ 27.20 × 1.602 × 10 –12 = R H × Z 2 × h × c 
....(i)
Similarly, total energy liberated during transition of electron from nth shell to second excited state, (i.e., 3rd shell)
= 4.25 + 5.95 = 10.20 eV
= 10.20 × 1.602 × 10 –12 erg
∴ 10.20 × 1.602 × 10 –12 = R H × Z 2 × h × c 
...(ii)
Dividing Equations (i) by (ii)
n = 6
On substituting the value of n in Equations (i)
Z = 3
So, n +Z = 6 + 3 = 9.
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