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Question

A hydrogen-like atom (atomic number Z) is in a higher excited state of the quantum number n. This excited atom can make a transition to the first excited state by successfully emitted to photons of energies 10.20eV and 17.00eV respectively.

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

A
Z=3,n=4
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B
Z=2,n=3
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C
Z=3,n=6
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D
Z=6,n=3
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Solution

The correct option is C Z=3,n=6
Total energy liberated during transition of electron from nth shell to first excited state (i.e., 2nd shell)
=10.20+17.0=27.20eV
=27.20×1.602×1012erg
hcλ=RH×Z2×hc[1221n2]

27.20×1.602×1012=RH×Z2×h×c[1221n2] ......(i)
Similarly, total energy liberated during transition of electron from nth shell to second excited state (i.e., 3rd shell)
=4.25×5.95=10.20eV
=10.20×1.602×1012erg
10.20×1.602×1012=RH×Z2×h×c[1321n2]...............(ii)
Dividing Eq.(i) by Eq. (ii),
n=6
On substituting the value of n in Eqs. (i) or (ii),
Z=3

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