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Question

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 state by successively emitting two photons of energies 10.20 eV and 17.00 eV. Alternatively, the atom from the same excited state can make a transition to the second excited state by emitting two photons of energies 4.24 eV and 5.95 eV. The hydrogen-like atom is :

A
Li2+
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B
He+
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C
H
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D
none
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Solution

The correct option is C Li2+
ΔE=13.6×Z2[1n121n22]

The electron makes a transition from state n2 is the first excited state (n1=2) releasing 10.2+17=27.2 eV energy.

27.2=Z2×13.6[141n22]........(i)

When the electron comes to the second excited state (n1=3) the energy is 4.25+5.95=10.2 eV

10.2=Z2×13.6[191n22]......(ii)

On solving both the equation (i) and (ii), we get Z=3 and n2=6

It is Li2+ ion.

Hence, option A is correct.

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