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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 Li$$^{2+}$$
$$\Delta E= 13.6\times Z^2 \left[\dfrac{1}{{n_1}^2} − \dfrac{1}{{n_2}^2} \right]$$ 

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.

$$27.2=Z^2\times 13.6 \left[\dfrac{1}{4} − \dfrac{1}{{n_2}^2} \right]$$........(i)

When the electron comes to the second excited state $$(n_1 = 3)$$ the energy is $$4.25 + 5.95 = 10.2\ eV$$ 

$$10.2 = Z^2\times 13.6 \left[\dfrac{1}{9} − \dfrac{1}{{n_2}^2} \right]$$......(ii)

On solving both the equation (i) and (ii), we get $$Z=3$$ and $$n_2=6$$

It is $$Li^{2+}$$ ion.

Hence, option A is correct.

Chemistry

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