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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 $$n^{th}$$ shell to first excited state (i.e., $$2$$nd shell)
$$=10.20+17.0=27.20eV$$
$$=27.20\times 1.602\times 10^{-12}erg$$
$$\because \frac{hc}{\lambda}=R_H\times Z^2\times hc\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$$

$$\because 27.20\times 1.602\times 10^{-12}=R_H\times Z^2\times h\times c\left[\frac{1}{2^2}-\frac{1}{n^2}\right]$$ ......(i)
Similarly, total energy liberated during transition of electron from $$n$$th shell to second excited state (i.e., $$3rd$$ shell)
$$=4.25\times 5.95=10.20eV$$
$$=10.20\times 1.602\times 10^{-12}erg$$
$$\therefore 10.20\times 1.602\times 10^{-12}=R_H\times Z^2\times h\times c\left[\frac{1}{3^2}-\frac{1}{n^2}\right]$$...............(ii)
Dividing Eq.(i) by Eq. (ii),
$$n=6$$
On substituting the value of $$n$$ in Eqs. (i) or (ii),
$$Z=3$$

Chemistry

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