  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  B
Z=2,n=3  C
Z=3,n=6  D
Z=6,n=3  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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