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Q.
The third line of Balmer series of an ion equivalnet to hydrogen atom has wavelength of $108.5\, nm$. The ground state energy of an electron of this ion will be
For third line of Balmer series $n_{1}=2, n_{2}=5$
$\therefore \frac{1}{\lambda}=R Z^{2}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]$
gives $Z^{2}=\frac{n_{1}^{2} n_{2}^{2}}{\left(n_{2}^{2}-n_{1}^{2}\right) \lambda R}$
On putting values $Z=2$ From
$E=-\frac{1.6 Z^{2}}{n^{2}}$
$=\frac{-12.6(2)^{2}}{(1)^{2}}=-54.4\, eV$