The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of $108.5\ nm$. The ground state energy of an electron of this ion will be

Question

Dr. Abubacker · Accepted Answer

(c) For third line of Balmer series $n_{1} = 2,n_{2} = 5$
$	herefore\frac{1}{\lambda} = RZ^{2}\left\lbrack \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} ightbrack$ gives $Z^{2} = \frac{n_{1}^{2}n_{2}^{2}}{(n_{2}^{2} - n_{1}^{2})\lambda R}$
On putting values $Z = 2$
From $E = - \frac{13.6Z^{2}}{n^{2}} = \frac{- 13.6(2)^{2}}{(1)^{2}} = - 54.4eV$

Q. The third line of Balmer series of an ion equivalent to hydrogen atom has wavelength of $108.5\ nm$. The ground state energy of an electron of this ion will be

Explanation

Reviewed by Dr. Abubacker

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