Question

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

• A. 3.4 eV
• B. 13.6 eV
• C. 54.4 eV
• D. 122. 4 eV

Answer 1

Answer: (C)

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
$E=-\frac{1.6Z^2}{n^2}$
$=\frac{-12.6(2{)}^2}{(1{)}^2}=-54.4eV$