Question

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

• A. $3.4eV$
• B. $13.6eV$
• C. $54.4eV$
• D. $122.4eV$

Answer 1

The correct option is C $54.4eV$

For third of Balmer series $n_1=2,n_2=5$

$\therefore \frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

gives $Z^2=\frac{n_1^2{n}_2^2}{(n_2^2-n_1^2)\lambda R}$

On putting values $Z=2$

From $E=-\frac{13.6Z^2}{n^2}=\frac{-13.6(2{)}^2}{(1{)}^2}=-54.4eV$