Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561\stackrel{\circ }{A}$. The wavelength of the second spectral line in the Balmer series of singly-ionized helium atom is

• A. $1215\stackrel{\circ }{A}$
• B. $1640\stackrel{\circ }{A}$
• C. $2430\stackrel{\circ }{A}$
• D. $4687\stackrel{\circ }{A}$

Answer 1

The correct option is A $1215\stackrel{\circ }{A}$

We know that
$\frac{1}{{\lambda }_P}=R{Z}_P^2[\frac{1}{n_f^2}-\frac{1}{n_i^2}]$
$\frac{1}{{\lambda }_H}=R{Z}_H^2[\frac{1}{4}-\frac{1}{9}]=R\left(1{)}^2\right[\frac{5}{36}]$
$\frac{1}{{\lambda }_{He}}=R{Z}_{He}^2[\frac{1}{4}-\frac{1}{16}]=R\left(2{)}^2\right[\frac{3}{16}]$

$\frac{{\lambda }_H}{{\lambda }_{He}}=\frac{1}{4}[\frac{16}{3}\times \frac{5}{36}(5\left)/\right(36\left)\right]=\frac{5}{27}$
${\lambda }_{He}=\frac{5}{27}\times 6561=1215\AA$