Question

The wavelength of the first spectral line in the Balmer series of the 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

Answer: (A)

For hydrogen or hydrogen type atoms
$\frac{1}{\lambda }={\left(RZ\right)}^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$
In the transition from $n_i\to n_f$
$\therefore \lambda \propto \frac{1}{Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}$
$\therefore \frac{{\lambda }_2}{{\lambda }_1}=\frac{Z_1^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_1}{Z_2^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_2}$
${\lambda }_2=\frac{{\lambda }_1{Z}_1^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_1}{Z_2^2{\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)}_2}$
Substituting the values, we have
$=\frac{(6561){\left(1\right)}^2\left(\frac{1}{2^2}-\frac{1}{3^2}\right)}{{\left(2\right)}^2\left(\frac{1}{2^2}-\frac{1}{4^2}\right)}$ $=1215\stackrel{{}^{\circ }}{A}$