Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is $6561\dot{\text{A}}.$ The wavelength of the second spectral line in the Balmer series of single ionized helium atom is ____ $\dot{\text{A}}$.

Answer 1

The Rydberg's formula for the wavelength of the spectral lines is given by,

$\frac{1}{\lambda }=Z^{2}R(\frac{1}{{n_f}^2}-\frac{1}{{n_i}^2})$

For $\text{H-}$ atom, $Z=1$

The first line of Balmer series corresponds to the transition from $n_i=3\to n_f=2$

$\frac{1}{{\lambda }_H}=R(1{)}^2(\frac{1}{4}-\frac{1}{9})$

$\frac{1}{{\lambda }_H}=R\left(\frac{5}{36}\right)..........\left(1\right)$

For single ionized $\text{He-}$ atom, $Z=2$

The second line of Balmer series corresponds to the transition from $n_i=4\to n_f=2$

$\frac{1}{{\lambda }_{\text{He}}}=R(2{)}^2(\frac{1}{4}-\frac{1}{16})$

$\frac{1}{{\lambda }_{\text{He}}}=R\left(4\right)\left(\frac{3}{16}\right)........\left(2\right)$

Dividing eq. $\left(1\right)$ by $\left(2\right)$,

$\frac{{\lambda }_{\text{He}}}{{\lambda }_{\text{H}}}=\frac{1}{4}[\frac{16}{3}\times \frac{5}{36}]=\frac{5}{27}$

$\therefore {\lambda }_{\text{He}}=\frac{5}{27}\times 6561\dot{\text{A}}=1215\dot{\text{A}}$