Question

A doubly ionised Li atom is excited from its ground state(w = 1) to n = 3 state. The wavelengths of the spectral lines are given by $\lambda_{32},\,\lambda_{31}$ and $\lambda_{21}$. The ratio $\lambda _{32}, / \lambda _{31}$ and $\lambda _{21}, / \lambda _{31}$ are, respectively

• A. 8.1,0.67
• B. 8.1,1.2
• C. 6.4,1.2
• D. 6.4,0.67

Answer 1

Answer: (C)

$\frac{1}{\lambda } = R \left(\frac{1}{n_{1}^{2}}-\frac{1}{n^{2}_{2}}\right)$ where R = Rydberg constant
$\frac{1}{\lambda_{32}} = \left(\frac{1}{4}-\frac{1}{9}\right) = \frac{5}{36}$
$\Rightarrow \quad\lambda_{32} = \frac{36}{5}$
Similarly solving for $\lambda_{31}$ and $\lambda_{21}$
$\lambda_{31} = \frac{9}{8}$ and $\lambda _{21} = \frac{4}{3}$
$\therefore \quad \frac{\lambda_{32}}{\lambda_{31}} = 6.4$ and $\frac{\lambda _{21}}{\lambda _{31}} \simeq 1.2$