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
$\frac{1}{{\lambda }_{32}}=\left(\frac{1}{4}-\frac{1}{9}\right)=\frac{5}{36}$
$\Rightarrow {\lambda }_{32}=\frac{36}{5}$
Similarly solving for ${\lambda }_{31}$ and ${\lambda }_{21}$
${\lambda }_{31}=\frac{9}{8}$ and ${\lambda }_{21}=\frac{4}{3}$
$\therefore \frac{{\lambda }_{32}}{{\lambda }_{31}}=6.4$ and $\frac{{\lambda }_{21}}{{\lambda }_{31}}\simeq 1.2$