Question

The ratio of wavelengths of first line of Lyman series in $L{i}^{+2}$ and first line of Lyman series in deuterium ${(}_1{H}^2)$ is:

• A. 1 : 9
• B. 9 : 1
• C. 1 : 4
• D. 4 : 1

Answer 1

Answer: (A)

1 : 9

Wavelength is given by

$\frac{1}{\lambda }=R{Z}^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

For first line of Lyman series, $n_1=1$ and $n_2=2$

$\frac{1}{\lambda }=R{Z}^2[\frac{1}{1^2}-\frac{1}{2^2}]$

$\frac{1}{\lambda }=R{Z}^2\times \frac{3}{4}$

$\lambda \propto \frac{1}{Z^2}$

$\frac{{\lambda }_{L{i}^{2+}}}{{\lambda }_{H^2}}=\frac{Z_{H^2}^2}{Z_{L{i}^{2+}}^2}=\frac{1}{3^2}=\frac{1}{9}$

Hence, the correct option is $A$