Question

If ${\lambda }_1\text{and}{\lambda }_2$ are the wavelengths of the first members of the Lyman and Paschen series respectively, then ${\lambda }_1:{\lambda }_2$ is-

• A. $7:108$
• B. $7:50$
• C. $1:30$
• D. $1:3$

Answer 1

The correct option is A $7:108$

In the Lyman series, the first line is,

$n_1=1\text{and}{n}_2=2$

$\frac{1}{{\lambda }_1}=R[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

$\frac{1}{{\lambda }_1}=R[1-\frac{1}{2^2}]=\frac{3R}{4}......\left(1\right)$

${\lambda }_1=\frac{4}{3R}........\left(1\right)$

Similarly, in Paschen series, the first line is,

$n_1=3\text{and}{n}_2=4$

$\frac{1}{{\lambda }_2}=R[\frac{1}{3^2}-\frac{1}{4^2}]=R[\frac{1}{9}-\frac{1}{16}]$

$\frac{1}{{\lambda }_2}=R\left[\frac{7}{144}\right]$

${\lambda }_2=\frac{144}{7R}........\left(2\right)$

From (1) $\&$ (2) we get,

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{4}{3R}\times \frac{7R}{144}=\frac{7}{3\times 36}=\frac{7}{108}$

$\therefore {\lambda }_1:{\lambda }_2=7:108$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(A\right)$ is the correct answer.