Question

The wavelength of the first line of Lyman series is $1215\mathring{A}$, the wavelength of first line of Balmer series will be

• A. $4545\mathring{A}$
• B. $5295\mathring{A}$
• C. $6561\mathring{A}$
• D. $6750\mathring{A}$

Answer 1

Answer: (C)

Here ${\lambda }_L=1215\mathring{A}$
For the first line of Lyman series.
$\frac{1}{{\lambda }_L}=R\left[\frac{1}{1^2}-\frac{1}{2^2}\right]$
$=R\left[1-\frac{1}{4}\right]=\frac{3R}{4}$
$\therefore {\lambda }_L=\frac{4}{3R}...\left(i\right)$
For first line of Balmer series
$\frac{1}{{\lambda }_B}=R\left[\frac{1}{2^2}-\frac{1}{3^2}\right]$
$=R\left[\frac{1}{4}-\frac{1}{9}\right]$
$\Rightarrow \frac{1}{{\lambda }_B}=R\left[\frac{5}{36}\right]$
$\therefore {\lambda }_B=\frac{36}{5R}...\left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$,
$\frac{{\lambda }_B}{{\lambda }_L}=\frac{36/5R}{4/3R}$
$=\frac{36\times 3}{4\times 5}$
$\therefore {\lambda }_B=\frac{108}{20}\times {\lambda }_L$
$=\frac{180}{20}\times 1215$
$=6561\mathring{A}$