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}$