Question

Ratio of the longest wavelengths corresponding to the Lyman and Balmer series in the hydrogen spectrum is

• A. $7/29$
• B. $9/31$
• C. $5/27$
• D. $3/23$

Answer 1

The correct option is C $5/27$

The wavelength of different spectral lines of Lyman series is given by
$\frac{1}{{\lambda }_L}=R[\frac{1}{1^2}-\frac{1}{n^2}]$ , where$n=2,3,4,\dots$
where subscript $L$ refer to Lyman.
For longest wavelength, $n=2$
$\therefore \frac{1}{{\lambda }_{L_{longest}}}=R[\frac{1}{1^2}-\frac{1}{2^2}]$
$\begin{array}{}=\frac{3}{4}R& \text{(i)}\end{array}$
The wavelength of different spectral lines of Balmer series is given by
$\frac{1}{{\lambda }_B}=R[\frac{1}{2^2}-\frac{1}{n^2}]$ , where$n=3,4,,5\dots$
where subscript B refers to Balmer.
For longest wavelength, $n=3$
$\therefore$ $\frac{1}{{\lambda }_{b_{longest}}}=R[\frac{1}{2^2}-\frac{1}{3^2}]=R[\frac{1}{4}-\frac{1}{9}]$
$\begin{array}{}=\frac{5}{36}R& \text{(ii)}\end{array}$
Divide (ii) by (i), we get
$\frac{{\lambda }_{L_{longest}}}{{\lambda }_{B_{longest}}}=\frac{5R}{36}\times \frac{4}{3R}=\frac{5}{27}$