Question

The first line of Balmer series has wavelength $6563\mathring{A}$. what will be the wavelength of the first memeber of lymais series.

• A. $1215\mathring{A}$
• B. $2500\mathring{A}$
• C. $7500\mathring{A}$
• D. $600\mathring{A}$

Answer 1

The correct option is A $1215\mathring{A}$

$\begin{array}{c}\text{Given,}\\ \text{First line of Balmer series has wavelength}=6563\hat{A}\end{array}$

$\begin{array}{c}\text{We know,}\\ \text{for Balmer series}\\ \text{line is from}{n}_1=2\text{to}{n}_2=3,4,5\dots \end{array}$

$\begin{array}{c}\text{and for laymen series}\\ \text{line is from}{n}_1=1\text{to}{n}_2=2,3,4\dots \end{array}$

$\begin{array}{c}\text{So, First line of balmer series mean}\\ n_1=2\text{to}{n}_2=3\end{array}$

$\text{we know}$

$\frac{1}{\lambda }=R{z}^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})$

$R{z}^2=\frac{36}{6563\times 5}\dots \text{(1)}$

$\begin{array}{c}\text{for first line of lymen series}\\ n=1\text{to}n=2\end{array}$

$\begin{array}{cc}\frac{1}{\lambda }& =R{z}^2(\frac{1}{(1{)}^2}-\frac{1}{4})\\ \frac{1}{\lambda }& =\frac{36}{6563\times 5}\left(\frac{3}{4}\right)\\ \lambda & =\frac{6563\times 5\times 4}{36\times 3}\\ \lambda & =1215.3\dot{A}\end{array}$