Question

If the first line of Lymann series has a wavelength $1215.4A$. The first line of Balmer series is approximately

• A. $4683\stackrel{0}{A}$
• B. $1025.5\stackrel{0}{A}$
• C. $6563\stackrel{0}{A}$
• D. $6400\stackrel{0}{A}$

Answer 1

The correct option is C $6563\stackrel{0}{A}$

The wavelength of different lines of Lymann series are given by ,

$\frac{1}{\lambda }=R{Z}^2[\frac{1}{1^2}-\frac{1}{n^2}]$

Let ${\lambda }_1$ be the wavelength of first line of Lymann series ($n=2$) , then

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

or $\frac{1}{{\lambda }_1}=R{Z}^2[\frac{1}{1}-\frac{1}{4}]$

or $\frac{1}{{\lambda }_1}=R{Z}^2\left[\frac{3}{4}\right]$

or $\frac{1}{1215.4}=R{Z}^2\left[\frac{3}{4}\right]$ ...........eq1 ( given $\lambda =1215.4A^o$)

The wavelength of different lines of Balmer series are given by ,

$\frac{1}{\lambda }=R{Z}^2[\frac{1}{2^2}-\frac{1}{n^2}]$

Now ,let ${\lambda }_1^{\prime }$ be the wavelength of first line of Balmer series ($n=3$) , then

$\frac{1}{{\lambda }_1^{\prime }}=R{Z}^2[\frac{1}{2^2}-\frac{1}{3^2}]$

or $\frac{1}{{\lambda }_1^{\prime }}=R{Z}^2[\frac{1}{4}-\frac{1}{9}]$

or $\frac{1}{{\lambda }_1^{\prime }}=R{Z}^2\left[\frac{5}{36}\right]$ ................eq2

Dividing eq1 by eq2 , we get

$\frac{{\lambda }_1^{\prime }}{1215.4}=\frac{3\times 36}{4\times 5}$

or ${\lambda }_1^{\prime }=\frac{3\times 36\times 1215.4}{4\times 5}=6563A^o$