Question

If the wavelength of first member of Balmer series in hydrogen spectrum is $6563\mathring{A}$ , the wavelength of second member of Balmer series will be :

• A. $1215\mathring{A}$
• B. $4861\mathring{A}$
• C. $6050\mathring{A}$
• D. data given is insufficient to calculate the value

Answer 1

Answer: (B)

Balmer found that the wavelength of all lines in hydrogen spectrum can be represented by
$\frac{1}{\lambda }=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$
where $R$ is Rydbergs constant.
Given, $n_2=3,n_1=2$
$\therefore \frac{1}{{\lambda }_1}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)=\frac{5R}{36}$
$\Rightarrow \frac{1}{{\lambda }_2}=R\left(\frac{1}{2^2}-\frac{1}{4^2}\right)=\frac{3R}{16}$
$\Rightarrow \frac{{\lambda }_2}{{\lambda }_1}=\frac{16}{3R}\times \frac{5R}{36}$
$\Rightarrow {\lambda }_2=\frac{80}{108}{\lambda }_1$
Given, ${\lambda }_1=6563\mathring{A}$
$\therefore {\lambda }_2=\frac{80}{180}\times 6563=4861\mathring{A}$