Question

The wavelength of the first spectral line in the Balmer series of hydrogen atom is 6561 A. The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is :

(in $A^o$)

• A. 1215
• B. 1640
• C. 2430
• D. 4687

Answer 1

Answer: (A)

1215

$\begin{array}{c}\text{Balmer series}=\text{transiston take place}\\ \text{from}n=2\text{to}n=3,4\text{and so on}\end{array}$

$\begin{array}{c}\text{For first line of balmer for the hydrogen}\\ n_1=2,\text{to}{n}_2=3,z=1\end{array}$

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

$\frac{1}{\lambda }=R{z}^2(\frac{1}{4}-\frac{1}{9})$

$R=\frac{36}{6561\times 5}\dot{A}-\text{(i)}$

$\begin{array}{c}\text{for second line of balmer series for Helium}\\ \begin{array}{cc}{n}_1& =2\text{to}{n}_2=4\\ & 2=2\\ \therefore \frac{1}{\lambda }& =R{z}^2(\frac{1}{4}-\frac{1}{16})\end{array}\end{array}$

$\begin{array}{c}\frac{1}{\lambda }=\frac{R\times 3}{4}\\ \text{from (i) value of}R\\ \lambda =\frac{4\times 6561\times 5}{3\times 36}\\ \lambda =1215A\\ \text{option. (A)}\end{array}$