Question

The wavelength of the first line of Lyman series is 1215 Å, the wavelength of first line of Balmer series will be

• A. 4545 Å
• B. 5295 Å
• C. 6561 Å
• D. 6750 Å

Answer 1

Answer: (C)

6561 Å

Answer 2

Here $\lambda_{L} = 1215Å$

For the first line of Lyman series

$\frac{1}{\lambda_{L}} = R\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack = R\left\lbrack 1 - \frac{1}{4} \right\rbrack = \frac{3R}{4}$

$\therefore\frac{1}{\lambda_{L}} = \frac{3R}{4}$

For first line of Balmer series

$\frac{1}{\lambda_{B}} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} \right\rbrack = R\left\lbrack \frac{1}{4} - \frac{1}{9} \right\rbrack \Rightarrow \frac{1}{\lambda_{B}} = R\left\lbrack \frac{5}{36} \right\rbrack$

$\therefore\frac{1}{\lambda_{B}} = \frac{36}{5R}$

From (i) and (ii)

$\frac{\lambda_{B}}{\lambda_{L}} = \frac{36/5R}{4/3R} = \frac{36 \times 3}{4 \times 5}$

$\therefore\lambda_{B} = \frac{108}{20} \times \lambda_{L} = \frac{108}{20} \times 1215 = 6561Å$