Question

The wavelength of first line of Balmer series is 6563 Å. The wavelength of first line of Lyman series will be–

• A. 1215.4 Å
• B. 2500 Å
• C. 7500 Å
• D. 600 Å

Answer 1

Answer: (A)

1215.4 Å

Answer 2

$\frac{1}{\lambda} = R\left( \frac{1}{n_{1}^{2}}–\frac{1}{n_{2}^{2}} \right)$

$\frac{1}{\lambda_{1}} = R\left\lbrack \frac{1}{2_{}^{2}}–\frac{1}{3_{}^{2}} \right\rbrack$

= $R\left( \frac{1}{4}–\frac{1}{9} \right)$= $\frac{5R}{36}$

$\frac{1}{\lambda_{2}} = R\left\lbrack \frac{1}{1}–\frac{1}{4} \right\rbrack$

= $R\left( \frac{3}{4} \right)$= $\frac{3}{4}$R

\ $\frac{\lambda_{2}}{\lambda_{1}} = \frac{5/36}{3/4}$ = $\frac{5}{36}$× $\frac{4}{3}$ = $\frac{5}{27}$

l₂ = $\frac{5}{27}\lambda_{1}$

l₂ = $\frac{5}{27} \times 6563Å$= 1215.4Å