Question

The first line of Balmer series has wavelength 6563 Å. What will be the wavelength of the first member of Lyman series.

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

Answer 1

Answer: (A)

1215.4 Å

Answer 2

$\frac{1}{\lambda_{Balmer}} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} \right\rbrack = \frac{5R}{36}$, $\frac{1}{\lambda_{Lyman}} = R\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack = \frac{3R}{4}$

$\therefore\lambda_{Lyman} = \lambda_{Balmer} \times \frac{5}{27} = 1215.4\mspace{6mu} Å$.