Question

The first line of Balmer series has wavelength $6563\stackrel{{}_{\circ }}{A}$. What will be the wavelength of the first member of Lyman series

• A. $1215.4\stackrel{{}_{\circ }}{A}$
• B. $2500\stackrel{{}_{\circ }}{A}$
• C. $7500\stackrel{{}_{\circ }}{A}$
• D. $600\stackrel{{}_{\circ }}{A}$

Answer 1

The correct option is A $1215.4\stackrel{{}_{\circ }}{A}$

$\frac{1}{{\lambda }_{Balmer}}=R[\frac{1}{2^2}-\frac{1}{3^2}]=\frac{5R}{36},$

$\frac{1}{{\lambda }_{Lyman}}=R[\frac{1}{1^2}-\frac{1}{2^2}]=\frac{3R}{4}$
$\therefore {\lambda }_{Lyman}={\lambda }_{Balmer}\times \frac{5}{27}=1215.4\stackrel{{}_{\circ }}{A}$