Question

If the first line of Lyman series has a wavelength $1215.4\mathring{A}$, the first line of Balmer series is approximately

• A. $4864\mathring{A}$
• B. $1025.5\mathring{A}$
• C. $6563\mathring{A}$
• D. $6400\mathring{A}$

Answer 1

The correct option is C $6563\mathring{A}$

Wavelength of first line in Lyman series ${\lambda }_L=\frac{3}{4R}$
$\therefore$ $\frac{4}{3R}=1215.4A^o$
We get $R=\frac{4}{3\times 1215.4}$
Wavelength of first line in Balmer series ${\lambda }_B=\frac{36}{5R}$
$⟹{\lambda }_B=\frac{36\times 3\times 1215.4}{5\times 4}=6563A^o$