Q. The first line of Balmer series has wavelength $6563\ \mathring{\mathrm{A}}$. What will be the wavelength of the first member of Lyman series
A) $1215.4\ \mathring{\mathrm{A}}$
B) $2500\ \mathring{\mathrm{A}}$
C) $7500\ \mathring{\mathrm{A}}$
D) $600\ \mathring{\mathrm{A}}$
Explanation
(a) $\frac{1}{\lambda_{Balmer}} = R\left\lbrack \frac{1}{2^{2}} - \frac{1}{3^{2}} \right\rbrack = \frac{5R}{36},\frac{1}{\lambda_{Lyman}} = R\left\lbrack \frac{1}{1^{2}} - \frac{1}{2^{2}} \right\rbrack = \frac{3R}{4}$
$\therefore\lambda_{Lyman} = \lambda_{Balmer} \times \frac{5}{27} = 1215.4\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.