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) $\frac{\lambda_{B}}{\lambda_{L}} = \frac{\left( \frac{1}{1^{2}} - \frac{1}{2\ ^{2}} \right)}{\left( \frac{1}{2^{2}} - \frac{1}{3^{2}} \right)} = \frac{3/4}{5/36} = \frac{27}{5}$
$\lambda_{L} = \frac{5}{27}\lambda_{B} = \frac{5}{27} \times 6563 = 1215.4\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.