Q. The first line in the Lyman series has wavelength $\lambda$. The wavelength of the first line in Balmer series is
A) $\frac{2}{9}\lambda$
B) $\frac{9}{2}\lambda$
C) $\frac{5}{27}\lambda$
D) $\frac{27}{5}\lambda$
Explanation
(d) For first line in Lyman series $\lambda_{L_{1}} = \frac{4}{3R}$ …(i)
For first line in Balmer series $\lambda_{B_{1}} = \frac{36}{5R}$ …(ii)
From equations (i) and (ii)
$\frac{\lambda_{B_{1}}}{\lambda_{L_{1}}} = \frac{27}{5} \Rightarrow \lambda_{B_{1}} = \frac{27}{5}\lambda_{L_{1}} \Rightarrow \lambda_{B_{1}} = \frac{27}{5}\lambda$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.