Q. If $\lambda_{\max}$ is $6563\ \mathring{\mathrm{A}}$, then wavelength of second line for Balmer series will be
A) $\lambda = \frac{16}{3R}$
B) $\lambda = \frac{36}{5R}$
C) $\lambda = \frac{4}{3R}$
D) None of the above
Explanation
(a) For Balmer series $\frac{1}{\lambda} = R\left( \frac{1}{2^{2}} - \frac{1}{n^{2}} \right)$ where $n = 3,\ 4,\ 5$
For second line $n = 4$
So $\frac{1}{\lambda} = R\left( \frac{1}{2^{2}} - \frac{1}{4^{2}} \right) = \frac{3}{16}R \Rightarrow \lambda = \frac{16}{3R}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.