Q. The shortest wavelength in the Lyman series of hydrogen spectrum is $912\mathring{\mathrm{A}}$ corresponding to a photon energy of $13.6\ eV$. The shortest wavelength in the Balmer series is about
A) $3648\ \mathring{\mathrm{A}}$
B) $8208\ \mathring{\mathrm{A}}$
C) $1228\ \mathring{\mathrm{A}}$
D) $6566\ \mathring{\mathrm{A}}$
Explanation
(a) In Lyman series $\left( \lambda_{\min} \right)_{L} = \frac{1}{R}$ and $\left( \lambda_{\min} \right)_{B} = \frac{4}{R}$
$\Rightarrow \left( \lambda_{\min} \right)_{B} = 4 \times \left( \lambda_{\min} \right)_{L} = 4 \times 912 = 3648\ \mathring{\mathrm{A}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.