Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The ratio of minimum wavelengths of Lyman and Balmer series will be
PhysicsAtomic PhysicsKCETKCET 2011
Options:
  • A $1.25$
  • B $0.25$
  • C 5
  • D 10
Solution:
2080 Upvotes Verified Answer
The correct answer is: $0.25$
For minimum wavelength,
$\therefore$
$$
\begin{gathered}
\lambda \propto n^{2} \\
\frac{\lambda_{\text {Lyman }}}{\lambda_{\text {Balmer }}}=\left(\frac{1}{2}\right)^{2} \\
\frac{\lambda_{L}}{\lambda_{B}}=\frac{1}{4}=0.25
\end{gathered}
$$

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.