KCET · Physics · Atomic Physics
The ratio of minimum wavelengths of Lyman and Balmer series will be
- A \(1.25\)
- B \(0.25\)
- C 5
- D 10
Answer & Solution
Correct Answer
(B) \(0.25\)
Step-by-step Solution
Detailed explanation
For minimum wavelength,
\(\therefore\)
\[
\begin{gathered}
\lambda \propto n^{2} \\
\frac{\lambda_{\text {Lyman }}}{\lambda_{\text {Balmer }}}=\left(\frac{1}{2}\right)^{2} \\
\frac{\lambda_{L}}{\lambda_{B}}=\frac{1}{4}=0.25
\end{gathered}
\]
\(\therefore\)
\[
\begin{gathered}
\lambda \propto n^{2} \\
\frac{\lambda_{\text {Lyman }}}{\lambda_{\text {Balmer }}}=\left(\frac{1}{2}\right)^{2} \\
\frac{\lambda_{L}}{\lambda_{B}}=\frac{1}{4}=0.25
\end{gathered}
\]
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