JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2023
- Medium
- Numerical
The ratio of wavelength of spectral lines \(H_\alpha\) and \(H _\beta\) in the Balmer series is \(\frac{x}{20}\). The value of \(x\) is \(...............\)
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(27\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\lambda}= R \left[\frac{1}{ n _1{ }^2}-\frac{1}{ n _2{ }^2}\right]\) for \(H\)-atom
For balmer series, \(n _1=2\)
\(\frac{1}{\lambda}= R \left[\frac{1}{4}-\frac{1}{ n _2^2}\right]\)
For \(H _{ i }, n _2=3\)
\(H _\beta, n _2=4\)
\(\frac{1}{\lambda_{ H _{ \alpha }}}= R \left[\frac{1}{4}-\frac{1}{9}\right]=\frac{5 R }{36}\)
\(\frac{1}{\lambda_{ H _{ \beta }}}= R \left[\frac{1}{4}-\frac{1}{16}\right]=\frac{3 R }{16}\)
\(\frac{\frac{1}{\lambda_{H_\alpha}}}{\frac{1}{\lambda_{H_\beta}}}=\frac{\frac{5 R }{36}}{\frac{3 R }{16}}\)
\(\frac{\lambda_{H_{\alpha}}}{\lambda_{ H _{\beta}}}=\frac{27}{20}=\frac{ x }{20}\)
\(x=27\)
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