Question:
In a hydrogen atom the electron makes a transition from \((n+1)^{th}\) level to the \(n^{th}\) level. If \(n \gg 1\), the frequency of radiation emitted is proportional to :
In a hydrogen atom the electron makes a transition from \((n+1)^{th}\) level to the \(n^{th}\) level. If \(n \gg 1\), the frequency of radiation emitted is proportional to :
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Use \(\nu = Rc\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)\) and approximate for large \(n\).
Updated On: Nov 5, 2026
- \(\frac{1}{n}\)
- \(\frac{1}{n^2}\)
- \(\frac{1}{n^3}\)
- \(\frac{1}{n^4}\)
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The Correct Option is C
Solution and Explanation
\(\nu = Rc\left[\frac{1}{n^2}-\frac{1}{(n+1)^2}\right] = Rc\,\frac{(n+1)^2-n^2}{n^2(n+1)^2} = Rc\,\frac{2n+1}{n^2(n+1)^2}\).
For \(n\gg1\): \(2n+1\approx 2n\) and \((n+1)^2\approx n^2\), so \(\nu \approx Rc\,\frac{2n}{n^4} = \frac{2Rc}{n^3}\).
Hence \(\nu \propto \frac{1}{n^3}\).
Answer: (3) \(\frac{1}{n^3}\)
For \(n\gg1\): \(2n+1\approx 2n\) and \((n+1)^2\approx n^2\), so \(\nu \approx Rc\,\frac{2n}{n^4} = \frac{2Rc}{n^3}\).
Hence \(\nu \propto \frac{1}{n^3}\).
Answer: (3) \(\frac{1}{n^3}\)
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