JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2023
- Hard
- Numerical
As per given figure \(A , B\) and \(C\) are the first, second and third excited energy level of hydrogen atom respectively. If the ratio of the two wavelengths \(\left(\right.\) i.e. \(\left.\frac{\lambda_1}{\lambda_2}\right)\) is \(\frac{7}{4 n}\), then the value of \(n\) will be

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Answer & Solution
Correct Answer
(A) \(5\)
Step-by-step Solution
Detailed explanation
As \(\frac{1}{\lambda}=R Z^2\left[\frac{1}{ n _1^2}-\frac{1}{ n _2^2}\right]\)
\(\frac{1}{\lambda_1}=R(1)^2\left[\frac{1}{(2)^2}-\frac{1}{(3)^2}\right]=R\left(\frac{5}{36}\right)\)
\(\frac{1}{\lambda_2}=R(1)^2\left[\frac{1}{(3)^2}-\frac{1}{(4)^2}\right]=R\left(\frac{7}{144}\right)\)
\((ii) \div (i)\) gives
\(\frac{\lambda_1}{\lambda_2}=\frac{7 / 144}{5 / 36}=\frac{7}{20}=\frac{7}{4 \times 5}\)
\(\therefore n =5\)
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