JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2019
- Hard
- MCQ
The electron in a hydrogen atom first jumps from the third excited state to the second excited state and subsequently to the first excited state. The ratio of the respective wavelengths, \(\lambda _1/\lambda _2\), of the photons emitted in this process is
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(20/7\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\lambda } = R\left( {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right);\) \(\frac{1}{{{\lambda _1}}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right)\)
\(\frac{1}{{{\lambda _1}}} = R\left( {\frac{7}{{9 \times 16}}} \right);\) \(\frac{1}{{{\lambda _2}}} = R\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)\)
\(=R\left(\frac{5}{4 \times 9}\right)\)
\(\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{5}{36}}{\frac{7}{9 \times 16}}=\frac{20}{7}\)
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