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JEE Mains · Physics · STD 12 - 12. atoms

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

  1. A \(20/7\)
  2. B \(7/5\)
  3. C \(9/7\)
  4. D \(27/5\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(20/7\)

Step-by-step Solution

Detailed explanation

\(\frac{1}{\lambda } = R\left( {\frac{1}{{n_f^2}} - \frac{1}{{n_i^2}}} \right);\) \(\frac{1}{{{\lambda _1}}} = R\left( {\frac{1}{{{3^2}}} - \frac{1}{{{4^2}}}} \right)\) \(\frac{1}{{{\lambda _1}}} = R\left( {\frac{7}{{9 \times 16}}} \right);\)…
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