COMEDK · Physics · 28. Atomic Physics
If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength \(\lambda\). When it jumps from the second to the first orbit, the corresponding wavelength of the photon will be
- A \(\dfrac{20 \lambda}{7}\)
- B \(\dfrac{7 \lambda}{20}\)
- C \(\dfrac{5 \lambda}{27}\)
- D \(\dfrac{16 \lambda}{9}\)
Answer & Solution
Correct Answer
(C) \(\dfrac{5 \lambda}{27}\)
Step-by-step Solution
Detailed explanation
The Rydberg formula for the wavelength of a photon emitted during an electronic transition in a hydrogen atom is given by \(\dfrac{1}{\lambda} = R \left( \dfrac{1}{n_f^2} - \dfrac{1}{n_i^2} \right)\), where \(R\) is the Rydberg constant. For the transition from the third orbit…
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