COMEDK · Physics · 28. Atomic Physics
If an electron from a hydrogen like atom with \(Z=2\) jumps from the \(4^{\text {th }}\) orbit to the \(2^{\text {nd }}\) orbit, it emits a photon of wavelength \(\lambda\). What is the wavelength of the photon emitted if an electron jumps from the \(5^{\text {th }}\) orbit to the \(4^{\text {th }}\) orbit?
- A \(\dfrac{25}{3} \lambda\)
- B \(\dfrac{9}{25} \lambda\)
- C \(\dfrac{16}{25} \lambda\)
- D \(\dfrac{9}{16} \lambda\)
Answer & Solution
Correct Answer
(A) \(\dfrac{25}{3} \lambda\)
Step-by-step Solution
Detailed explanation
The wavelength \(\lambda\) of a photon emitted during a transition from orbit \(n_2\) to \(n_1\) in a hydrogen-like atom with atomic number \(Z\) is given by the Rydberg formula: \(\dfrac{1}{\lambda} = R Z^2 \left( \dfrac{1}{n_1^2} - \dfrac{1}{n_2^2} \right)\) For the first…
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