Atomic Structure: Practice Problem & Solution
If an electron in a hydrogen atom jumps from the 3rd orbit to the 2nd orbit, it emits a photon of wavelength $\lambda$. When it jumps from the 4th orbit to the 3rd orbit, the corresponding wavelength of the photon will be: (2016 - II)
Solution Explained:
To solve this problem, we apply the core principles of Atomic Structure. Understanding the underlying formula is key to arriving at the correct answer below:
For the first transition, $1/\lambda = R(\frac{1}{2^2} - \frac{1}{3^2}) = \frac{5R}{36}$. For the second transition, $1/\lambda' = R(\frac{1}{3^2} - \frac{1}{4^2}) = \frac{7R}{144}$. Dividing the two gives $\lambda' = \lambda \times \frac{144/7}{36/5} = \frac{20}{7}\lambda$.
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