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QPages 384-386: Q7: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q7 Solution: A hydrogen atom emits ultraviolet radiation of wavelength 102.5 …
⚠️ Common Exam Traps & Conceptual Pitfalls
Students often fail to deduce that ultraviolet emission in hydrogen corresponds exclusively to the Lyman series, where the final energy level is $n_f = 1$. If the final state is not correctly identified as the ground state, the Rydberg equation cannot be simplified to solve for the initial quantum number $n_i$.
A hydrogen atom emits ultraviolet radiation of wavelength 102.5 nm. What are the quantum numbers of the states involved in the transition?
Step-by-Step Verified Solution
**Answer:** The quantum numbers are n_ = 1 and n_ = 3.
**Step-by-Step Explanation:**
The wavelength 102.5 nm lies in the ultraviolet region, so it belongs to the Lyman series (n_ = 1).
Using Rydberg's formula: .
.
.
.
So the transition is from n = 3 to n = 1 [citation:3][citation:11].
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