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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?

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**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: 1λ=R(1−1n22)\frac{1}{\lambda} = R\left(1 - \frac{1}{n_2^2}\right). 1/(102.5×10−9)=1.097×107(1−1/n22)1/(102.5 \times 10^{-9}) = 1.097 \times 10^7 (1 - 1/n_2^2). 9.76×106=1.097×107(1−1/n22)9.76 \times 10^6 = 1.097 \times 10^7 (1 - 1/n_2^2). 1−1/n22=0.889⇒1/n22=0.111⇒n22=9⇒n2=31 - 1/n_2^2 = 0.889 \Rightarrow 1/n_2^2 = 0.111 \Rightarrow n_2^2 = 9 \Rightarrow n_2 = 3. So the transition is from n = 3 to n = 1 [citation:3][citation:11].

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