HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q13 Solution: Whenever a photon is emitted by hydrogen in Balmer series, it…

QPages 384-386: Q13: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q13 Solution: Whenever a photon is emitted by hydrogen in Balmer series, it…

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Students often assume the Balmer emission starts from the ground state or misidentify the final energy level, but the defining characteristic of the Balmer series is that the electron always transitions to the $n=2$ level. Consequently, the subsequent Lyman series photon must originate from this $n=2$ state, making the transition $n=2 \to n=1$ (Lyman-alpha) the only possible next step.

Whenever a photon is emitted by hydrogen in Balmer series, it is followed by another photon in Lyman series. What wavelength does this latter photon correspond to?

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**Answer:** 121.6 nm. **Step-by-Step Explanation:** When a photon is emitted in the Balmer series (n _ 2), the atom ends in n = 2. The subsequent Lyman series photon is emitted when the atom goes from n = 2 to n = 1. This is the Lyman _ line:

\frac{1}{\lambda} = R(1 - 1/4) = 3R/4

\lambda = 4/(3 \times 1.097 \times 10^7) = 1.216 \times 10^{-7}

m = 121.6 nm [citation:3].

QPages 384-386: Q13: HC Verma: Concepts of Physics - Volume 2 Bohr's Model and Physics of the Atom Pages 384-386: Q13 Solution: Whenever a photon is emitted by hydrogen in Balmer series, it…

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