Hydrogen Spectrum - Balmer Series

Question

When the electron in a hydrogen atom drops from the third energy level to the second energy level, a visible red photon is emitted. What is the wavelength of this spectral line, taking the Rydberg constant as 1.097 × 10^7 per metre?

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Spectral wavelengths in hydrogen follow the Rydberg relation $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}ight)$, and a transition ending at $n_1 = 2$ belongs to the visible Balmer series. For the $n_2 = 3 	o n_1 = 2$ jump (the H-alpha line), $\dfrac{1}{\lambda} = R\left(\dfrac{1}{4} - \dfrac{1}{9}ight) = R\left(\dfrac{9 - 4}{36}ight) = \dfrac{5R}{36}$. Hence $\lambda = \dfrac{36}{5R} = \dfrac{36}{5 	imes 1.097	imes10^7} \approx 6.56	imes10^{-7}$ m $= 656$ nm. The value 486 nm is the H-beta line ($4 	o 2$), not the requested transition. The value 121 nm is the Lyman-alpha line ($2 	o 1$), which lies in the ultraviolet, not the visible. The value 365 nm is the Balmer series limit ($\infty 	o 2$). The Balmer series, lying in the visible region, was historically the first hydrogen series discovered and gave early support to the Bohr model. Because the lower level is fixed at $n_1 = 2$, the longest-wavelength line of the series arises from the smallest energy gap, namely the $3 	o 2$ transition. This matches the NCERT Balmer-series description. A plausibility check confirms 656 nm falls in the red part of the visible spectrum, exactly where H-alpha is observed experimentally.

Hydrogen Spectrum - Balmer Series

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