Discussion
The ratio of the longest wavelength of the Lyman series to that of Ballmer series of the hydrogen spectrum is
Answer: C
The ratio is 5:27.
The longest wavelength in each series comes from the smallest energy jump into its lowest level.
\[ \begin{aligned} \text{Lyman } (2 \to 1): \frac{1}{\lambda_{L}} &= R\left(1 - \tfrac{1}{4}\right) = \tfrac{3R}{4} \\ \text{Balmer } (3 \to 2): \frac{1}{\lambda_{B}} &= R\left(\tfrac{1}{4} - \tfrac{1}{9}\right) = \tfrac{5R}{36} \end{aligned} \]
\[ \begin{aligned} \frac{\lambda_{L}}{\lambda_{B}} &= \frac{5R/36}{3R/4} \\ &= \frac{5}{27} \end{aligned} \]
Inverting the ratio by mistake gives 27 : 5. The options with 29 do not come from the Rydberg formula.
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