Discussion
If the wavelength of the first line of the Balmer series in the hydrogen spectrum is λ, then the wavelength of the first line of the Lyman series is :
Answer: C
The Lyman wavelength is \(\frac{5}{27}\lambda\).
Use the Rydberg formula for the first line of each series:
\[ \begin{aligned} \frac{1}{\lambda_{B}} &= R\left(\frac{1}{4} - \frac{1}{9}\right) \\ &= \frac{5R}{36} \\ \frac{1}{\lambda_{L}} &= R\left(1 - \frac{1}{4}\right) \\ &= \frac{3R}{4} \\ \frac{\lambda_{L}}{\lambda_{B}} &= \frac{5/36}{3/4} \\ &= \frac{5}{27} \end{aligned} \]
The Lyman line is in the ultraviolet, so it must be shorter than the Balmer line; that rules out 27/5 and 32/27. 27/32 is the inverse of the ratio for a different pair of lines.
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