What is the exact ratio of the minimum frequency of the Lyman series to that of the Balmer series? β Atomic Structure Chemistry Question
Question
What is the exact ratio of the minimum frequency of the Lyman series to that of the Balmer series?
Answer: D
π‘ Solution & Explanation
Minimum frequency corresponds to the longest wavelength transition. Lyman ($n=2 \to 1$) $\propto R(1 - 1/4) = 3R/4$. Balmer ($n=3 \to 2$) $\propto R(1/4 - 1/9) = 5R/36$. Ratio = $(3/4) / (5/36) = 27/5 = 5.4$.
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