TS EAMCET · Physics · Atomic Physics
The ratio of wavelengths of second line in Balmer series and the first line in Lyman series of hydrogen atom is
- A \(2: 1\)
- B \(9: 4\)
- C \(4: 1\)
- D \(3: 2\)
Answer & Solution
Correct Answer
(C) \(4: 1\)
Step-by-step Solution
Detailed explanation
\( \frac{1}{\lambda_{Balmer, 2}} = R \left( \frac{1}{2^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{4} - \frac{1}{16} \right) = R \left( \frac{4-1}{16} \right) = \frac{3R}{16} \implies \lambda_{Balmer, 2} = \frac{16}{3R} \)…
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