TS EAMCET · Physics · Atomic Physics
If the first line in the Lyman series has wavelength \(\lambda\), then the first line in Balmer series has the wavelength
- A \(\frac{27}{5} \lambda\)
- B \(\frac{32}{27} \lambda\)
- C \(\frac{28}{21} \lambda\)
- D \(\frac{15}{4} \lambda\)
Answer & Solution
Correct Answer
(A) \(\frac{27}{5} \lambda\)
Step-by-step Solution
Detailed explanation
\(\lambda=\frac{4}{3 R} \Rightarrow R=\frac{4}{3 \lambda}\)Wavelength for Lyman series is calculated as \(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)\) For first line of Lyman series, \(n=2\)…
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