TS EAMCET · Physics · Atomic Physics
If the first line of Lyman series has a wavelength \(1215.4 Å\), the first line of Balmer series is approximately
- A \(4864 Å\)
- B \(1025.5 Å\)
- C \(6563 Å\)
- D \(6400 Å\)
Answer & Solution
Correct Answer
(C) \(6563 Å\)
Step-by-step Solution
Detailed explanation
From hydrogen spectrum, when electron transist from \(n_2\) orbit to \(n_1\) orbit, the emitted wavelength is given by the formula \(\frac{1}{\lambda_L}=R Z^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)\) Here, \(R=\) Rydberg constant. For Lyman series, \(n_1=1, n_2=2\) For…
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