TS EAMCET · Physics · Atomic Physics
The shortest wavelength in Lyman series is \(912 Å\). Then, the longest wavelength in the series must be
- A \(9120 Å ̊\)
- B \(1824 Å\)
- C \(1216 Å\)
- D \(2432 Å\)
Answer & Solution
Correct Answer
(C) \(1216 Å\)
Step-by-step Solution
Detailed explanation
For \(\mathrm{H}\)-atom (Lyman series formula) \(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n_2^2}\right)\) \(\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right) \Rightarrow \frac{1}{912 Å}=R\left(\frac{1}{1}-\frac{1}{\infty}\right)\)…
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