WBJEE · Physics · Atomic Physics
The wavelength of second Balmer line in hydrogen &spectrum is 600 nm The wavelength for its third line in Lyman series is
- A \(800 \mathrm{nm}\)
- B \(600 \mathrm{nm}\)
- C \(400 \mathrm{nm}\)
- D None of the above
Answer & Solution
Correct Answer
(D) None of the above
Step-by-step Solution
Detailed explanation
\(\frac{1}{\lambda}=R\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\) For Balmer series, \(\begin{array}{l} \therefore n_{1}=2, n_{2}=3,4,5 \ldots \\ \frac{1}{\lambda_{B}}=R\left[\frac{1}{2^{2}}-\frac{1}{4^{2}}\right] \end{array}\)…
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