JEE Mains · Physics · STD 12 - 12. atoms
The ratio of the shortest wavelength of Balmer series to the shortest wavelength of Lyman series for hydrogen atom is _______.
- A \(4: 1\)
- B \(1: 2\)
- C \(1: 4\)
- D \(2: 1\)
Answer & Solution
Correct Answer
(A) \(4: 1\)
Step-by-step Solution
Detailed explanation
\(\frac{1}{\lambda}=\mathrm{Rz}^2\left(\frac{1}{\mathrm{n}_1^2}-\frac{1}{\mathrm{n}_2^2}\right)\) \(\frac{1}{\lambda_{\mathrm{L}}}=\mathrm{Rz}^2\left(\frac{1}{1^2}\right)\) \(\frac{1}{\lambda_{\mathrm{B}}}=\mathrm{Rz}^2\left(\frac{1}{2^2}\right)\)…
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