JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2023
- Medium
- Numerical
If \(917 \mathring A\) be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be .......\(\mathring A\).
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(3668\)
Step-by-step Solution
Detailed explanation
For lowest wavelength of Lyman series
\(\frac{1}{\lambda}=R Z^2\left[\frac{1}{1^2}-\frac{1}{\infty^2}\right]=R Z^2\)
For lowest wavelength of Balmer series
\(\frac{1}{\lambda^{\prime}} = RZ ^2\left[\frac{1}{2^2}-\frac{1}{\infty^2}\right]=\frac{ RZ ^2}{4}\)
\(\lambda^{\prime} =\frac{4}{R Z^2}=4 \times 917\)
\(=3668 \mathring A\)
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