JEE Mains · Physics · STD 12 - 12. atoms
If \(917 \mathring A\) be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be .......\(\mathring A\).
- A \(3667\)
- B \(3365\)
- C \(3668\)
- D \(3658\)
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 \(\frac{1}{\lambda^{\prime}} = RZ ^2\left[\frac{1}{2^2}-\frac{1}{\infty^2}\right]=\frac{ RZ ^2}{4}\)…
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