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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 .......\(\mathring A\).

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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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