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

  1. A \(3667\)
  2. B \(3365\)
  3. C \(3668\)
  4. D \(3658\)
Verified Solution

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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