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TS EAMCET · Physics · Atomic Physics

In hydrogen spectrum, if the shortest wavelength in Balmer series is \(\lambda\), the shortest wavelength in Brackett series is

  1. A \(\lambda\)
  2. B \(\lambda / 2\)
  3. C \(4 \lambda\)
  4. D \(9 \lambda\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4 \lambda\)

Step-by-step Solution

Detailed explanation

For Balmer series, \(\frac{1}{\lambda_{\text {Balmer }}}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)\), where \(n=3,4,5 \ldots\) For shortest wavelength, \(n=\infty\)…
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