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JEE Mains · Physics · STD 12 - 12. atoms

In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is \(304\,\mathring {A}.\) The corresponding difference for the Paschan series in \(\mathring {A}\) is

  1. A \(10553\)
  2. B \(10560\)
  3. C \(10555\)
  4. D \(10165\)
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Answer & Solution

Correct Answer

(A) \(10553\)

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Detailed explanation

\(\lambda=\frac{ c }{\left(\frac{1}{ n _{1}^{2}}-\frac{1}{ n _{2}^{2}}\right)}\) for lyman series \(\lambda_{1}=\frac{ c }{\frac{1}{1^{2}}-\frac{1}{\infty^{2}}}= c ( n =\infty\) to \(n =1)\) \(\lambda_{2}=\frac{c}{\frac{1}{1^{2}}-\frac{1}{2^{2}}}=\frac{4 c}{3}(n=2\) to \(n=1)\)…
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