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JEE Mains · Physics · STD 12 - 12. atoms

If \(\lambda_{1}\) and \(\lambda_{2}\) are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of \(\lambda_{1}: \lambda_{2}\) is

  1. A \(1: 9\)
  2. B \(7: 108\)
  3. C \(7: 135\)
  4. D \(1: 3\)
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Answer & Solution

Correct Answer

(C) \(7: 135\)

Step-by-step Solution

Detailed explanation

\(\frac{1}{\lambda_{1}}= R \left[\frac{1}{1^{2}}-\frac{1}{4^{2}}\right]\) \(\frac{1}{\lambda_{2}}= R \left[\frac{1}{3^{2}}-\frac{1}{4^{2}}\right]\) \(\frac{\lambda_{1}}{\lambda_{2}}=\frac{\left[\frac{1}{9}-\frac{1}{16}\right]}{\left[1-\frac{1}{16}\right]}=\frac{7}{9 \times 15}\)…
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