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JEE Main 2021 (26 Feb, Shift 1) · Hydrogen Spectrum & Spectral Series

Hydrogen Spectrum & Spectral Series question from JEE Main 2021 (26 Feb, Shift 1)

Q44JEE Main 2021 · 26 FEB · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series

If λ\lambda1 and λ\lambda2 are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of λ\lambda1: λ\lambda2 is:

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Answer: (a)

  1. Lyman third member (4→14 \to 1): 1λ1=R(1−116)=15R16\frac{1}{\lambda_1} = R\left(1-\frac{1}{16}\right) = \frac{15R}{16}
  2. Paschen first member (4→34 \to 3): 1λ2=R(19−116)=7R144\frac{1}{\lambda_2} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}
  3. λ1λ2=7/14415/16=1122160=7135\frac{\lambda_1}{\lambda_2} = \frac{7/144}{15/16} = \frac{112}{2160} = \frac{7}{135}
  4. So the answer is (a) 7 : 135.
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