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JEE Main 2024 (27 Jan, Shift 2) · Hydrogen Spectrum & Spectral Series

Hydrogen Spectrum & Spectral Series question from JEE Main 2024 (27 Jan, Shift 2)

Q26JEE Main 2024 · 27 JAN · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series

If Rydberg's constant is RR, the longest wavelength of radiation in Paschen series will be α7R\frac{\alpha}{7R}, where α=\alpha = ________.

Numerical answer type. Work it out, then open the answer.

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Answer: 144

  1. Paschen longest (4→34 \to 3): 1λ=R(19−116)=7R144\frac{1}{\lambda} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}
  2. λ=1447R\lambda = \frac{144}{7R}
  3. Answer: α=144\alpha = 144
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