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JEE Main 2023 (10 Apr, Shift 2) · Hydrogen Spectrum & Spectral Series

Hydrogen Spectrum & Spectral Series question from JEE Main 2023 (10 Apr, Shift 2)

Q36JEE Main 2023 · 10 APR · SHIFT 2Medium

Hydrogen Spectrum & Spectral Series

If 917 A˚\text{Å} be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be ___________ A˚\text{Å}.

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

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Answer: 3668

  1. Lowest wavelength = series limit: λ=n12R\lambda = \frac{n_1^2}{R}
  2. Lyman: 1R=917\frac{1}{R} = 917 Å
  3. Balmer: λ=4R=4×917=3668\lambda = \frac{4}{R} = 4\times 917 = 3668 Å
  4. Answer: 3668 Å
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