In the line spectra of hydrogen atoms, difference between the largest and the shortest wavelengths of the Lyman series is 304 Å. The corresponding difference for the Paschan series in Å is: ___________. Numerical answer type. Work it out, then open the answer.

Answer: 10553. Lyman: λmax = 4/3R, λmin = 1/R, difference = 1/3R = 304 Å, so 1/R = 912 Å Paschen: λmax = 144/7R, λmin = 9/R Difference = (144/7-9)1/R = 81/7× 912 Answer: ≈ 10553 Å

In the line spectra of hydrogen atoms, difference between the largest… | JEE Main 2020 (4 Sep, Shift 1)

Q46JEE Main 2020 · 4 SEP · SHIFT 1Medium

Hydrogen Spectrum & Spectral Series

In the line spectra of hydrogen atoms, difference between the largest and the shortest wavelengths ${}$ of the Lyman series is 304 $\text{Å}$ . The corresponding difference for the Paschan series in $\text{Å}$ is: ___________.

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

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Answer: 10553

Lyman: $\lambda_{max} = \frac{4}{3R}$ , $\lambda_{min} = \frac{1}{R}$ , difference $= \frac{1}{3R} = 304$ Å, so $\frac{1}{R} = 912$ Å
Paschen: $\lambda_{max} = \frac{144}{7R}$ , $\lambda_{min} = \frac{9}{R}$
Difference $= \left(\frac{144}{7}-9\right)\frac{1}{R} = \frac{81}{7}\times 912$
Answer: $\approx 10553$ Å

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Hydrogen Spectrum & Spectral Series question from JEE Main 2020 (4 Sep, Shift 1)

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