Question:
In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is 304 \(\mathrm{\AA}\). The corresponding difference for the Paschen series (in \(\mathrm{\AA}\)) is ____.
In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is 304 \(\mathrm{\AA}\). The corresponding difference for the Paschen series (in \(\mathrm{\AA}\)) is ____.
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Use \(\frac{1}{\lambda} = R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\). Lyman gives \(\frac{1}{3R} = 304\ \mathrm{\AA}\), so \(\frac{1}{R} = 912\ \mathrm{\AA}\); then compute the Paschen difference.
Updated On: Nov 5, 2026
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Correct Answer: 10553.14
Solution and Explanation
Lyman series (\(n_1 = 1\)): longest wavelength (n=2): \(\lambda_{max} = \frac{4}{3R}\); shortest (n = infinity): \(\lambda_{min} = \frac{1}{R}\).
Difference: \(\frac{4}{3R} - \frac{1}{R} = \frac{1}{3R} = 304\ \mathrm{\AA}\) so \(\frac{1}{R} = 912\ \mathrm{\AA}\).
Paschen series (\(n_1 = 3\)): longest (n=4): \(\frac{1}{\lambda} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}\) so \(\lambda_{max} = \frac{144}{7R}\); shortest (n = infinity): \(\lambda_{min} = \frac{9}{R}\).
Difference: \(\frac{1}{R}\left(\frac{144}{7} - 9\right) = \frac{81}{7R} = \frac{81}{7}\times 912 = 10553.14\ \mathrm{\AA}\).
Answer: 10553.14
Difference: \(\frac{4}{3R} - \frac{1}{R} = \frac{1}{3R} = 304\ \mathrm{\AA}\) so \(\frac{1}{R} = 912\ \mathrm{\AA}\).
Paschen series (\(n_1 = 3\)): longest (n=4): \(\frac{1}{\lambda} = R\left(\frac{1}{9}-\frac{1}{16}\right) = \frac{7R}{144}\) so \(\lambda_{max} = \frac{144}{7R}\); shortest (n = infinity): \(\lambda_{min} = \frac{9}{R}\).
Difference: \(\frac{1}{R}\left(\frac{144}{7} - 9\right) = \frac{81}{7R} = \frac{81}{7}\times 912 = 10553.14\ \mathrm{\AA}\).
Answer: 10553.14
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