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In the line spectra of hydrogen atom, difference between the largest and the shortest wavelengths of the Lyman series is 304Å. The corresponding difference for the Paschan series in Å is:

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    Answered by

    alok kumar singh | Contributor-Level 10

    a month ago

    1/λmin = R [1/1² - 1/∞²] = R [n=∞ → n=1]
    1/λmax = R [1/1² - 1/2²] = 3R/4 [n=2 → n=1]
    ⇒ Δλ ⇒ 4/3R - 1/R ⇒ 1/3R ⇒ 340
    For Paschan ⇒ 1/λmin = R [1/9] [n=∞ → h=3]
    ⇒ 1/λmax = R [1/9 - 1/16] = 7R/144
    ⇒ Δλ = 81/7R

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