The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearl $y ____$ nm.
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The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearl y ____ nm.

G1 Physics · Accepted Answer

For the Lyman-series limit, 1/λ(min,Lyman) = R. Given, λ(min,Lyman) = 91 nm. Therefore, R = 1/(91 nm). For the largest Balmer wavelength, the transition is n = 3 → n = 2. Hence, 1/λ(B,max) = R[1/2² - 1/3²] = R(1/4 - 1/9) = 5R/36. Thus, λ(B,max) = 36/(5R) = 36×91/5 = 655.2 nm. For the largest Paschen wavelength, the transition is n = 4 → n = 3. Therefore, 1/λ(P,max) = R[1/3² - 1/4²] = R(1/9 - 1/16) = 7R/144. Hence, λ(P,max) = 144/(7R) = 144×91/7 = 1872 nm. Therefore, Δλ = λ(P,max) - λ(B,max) = 1…