If 917 Å is the lowest wavelength of the Lyman series, find the lowest wavelength of the Balmer series in Å.

Question

G1 Physics · Accepted Answer

The shortest wavelength of a series corresponds to transition from n = ∞ to the lower level. For Lyman series: 1/λ(L) = R So, λ(L) = 1/R = 917 Å For Balmer series: 1/λ(B) = R/4 Therefore, λ(B) = 4/R = 4λ(L) = 4×917 = 3668 Å Final value: 3668