The ratio of minimum wavelengths of Lyman and Balmer series will be

Step-by-step solution

The series end of Lyman series corresponds to transition from n i = ∞ to n f = 1 , corresponding to the wavelength 1 λ m i n L = R 1 1 - 1 ∞ = R ⇒ ( λ m i n ) L = 1 R = 912 Å ...(i) For last line of Balmer series 1 ( λ m i n ) B = R 1 2 2 - 1 ∞ 2 = R 4 ⇒ ( λ m i n ) B = 4 R = 3 6 4 8 Å ...(ii) Dividing Eq.(i) by Eq. (ii) .we get ( λ m i n ) L ( λ m i n ) B = 0.25