the ratio of minimum frequency of lyman and balmer series will be?
Ans= 5.4

Dear student.... In Lyman series, the electron returns to the first state while in Balmer series, it returns to the second state (first excited state). ΔE = h(f₂-f₁) ; f ⇒ frequency Hence, smaller the value of ΔE, smaller would be the frequency. Hence, to have minimum frequency for any particular spectral line, we'll have to find the minimum value of ΔE. For that to happen in Lyman series, the electron must jump from n=2 to n=1, and from n=3 to n=2 in Balmer series. Ratio of minimum frequency of Lyman and Balmer series is equal to ratio of ΔE of Lyman to ΔE of Balmer. Minimum value of |ΔE| for Lyman series would be K(1-1/4)=3K/4 and that for Balmer series would be K(1/4-1/9) = 5K/36. Hence, ratio of minimum frequency of Lyman and Balmer series would be (3K/4)÷(5K/36), i.e. ratio = 27/5. Regards