Structure of Atom Part 2

Concept:The energy of a spectral line in the hydrogen atom is inversely proportional to its wavelength.The Rydberg formula gives $1/\lambda = R(1/n_1^2 - 1/n_2^2)$ where $n_2 > n_1$.Thus, the photon energy is proportional to $\Delta = (1/n_1^2 - 1/n_2^2)$.Explanation:Paschen series has $n_1 = 3$.First line (A) is $n_2 = 4 \to 3$: $\Delta_A = 1/9 - 1/16 = 7/144 \approx 0.04861$.Third line (C) is $n_2 = 6 \to 3$: $\Delta_C = 1/9 - 1/36 = 1/12 \approx 0.08333$.Balmer series has $n_1 = 2$.Second line (B) is $n_2 = 4 \to 2$: $\Delta_B = 1/4 - 1/16 = 3/16 = 0.1875$.Brackett series has $n_1 = 4$.Fourth line (D) is $n_2 = 8 \to 4$: $\Delta_D = 1/16 - 1/64 = 3/64 = 0.046875$.Ascending order of energy means smallest $\Delta$ first: $D < A < C < B$.Shortcut:Energy decreases as the lower principal quantum number $n_1$ increases.Balmer ($n_1=2$) gives the highest energy, then Paschen ($n_1=3$), then Brackett ($n_1=4$).Within Paschen, higher $n_2$ gives higher energy, so third line > first line.Thus, ascending order: D (Brackett 4th) < A (Paschen 1st) < C (Paschen 3rd) < B (Balmer 2nd).Answer:Option C: $D < A < C < B$