Lyman vs Balmer Series — Wavelength Ratio Trick in 60 Sec! 🔥
❓ Question For a hydrogen atom , the ratio of the largest wavelength of the Lyman series to that of the Balmer series is equal to ? 🖼️ Question Image ✍️ Short Solution This is a pure concept + formula-based JEE question . The key is to remember: Largest wavelength ⇔ smallest energy difference So we must identify the closest transition in each series. 🔹 Step 1 — Rydberg formula (foundation 💯)** For hydrogen spectrum: 1 λ = R ( 1 n 1 2 − 1 n 2 2 ) , n 2 > n 1 Where: n 1 n_1 = lower energy level (series identifier) n 2 n_2 = higher energy level 📌 Larger wavelength ⇒ smaller value of 1 λ \frac{1}{\lambda} . 🔹 Step 2 — Largest wavelength of Lyman series Lyman series: n 1 = 1 Largest wavelength occurs for smallest jump : n 2 = 2 → 1 So: 1 λ L = R ( 1 − 1 4 ) = 3 R 4 Thus: λ L = 4 3 R 🔹 Step 3 — Largest wavelength of Balmer series Balmer series: n 1 = 2 Largest wavelength ⇒ nearest upper level : n 2 = 3 → 2 So: 1 λ B = R ( 1 4 − 1 9 ...