Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

❓ Concept 🎬 Largest Wavelength: Lyman vs Balmer — Hydrogen Spectrum in 59 Sec Hydrogen atom mein itni saari spectral lines hoti hain… par exam mein sawal hota hai 👇 👉 Sabse badi wavelength kis series se aati hai? 🌈🤔 Is concept ko samajh liya, toh ratio-based spectrum questions instant ho jaate hain ⚡ 🖼️ Concept Image ✍️ Short Explanation Yeh question ratta nahi , energy-gap logic ka test hota hai. Bas yeh yaad rakho: 👉 Badi wavelength = chhota energy gap . 🔹 Step 1 — Hydrogen Spectral Series Basics Har spectral series ka final energy level fixed hota hai: Lyman series → final level n = 1 n = 1 Balmer series → final level n = 2 n = 2 📌 Transitions higher level → final level ke beech hote hain. 🔹 Step 2 — Energy–Wavelength Relation (FOUNDATION 💯)** Photon energy: E = h c λ​ Iska matlab: λ ∝ 1 Δ E​ 📌 Largest wavelength ⇔ smallest energy difference 🔹 Step 3 — Smallest Energy Gap in Any Series Har spectral series mein: Sabse chhota...

❓ 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 ...