JEE Main 2026 Physics Solution | 22 Jan Shift 2 | Atoms & Hydrogen Spectrum (Q.34) скачать в хорошем качестве

JEE Main 2026 Physics Solution | 22 Jan Shift 2 | Atoms & Hydrogen Spectrum (Q.34)

Welcome back to Physics Unlocked ⚛🔓! In this video, we are solving Question 34 from the JEE Main 2026 Physics paper (22nd January, Shift 2). This problem comes from the 'Atoms' chapter and tests your understanding of the Hydrogen emission spectrum and Rydberg's formula. We break down how to interpret "smallest" and "largest" wavelengths in terms of electron transitions, calculate the Rydberg constant using the Lyman series data, and then find the precise wavelength difference between the Paschen and Balmer series. 📝 Question: The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly: Options: 1. 1784 nm 2. 1875 nm 3. 1217 nm 4. 1550 nm ⏱️ Timestamps: 0:00 - Question Reading & Analysis 0:23 - Smallest Wavelength in Lyman Series (Finding R) 0:59 - Largest Wavelength in Paschen Series 1:17 - Largest Wavelength in Balmer Series 1:36 - Calculating the Difference (Δλ) 2:05 - Final Answer & Conclusion If you found this step-by-step derivation helpful, make sure to Like, Share, and Subscribe for more detailed JEE Main and Advanced physics problem-solving! #JEEMain2026 #PhysicsUnlocked #JEEPhysics #Atoms #HydrogenSpectrum #RydbergFormula #JEEPreparation #IITJEE #Class12Physics