Q. Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelengths λ₁ : λ₂ emitted in the two cases is
A) 20/7
B) 7/5
C) 27/20
D) 27/5
Explanation
[Solution from source] Third excited state n = 4 → second excited n = 3: 1/λ₁ = R(1/9 − 1/16) = 7R/144. Second excited n = 3 → first excited n = 2: 1/λ₂ = R(1/4 − 1/9) = 5R/36 = 20R/144. Ratio λ₁/λ₂ = 20/7. Taking excited state numbers as n values directly gives a wrong ratio. Answer (1).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.