JEE Mains · Physics · STD 12 - 12. atoms
The recoil speed of a hydrogen atom after it emits a photon in going from \(n =5\) state to \(n =1\) state will be ..... \(m/s\)
- A \(4.17\)
- B \(2.19\)
- C \(3.25\)
- D \(4.34\)
Answer & Solution
Correct Answer
(A) \(4.17\)
Step-by-step Solution
Detailed explanation
\((\Delta E )\) Releases when photon going from \(n =5\) to \(n =\Delta E =(13.6-0.54)\, eV =13.06\, eV\) \(P _{ i }= P _{ f }\) (By linear momentum conservation) \(0=\frac{ h }{\lambda}- Mv = V _{ Recoil }=\frac{ h }{\lambda M }.....(i)\) And…
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