JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2021
- Hard
- MCQ
The recoil speed of a hydrogen atom after it emits a photon in going from \(n =5\) state to \(n =1\) state will be ..... \(m/s\)
Pick an option to check your answer.
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 \(\Delta E =\frac{ hc }{\lambda}=\frac{ hc }{\lambda M } \times M \Rightarrow McV _{ Recoil }\)
\(V _{ Recoil }=\frac{\Delta E }{ Mc }=\frac{13.06 \times 1.6 \times 10^{-19}}{1.67 \times 10^{-27} \times 3 \times 10^{8}}=4.17 \,m / sec\)
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