JEE Mains · Physics · STD 12 - 12. atoms
A free electron of \(2.6\, {eV}\) energy collides with a \({H}^{+}\) ion. This results in the formation of a hydrogen atom in the first excited state and a photon is released. Find the frequency of the emitted photon. \(\left({h}=6.6 \times 10^{-34}\, {J} {s}\right)\)
- A \(1.45 \times 10^{16}\, {MHz}\)
- B \(0.19 \times 10^{15} \,{MHz}\)
- C \(1.45 \times 10^{9}\, {MHz}\)
- D \(9.0 \times 10^{27} \,{MHz}\)
Answer & Solution
Correct Answer
(C) \(1.45 \times 10^{9}\, {MHz}\)
Step-by-step Solution
Detailed explanation
For every large distance \(P.E.\) \(=0\) \(\&\) total energy \(=2.6+0=2.6\, {eV}\) Finally in first excited state of \({H}\) atom total energy \(=-3.4 \,{eV}\) Loss in total energy \(=2.6-(-3.4)\) \(\quad\quad\quad\quad\quad\quad\quad\quad=6\,eV\) It is emitted as photon…
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