JEE Mains · Physics · STD 12 - Atoms
- JEE Mains 2021
- Hard
- MCQ
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. \(\left({h}=6.6 \times 10^{-34}\, {J} {s}\right)\)
Pick an option to check your answer.
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
\(\lambda=\frac{1240}{6}=206\, {nm}\)
\({f}=\frac{3 \times 10^{8}}{206 \times 10^{-9}}=1.45 \times 10^{15} \,{Hz}\)
\(=1.45 \times 10^{9}\, {Hz}\)
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