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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. \(\left({h}=6.6 \times 10^{-34}\, {J} {s}\right)\)

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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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