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  • The total energy of an electron in the first excited state of hydrogen is about  <img alt="\mathrm{-3.4\: eV}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B-3.4%5C%3A%20eV%

The total energy of an electron in the first excited state of hydrogen is about  \mathrm{-3.4\: eV}. Its kinetic energy in this state is
 

Option: 1

\mathrm{ -3.4 \: eV}
 


Option: 2

\mathrm{ -6.8 \: eV}
 


Option: 3

\mathrm{ 6.8 \: eV}
 


Option: 4

\mathrm{ 3.4 \: eV}


Answers (1)

best_answer

Kinetic energy of electron, \mathrm{K=\frac{Z e^2}{8 \pi \varepsilon_{0^{r}}}}

Potential energy of electron, \mathrm{U=-\frac{1}{4 \pi \varepsilon_0} \frac{Z e^2}{r}}

\mathrm{\therefore} Total energy,\mathrm{E =K+U }

\mathrm{ =\frac{Z e^2}{8 \pi \varepsilon_0 r}-\frac{Z e^2}{4 \pi \varepsilon_0 r}\text { or } E=-\frac{Z e^2}{8 \pi \varepsilon_0 r}}

\mathrm{or\: K=-E=-(-3.4)=3.4 \mathrm{eV} }

Hence option 4 is correct.

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