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  • The total energy of electron in the ground state of hydrogen atom is <img alt="\mathrm{ -13.6\: eV.}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%20-13.6%5C%3A%20eV.%7D"

The total energy of electron in the ground state of hydrogen atom is \mathrm{ -13.6\: eV.} The kinetic energy of an electron in the first excited state is
 

Option: 1

\mathrm{3.4\: eV}
 


Option: 2

\mathrm{6.8\: eV}
 


Option: 3

\mathrm{13.6\: eV}


Option: 4

\mathrm{1.7\: eV}


Answers (1)

best_answer

The energy of hydrogen atom when the electron revolves in \mathrm{\text{nth}} orbit, is

\mathrm{\quad E=\frac{-13.6}{n^2} \mathrm{eV}}
In the ground state,\mathrm{ n=1}

\mathrm{ E=\frac{-13.6}{1^2}=-13.6 \mathrm{eV} }

For \mathrm{ n=2, \quad, E=\frac{-13.6}{2^2}=-3.4 \mathrm{eV} }

So, kinetic energy, of electron in the first excited state (i.e. for \mathrm{ n=2 } ), is

\mathrm{ \mathrm{KE}=-E=-(-3.4)=3.4 \mathrm{eV} }

Hence option 1 is correct.
 

Posted by

vishal kumar

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