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(C) For a hydrogen-like species,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \, eV$.For $He^{+}$ io

Vedclass · Accepted Answer

$-6.8$

Vedclass · Answer

(C) For a hydrogen-like species,the energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6 	imes \frac{Z^2}{n^2} \, eV$.For $He^{+}$ ion,the atomic number $Z = 2$.The third excited state corresponds to $n = 4$ (since ground state is $n=1$,first excited is $n=2$,second is $n=3$,and third is $n=4$).$E_4 = -13.6 	imes \frac{2^2}{4^2} = -13.6 	imes \frac{4}{16} = -13.6 	imes 0.25 = -3.4 \, eV$.The potential energy $(P.E.)$ is related to total energy $(E)$ by the relation $P.E. = 2 	imes E$.Therefore,$P.E. = 2 	imes (-3.4 \, eV) = -6.8 \, eV$.

The potential energy of an electron in the third excited state of $He^{+}$ ion is ............. $eV$.

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