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(B) For a hydrogen atom,the potential energy $(P.E.)$ and total energy $(T.E.)$ are related as $P.E. = 2 	imes T.E.$Given $P.E. = -3.02 \ eV$,so $T.E

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$2^{nd}$

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(B) For a hydrogen atom,the potential energy $(P.E.)$ and total energy $(T.E.)$ are related as $P.E. = 2 	imes T.E.$Given $P.E. = -3.02 \ eV$,so $T.E. = \frac{-3.02}{2} = -1.51 \ eV$.The total energy of an electron in the $n^{th}$ orbit is given by $E_n = -13.6 / n^2 \ eV$.Setting $-13.6 / n^2 = -1.51$,we get $n^2 = 13.6 / 1.51 \approx 9$,so $n = 3$.Since the electron is in the $n = 3$ shell,it corresponds to the $2^{nd}$ excited state (as $n=1$ is ground state,$n=2$ is $1^{st}$ excited state,and $n=3$ is $2^{nd}$ excited state).

If the potential energy of an electron in a hydrogen atom is $-3.02 \ eV$,then the electron is present in which of the following excited states?

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