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The total energy of electron in the ground state of hydrogen atom is $-13.6 \mathrm{eV}$. The kinetic energy of an electron in the first excited state is:
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The correct answer is:
$-3.4 \mathrm{eV}$.
Energy of $n^{\text {th }}$ orbit of hydrogen atom is given by
$E_n=-\frac{13.6}{n^2} e V$
for ground state $n=1$
$E_1=-\frac{13.6}{12}=-13.6 \mathrm{eV}$
for excited state $n=2$
$E_2=-\frac{13.6}{22}=-3.4 \mathrm{eV}$
$E_n=-\frac{13.6}{n^2} e V$
for ground state $n=1$
$E_1=-\frac{13.6}{12}=-13.6 \mathrm{eV}$
for excited state $n=2$
$E_2=-\frac{13.6}{22}=-3.4 \mathrm{eV}$
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