The total energy of electron in the ground state of hydrogen atom is $-13.6 \mathrm{eV}$. The kinetic energy…
- $6.8 \mathrm{eV}$
- $13.6 \mathrm{eV}$
- $1.7 \mathrm{eV}$
- $-3.4 \mathrm{eV}$.
Solution
$E_n=-\frac{13.6}{n^2} e V$
for ground state $n=1$
$E_1=-\frac{13.6}{12}=-13.6 \mathrm{eV}$
for excited state $n=2$
$E_2=-\frac{13.6}{22}=-3.4 \mathrm{eV}$
Asked in: NEET 2007
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