Answer
$13.57\;eV$
Work Step by Step
The electron is one Bohr radius $(a)$ from the central nucleus.
$a = 5.291772\times 10^{-11}\;m$
Now, the potential energy of the hydrogen atom at the ground state is:
$U=-\frac{e^2}{4\pi \epsilon_0 a}$
Putting known values we obtain
$U=-\frac{(1.6\times 10^{-19})^2}{4\pi\times 8.854\times 10^{-12}\times 5.291772\times 10^{-11}}\;J\\
or, U=-4.35\times 10^{-18}\;J\\
or, U=-27.17\;eV$
Total energy of the hydrogen atom at the ground state is: $E=-13.6\;eV$
Therefore, the kinetic energy of the hydrogen atom at the ground state is:
$E_k=E-U\\
or, E_k=-13.6-(-27.17)\;eV\\
or, \boxed {E_k=13.57\;eV} $
