Answer
PE = -27.2 eV, KE = +13.6 eV.
Work Step by Step
The potential energy associated with the ground state of hydrogen is the electron’s charge multiplied by the electric potential at its location, due to the proton.
$$PE=qV=(-e)\frac{1}{4\pi \epsilon_0}\frac{e}{r}$$
$$-(9.00\times10^9Nm^2/C^2) \frac{(1.60\times10^{-19}C)^2(eV/1.60\times10^{-19}J)}{0.529\times10^{-10}m}$$
$$=-27.2eV$$
The kinetic energy is the total energy minus the potential energy.
$$KE=E_1-PE=-13.6eV-(-27.2eV)=+13.6eV$$
