Answer
See the detailed answer below.
Work Step by Step
We know that the electrostatic potential energy of an electron is given by
$$PE=-\dfrac{1}{4\pi\varepsilon_0}\times \dfrac{\pi Z^2e^4m}{n^2h^2\varepsilon_0 }$$
The charge is negative;
$$PE= \dfrac{- Z^2e^4m}{4n^2h^2\varepsilon_0^2 }\tag 1$$
We know its total energy is given by
$$E=KE+PE$$
Thus,
$$KE=E-PE= \dfrac{- Z^2e^4m}{8n^2h^2\varepsilon_0^2 }- \dfrac{- Z^2e^4m}{4n^2h^2\varepsilon_0^2 }$$
$$KE= \dfrac{- Z^2e^4m}{8n^2h^2\varepsilon_0^2 }+ \dfrac{ Z^2e^4m}{4n^2h^2\varepsilon_0^2 }=\dfrac{- Z^2e^4m+2 Z^2e^4m}{8n^2h^2\varepsilon_0^2 }$$
$$KE =\dfrac{ Z^2e^4m}{8n^2h^2\varepsilon_0^2 }\tag 2$$
Dividing (1) by (2);
$$\bigg|\dfrac{PE}{KE}\bigg|=\left|\dfrac{ \dfrac{- Z^2e^4m}{4n^2h^2\varepsilon_0^2 }}{\dfrac{ Z^2e^4m}{8n^2h^2\varepsilon_0^2 }}\right|$$
$$ \dfrac{PE}{KE} = \dfrac{ Z^2e^4m}{4n^2h^2\varepsilon_0^2 }\times \dfrac{8n^2h^2\varepsilon_0^2 }{ Z^2e^4m}=2 $$
Hence,
$$\boxed{PE=\color{red}{\bf 2}KE}$$
Therefore, the electrostatic potential energy of an electron in any Bohr orbit of a hydrogen atom is twice the magnitude of its kinetic energy in the same orbit.
