In the Bohr’s model of the hydrogen atom

In Bohr’s model of the hydrogen atom, Coulomb’s force provides the centripetal acceleration to the electron of mass m moving with velocity $\upsilon$ in an orbit of radius r i.e.,
$m{v}^2/r=Z{e}^2/\left(4\pi {\in }_0{r}^2\right).$            (1)
Also, angular momentum of the electron in $n^{th}$ orbit is quantized i.e.,
$mvr=nh/\left(2\pi \right).$                (2)
Eliminate $\upsilon$ form equations (1) and (2) to get the radius of the $n^{th}$ orbit
$r_n={\in }_0{h}^2{n}^2/\left(\pi mZ{e}^2\right).$                 (3)
Total energy of the electron in $n^{th}$ orbit is the sum of its kinetic energy K and potential energy U i.e.,
$E_n=K+U=\frac{1}{2}m{v}^2-\frac{Z{e}^2}{4\pi {\in }_{0}r}=-\frac{m{Z}^2{e}^4}{8{\in }_0^2{h}^2{n}^2},$
Where we have used equations (1)_(3) to eliminate r and  $\upsilon$. The ratio of kinetic energy to potential energy is given by 

$\frac{K}{U}=\frac{\frac{1}{2}m{\upsilon }^2}{-\frac{Z{e}^2}{4\pi {\in }_{0}r}}=-\frac{1}{2}\frac{m{\upsilon }^2}{{\left(\frac{Z{e}^2}{4\pi {\in }_0{r}^2}\right)}^r}=-\frac{1}{2}\frac{m{\upsilon }^2}{(m{\upsilon }^2/r)r} =-\frac{1}{2}.$