Answer
$6.56\times 10^{15}$ revolutions/s
Work Step by Step
The force between the proton and the electron provides the centripetal force to keep the electron moving in a circle.
We can find the electron's speed:
$\frac{kq^2}{r^2} = \frac{mv^2}{r}$
$v^2 = \frac{kq^2}{m~r}$
$v = \sqrt{\frac{kq^2}{m~r}}$
$v = \sqrt{\frac{(9.0\times 10^9~N~m^2/C^2)(1.6\times 10^{-19}~C)^2}{(9.109\times 10^{-31}~kg)~(0.053\times 10^{-9}~m)}}$
$v = 2.185\times 10^6~m/s$
We can find the number of revolutions per second the electron makes:
$N = \frac{v}{2\pi~r}$
$N = \frac{2.185\times 10^6~m/s}{(2\pi)~(0.053\times 10^{-9}~m)}$
$N = 6.56\times 10^{15}~rev/s$
