Answer
${\bf 1.40\times 10^{-10}}\;\rm m$,
${\bf 1.344\times10^6}\;\rm m/s$
Work Step by Step
As an abstract, we have an electron $(-e)$ that rotates around a point that has a net charge of $(+e)$.
Thus, the electron energy in this case is given by
$$E_e=\frac{1}{2}m v^2+\dfrac{k(+e)(-e)}{r}$$
$$E_e=\frac{1}{2}m v^2-\dfrac{ke^2}{r} $$
To ionize this atom, we need to add an amount of energy to this electron of $\rm 5.14 eV= 5.14\times 1.6\times 10^{-19}=8.224\times10^{-19}\;\rm J$
And since we need to remove the electron from the atom to infinity, then the energy of the electron is then
$$-8 .224\times10^{-19} =\frac{1}{2}m v^2-\dfrac{ke^2}{r} \tag 1$$
We know that the force exerted on the electron that makes it rotate around the atom is given by
$$F_e=ma_r=\dfrac{mv^2}{r}$$
Hence,
$$\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}$$
Hence,
$$mv^2=\dfrac{ke^2}{r}\tag 2$$
Plug into (1);
$$-8 .224\times10^{-19} =\dfrac{ke^2}{2r} -\dfrac{ke^2}{r} =\dfrac{ke^2-2ke^2}{2r}$$
$$-8 .224\times10^{-19} =\dfrac{-ke^2}{2r}$$
Solving for $r$;
$$r =\dfrac{ ke^2}{2( 8 .224\times10^{-19} )}$$
Plug the known;
$$r =\dfrac{ (9.0\times 10^9)(1.6\times 10^{-19})^2}{2( 8 .224\times10^{-19} )}$$
$$r=\color{red}{\bf 1.40\times 10^{-10}}\;\rm m$$
Solving (2) for $v$;
$$v =\sqrt{\dfrac{ke^2}{mr}}$$
Plug the known;
$$v =\sqrt{\dfrac{(9.0\times 10^9)(1.6\times 10^{-19})^2}{(9.11\times 10^{-31})(1.40×10^{-10})}}$$
$$r=\color{red}{\bf 1.344\times10^6}\;\rm m/s$$
