According to Bohr's postulate for the angular momentum of an electron in a stable orbit: $$ \begin{aligned} L &= I\omega = \frac{nh}{2\pi} \end{aligned} $$ where $I = mr^2$ is the moment of inertia of the electron about the nucleus, $m$ is the mass of the electron, $r$ is the radius of the orbit, and $\omega$ is the angular velocity. For the ground state, we take $n=1$. Rearranging the formula to solve for $\omega$: $$ \begin{aligned} mr^2\omega &= \frac{h}{2\pi} \\ \omega &= \frac{h}{2\pi mr^2} \end{aligned} $$ Given: $h = 6.63 \times 10^{-34} \text{ Js}$ $m = 9.1 \times 10^{-31} \text{ kg}$ $r = 0.53 \times 10^{-10} \text{ m}$ Substituting the values: $$ \begin{aligned} \omega &= \frac{6.63 \times 10^{-34}}{2 \times 3.1416 \times 9.1 \times 10^{-31} \times (0.53 \times 10^{-10})^2} \\ &= \frac{6.63 \times 10^{-34}}{2 \times 3.1416 \times 9.1 \times 10^{-31} \times 0.2809 \times 10^{-20}} \\ &= \frac{6.63 \times 10^{-34}}{1.6067 \times 10^{-50}} \\ &\approx 4.126 \times 10^{16} \text{ rad/s} \end{aligned} $$ Thus, the angular velocity of the electron is approximately $4.1 \times 10^{16} \text{ rad/s}$.