Answer
$2.19\times 10^{6}m/s$.
Work Step by Step
The uncertainty in position is given. Use Eq. 28–1 to find the uncertainty in the speed. Note that the uncertainty in momentum is the electron mass multiplied by the uncertainty in the velocity.
$$\Delta p= m \Delta v \geq \frac{\hbar}{\Delta x}$$
$$\Delta v \geq \frac{\hbar}{m \Delta x}$$
$$=\frac{1.055\times 10^{-34}J \cdot s}{(9.11\times10^{-31}kg)(0.529\times10^{-10}m)}$$
$$= 2.19\times 10^{6}m/s $$