Answer
$4.8\times 10^{3}m/s$
Work Step by Step
The uncertainty in position is given. Use Eq. 28–1 to find the uncertainty in the speed.
$$\Delta p= m \Delta v \geq \frac{\hbar}{\Delta x}$$
$$\Delta v \geq \frac{\hbar}{m \Delta x}$$
$$=\frac{1.055\times 10^{-34}J \cdot s}{(9.11\times10^{-31}kg)(2.4\times10^{-8}m)}$$
$$=4825 m/s \approx 4.8\times 10^{3}m/s $$
