Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
We are asked to model the nucleus as a one-dimensional box with a size equal to the nucleus's diameter, and then find the minimum speed range of an electron confined in this "box."
The Heisenberg Uncertainty Principle relates the uncertainty in position ($ \Delta x $) to the uncertainty in momentum ($ \Delta p_x $):
$$ \Delta x \Delta p_x \geq \frac{h}{2} $$
Solving for $\Delta p_x$ to find the uncertainty in momentum:
$$ \Delta p_x\approx \frac{h}{2 \Delta x} $$
Plug the known;
$$ \Delta p_x \approx \frac{(6.63\times 10^{-34})}{2 (10\times 10^{-15})}=\bf 3.315\times 10^{-20}\;\rm kg\cdot m/s $$
Recalling that $p=mv$, so
$$v_x=\dfrac{\Delta p_x}{m}=\dfrac{3.315\times 10^{-20}}{(9.11\times 10^{-31})}=\bf 3.64\times 10^{10}\;\rm m/s$$
This means that the range of possible velocities is given by
$$-1.82\times 10^{10}{\;\rm m/s}\leq v_x\leq1.82\times 10^{10} \;\rm m/s$$
Therefore, the minimum speed range is from $0$ m/s to $1.82\times 10^{10}$ m/s.
$$\color{blue}{\bf [b]}$$
The electron's speed in this case exceeds the speed of light, which, if true, would mean that no electrons could exist around the atom.
