Answer
See the detailed answer below.
Work Step by Step
In this problem, the electron is initially moving in the $z $-direction with no velocity component in the $x $-direction ($v_x = 0 \;\rm{m/s} $). The question asks whether $v_x $ will still be zero after the electron passes through the hole in the barrier or not.
This is a case involving the Heisenberg Uncertainty Principle, which relates the uncertainty in position $\Delta x $ and the uncertainty in momentum $\Delta p_x $ where
$$ \Delta x \cdot \Delta p_x \geq \frac{h}{2}\tag 1 $$
The electron passes through a 10-$\mu$m-diameter circular hole. This means that the uncertainty in the electron's position in the $x $-direction ($\Delta x $) is approximately the diameter of the hole:
$$ \Delta x \approx 10 \;\mu\rm{m} = \bf 10 \times 10^{-6} \;\rm{m}\tag2 $$
Solving (1) for $\Delta p_x$
$$ \Delta p_x \geq \frac{h}{2 \Delta x}\tag 3 $$
The uncertainty in velocity $\Delta v_x $ can be found by
$$ \Delta p_x = m_e \Delta v_x $$
Plug into (3);
$$m_e \Delta v_x\geq \frac{h}{2 \Delta x} $$
Hence,
$$ \Delta v_x \geq \frac{h}{2m_e \Delta x} $$
Plug the known;
$$ \Delta v_x \geq \frac{(6.63\times 10^{-34})}{2(9.11\times 10^{-31}) (10 \times 10^{-6})}=\bf 36.4\;\rm m/s $$
Thus,
$$ \Delta v_x \geq\bf 36\;\rm m/s $$
After the electron passes through the hole, it is not certain that $v_x $ is still zero. And since the the average velocity in the $x$-directionis zero, So
$$\boxed{-18\;{\rm m/s}\leq 18\leq \;{\rm m/s}}$$
