Answer
${\bf 18.2} \; \rm{ \mu m} $
Work Step by Step
This problem involves the Heisenberg Uncertainty Principle, which relates the uncertainty in position ($ \Delta x $) to the uncertainty in momentum ($ \Delta p $) of a particle. The uncertainty principle is given by:
$$ \Delta x \cdot \Delta p_x \geq \frac{h}{2} \tag 1$$
Where $ \Delta x $ is the uncertainty in the electron's position (which will be the size of the box), $ \Delta p_x $ is the uncertainty in momentum, and $h$ is Planck’s constant.
We want to find the smallest size of the box such that the electron's speed is no more than $ 10 \; \text{m/s} $.
Recalling that the linear momentum is given by
$$ p_x = mv_x $$
So,
$$ \Delta p = m \Delta v_x \tag 2$$
Now, we need to solve (1) for $\Delta x$ to find the smallest box size.
$$ \Delta x \geq \frac{h}{2 \Delta p_x}$$
Plug from (2);
$$ \Delta x \geq \frac{h}{2 m \Delta v_x }$$
Plug the known;
$$ \Delta x \geq \frac{(6.63\times 10^{-34}) }{2 (9.11 \times 10^{-31}) \Delta v_x } $$
where we know that the electron is moving at a speed that is less than 10 m/s, so we can say that the range of the electron velocities is $-10\lt v_x\lt 10$. Therefore, the change in the electron velocity $\Delta v_x=10-(-10)=20\;\rm m/s$.
Plug that into the previous formula above.
$$ \Delta x \geq \frac{(6.63\times 10^{-34}) }{2 (9.11 \times 10^{-31}) (20) } =18.2\times 10^{-6}\;\rm m$$
$$ \Delta x\approx \color{red}{\bf 18.2} \; \rm{ \mu m} $$
