Determine the velocity of an electron with a wavelength of 1 imes 10⁻¹⁰ m, and calculate the minimum uncertainty in momentum if position uncertainty is 2.0 imes 10⁻¹¹ m.

$$v = \frac{6.626 \times 10^{-34}}{9.11 \times 10^{-41}} \text{ m/s}$$ $$v = 0.72733 \times 10^{7} \text{ m/s}$$ $$v = \boxed{\mathbf{7.27 \times 10^{6} \text{ m/s}}}$$

Now, let's solve the two numbered questions.

1. Calculate the minimum uncertainty in momentum if position uncertainty is $2.0 \times 10^{-11}$ m.

Rearrange to solve for the uncertainty in momentum ($\Delta p$).

$$\Delta p = \frac{h}{4\pi \Delta x}$$

Planck's constant $h = 6.626 \times 10^{-34}$ J s and uncertainty in position $\Delta x = 2.0 \times 10^{-11}$ m.

$$\Delta p = \frac{6.626 \times 10^{-34} \text{ J s}}{4\pi \times (2.0 \times 10^{-11} \text{ m})}$$

$$\Delta p = \frac{6.626 \times 10^{-34}}{25.1327 \times 10^{-11}} \text{ kg m/s}$$ $$\Delta p = 0.26356 \times 10^{-23} \text{ kg m/s}$$ $$\Delta p = \boxed{\mathbf{2.64 \times 10^{-24} \text{ kg m/s}}}$$

2. Determine the momentum of a photon with wavelength 400 nm.

$h = 6.626 \times 10^{-34}$ J s and $\lambda = 400 \text{ nm} = 400 \times 10^{-9} \text{ m} = 4.00 \times 10^{-7} \text{ m}$.

$$p = \frac{6.626 \times 10^{-34} \text{ J s}}{4.00 \times 10^{-7} \text{ m}}$$

$$p = 1.6565 \times 10^{-27} \text{ kg m/s}$$ $$p = \boxed{\mathbf{1.66 \times 10^{-27} \text{ kg m/s}}}$$