Assuming an electron is confined to a $1 \, nm$ wide region,find the uncertainty in momentum using Heisenberg's uncertainty principle. You can assume the uncertainty in position $\Delta x = 1 \, nm$. Assuming $\Delta p = p$,find the energy of the electron in electron volts $(eV)$.

(N/A) According to Heisenberg's uncertainty principle,$(\Delta x)(\Delta p) \approx \frac{h}{4\pi}$.
Using the relation $\Delta x \Delta p \approx \frac{h}{2\pi}$ for a confined particle:
$\Delta p = \frac{h}{2\pi \Delta x} = \frac{6.626 \times 10^{-34}}{2 \times 3.1416 \times 10^{-9}} \approx 1.055 \times 10^{-25} \, kg \cdot m/s$.
Given $\Delta p = p$,the momentum $p = 1.055 \times 10^{-25} \, kg \cdot m/s$.
The kinetic energy $E$ is given by $E = \frac{p^2}{2m}$.
$E = \frac{(1.055 \times 10^{-25})^2}{2 \times 9.11 \times 10^{-31}} = \frac{1.113 \times 10^{-50}}{1.822 \times 10^{-30}} \approx 6.109 \times 10^{-21} \, J$.
To convert to $eV$,divide by $1.602 \times 10^{-19} \, J/eV$:
$E = \frac{6.109 \times 10^{-21}}{1.602 \times 10^{-19}} \approx 0.0381 \, eV$.