Question

Assuming an electron is confined to a $1nm$ wide region, find the uncertainty in momentum using Heisenberg Uncertainty principle $\Delta x\times \Delta p=\frac{h}{2\pi }$. You can assume the uncertainty in position $\Delta x\text{as}1nm$. Assuming $p\cong \Delta p$, find the energy of the electron in electron volts.

Answer 1

Given, $\Delta x$ is equal to $1nm={10}^{-9}$

According to the Heisenberg's Uncertainity principle,
$\Delta x\times \Delta p=\frac{h}{2\pi }$
$h$ is Plank's constant,
$\Delta x$ is the uncertainty in the position.

$\Delta p$ is the uncertainity in the momentum of the electron.

$\Delta p=\frac{h}{2\pi \times \Delta x}=\frac{6.626\times {10}^{-34}}{2\times 3.14\times {10}^{-9}}$

$\Delta p=1.055\times {10}^{-25}kgm/s$

Find the energy of electron.

We can treat $\Delta p\text{as}p$ to find the energy of the electron,
$\text{Energy}=\frac{p^2}{2\times m}=\frac{\Delta p^2}{2\times m}$

$\text{Energy}=\frac{(1.055\times {10}^{-25}{)}^2}{2\times 9.1\times {10}^{-31}}=6.1\times {10}^{-21}J$

Converting to $eV$ by dividing by $1.6\times {10}^{-19}$,
Energy$=\frac{6.1\times {10}^{-21}}{1.6\times {10}^{-19}}=3.8\times {10}^{-2}eV$

Final Answer: $3.8\times {10}^{-2}eV$