Uncertainty in momentum in a 0.1 nm well

The Heisenberg uncertainty principle says that position and momentum cannot both be known exactly. In the form used in Kasap's textbook,

$$\Delta x\,\Delta p_x \ge \hbar$$

For an electron confined in a well of width $L$, its position is uncertain by at most the width of the well, so take $\Delta x = L$.

Given: $L = 0.1\ \text{nm} = 1\times10^{-10}$ m, $\hbar = 1.055\times10^{-34}$ J s, $m = 9.1\times10^{-31}$ kg.

Uncertainty in momentum

$$\begin{aligned} \Delta p_x &\approx \frac{\hbar}{\Delta x} = \frac{1.055\times10^{-34}}{1\times10^{-10}} \\ &= 1.055\times10^{-24}\ \text{kg m s}^{-1} \end{aligned}$$

Kinetic energy

The momentum of the electron must be at least of the order of its uncertainty, so the corresponding kinetic energy is

$$\begin{aligned} KE &\approx \frac{(\Delta p_x)^2}{2m} = \frac{(1.055\times10^{-24})^2}{2\times9.1\times10^{-31}} \\ &= 6.11\times10^{-19}\ \text{J} = \frac{6.11\times10^{-19}}{1.6\times10^{-19}} = 3.82\ \text{eV} \end{aligned}$$

Comment

• A confined electron can never be at rest: confining it to 0.1 nm (about the size of an atom) forces it to have a kinetic energy of a few eV. This is why electrons in atoms have energies in the eV range.
• This is an order-of-magnitude estimate. With the stricter form $\Delta x\,\Delta p \ge \hbar/2$, the values are half and one quarter: $\Delta p = 5.27\times10^{-25}$ kg m/s and $KE \approx 0.95$ eV. The exact ground-state energy of a 0.1 nm infinite well is $h^2/8mL^2 = 37.7$ eV.

Answer: $\Delta p \approx 1.06\times10^{-24}$ kg m s⁻¹, $KE \approx 6.11\times10^{-19}$ J $\approx 3.8$ eV.