Question

6. Assume that an electron is located somewhere within a region of atomic size. Estimate the minimum uncertainty in its momentum. By assuming that this uncertainty is comparable with its average momentum, estimate the average kinetic energy of the electron.

Answer 1

According to the Heisenberg's uncertainty principle, the minimum uncertainty in momentum will be:

$$\Delta p = \dfrac{\hbar}{2\Delta x}$$

where $\hbar = 1.05\cdot 10^{-34}J\cdot s$ is the reduced Planck constant and $\Delta x = 1.06\cdot 10^{-10}m$ is the atom diameter (uncertainty of the coordinate). Thus:

$$\Delta p = \dfrac{1.05\cdot 10^{-34}J\cdot s}{2\cdot 1.06\cdot 10^{-10}m} = 5\cdot 10^{-25} kg\cdot m/s$$

The average kinetic energy of the electron will be:

$$K = \dfrac{\Delta p^2}{2m_e}$$

where $m_e = 9.11\cdot 10^{-31}kg$ is the mass of electon.

$$K = \dfrac{(5^\cdot 10^{-25} kg\cdot m/s)^2}{2\cdot 9.11\cdot 10^{-31}kg} = 1.37\cdot 10^{19}J$$