5.7 Probability: the heisenberg uncertainty principle  (Page 3/11)

Heisenberg uncertainty principle in position and momentum for an atom

(a) If the position of an electron in an atom is measured to an accuracy of 0.0100 nm, what is the electron’s uncertainty in velocity? (b) If the electron has this velocity, what is its kinetic energy in eV?

Strategy

The uncertainty in position is the accuracy of the measurement, or $\mathrm{\Delta }x=0\text{.}\text{0100}\text{nm}$ . Thus the smallest uncertainty in momentum $\mathrm{\Delta }p$ can be calculated using $\mathrm{\Delta }x\mathrm{\Delta }p\ge h\text{/4}\pi$ . Once the uncertainty in momentum $\mathrm{\Delta }p$ is found, the uncertainty in velocity can be found from $\mathrm{\Delta }p=m\mathrm{\Delta }v$ .

Solution for (a)

Using the equals sign in the uncertainty principle to express the minimum uncertainty, we have

Solving for $\mathrm{\Delta }p$ and substituting known values gives

Thus,

Solving for $\mathrm{\Delta }v$ and substituting the mass of an electron gives

Solution for (b)

Although large, this velocity is not highly relativistic, and so the electron’s kinetic energy is

Discussion

Since atoms are roughly 0.1 nm in size, knowing the position of an electron to 0.0100 nm localizes it reasonably well inside the atom. This would be like being able to see details one-tenth the size of the atom. But the consequent uncertainty in velocity is large. You certainly could not follow it very well if its velocity is so uncertain. To get a further idea of how large the uncertainty in velocity is, we assumed the velocity of the electron was equal to its uncertainty and found this gave a kinetic energy of 95.5 eV. This is significantly greater than the typical energy difference between levels in atoms (see [link] ), so that it is impossible to get a meaningful energy for the electron if we know its position even moderately well.