<< Chapter < Page | Chapter >> Page > |
This is known as the Heisenberg uncertainty principle . It is impossible to measure position $x$ and momentum $p$ simultaneously with uncertainties $\mathrm{\Delta}x$ and $\mathrm{\Delta}p$ that multiply to be less than $h/\mathrm{4\pi}$ . Neither uncertainty can be zero. Neither uncertainty can become small without the other becoming large. A small wavelength allows accurate position measurement, but it increases the momentum of the probe to the point that it further disturbs the momentum of a system being measured. For example, if an electron is scattered from an atom and has a wavelength small enough to detect the position of electrons in the atom, its momentum can knock the electrons from their orbits in a manner that loses information about their original motion. It is therefore impossible to follow an electron in its orbit around an atom. If you measure the electron’s position, you will find it in a definite location, but the atom will be disrupted. Repeated measurements on identical atoms will produce interesting probability distributions for electrons around the atom, but they will not produce motion information. The probability distributions are referred to as electron clouds or orbitals. The shapes of these orbitals are often shown in general chemistry texts and are discussed in The Wave Nature of Matter Causes Quantization .
(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 $\mathrm{\Delta}x=0\text{.}\text{0100}\phantom{\rule{0.25em}{0ex}}\text{nm}$ . 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.
Notification Switch
Would you like to follow the 'College physics' conversation and receive update notifications?