Question

Motion of an election in a hydrogen atom corresponds to the potential energy U(r) = − e^2/(4πε0r) ,

4πε0 r which comes from the Coulomb attraction between the electron and the proton. Using the uncertainty

relation between the momentum of the electron, and its position, estimate the size of the hydrogen atom.

Answer 1

Suppose we have a hydrogen atom, and measure the position of the electron; we must not be able to predict exactly where the electron will be, or the momentum spread will then turn out to be infinite. Every time we look at the electron, it is somewhere, but it has an amplitude to be in different places so there is a probability of it being found in different places. These places cannot all be at the nucleus; we shall suppose there is a spread in position of order a

. That is, the distance of the electron from the nucleus is usually about a. We shall determine a by minimizing the total energy of the atom.

The spread in momentum is roughly ℏ/a

because of the uncertainty relation, so that if we try to measure the momentum of the electron in some manner, such as by scattering x-rays off it and looking for the Doppler effect from a moving scatterer, we would expect not to get zero every time—the electron is not standing still—but the momenta must be of the order p≈ℏ/a. Then the kinetic energy is roughly 12mv2=p2/2m=ℏ2/2ma2. (In a sense, this is a kind of dimensional analysis to find out in what way the kinetic energy depends upon the reduced Planck constant, upon m, and upon the size of the atom. We need not trust our answer to within factors like 2,π, etc. We have not even defined a very precisely.) Now the potential energy is minus e2 over the distance from the center, say −e2/a, where, as defined in Volume I, e2 is the charge of an electron squared, divided by 4πϵ0. Now the point is that the potential energy is reduced if a gets smaller, but the smaller a is, the higher the momentum required, because of the uncertainty principle, and therefore the higher the kinetic energy. The total energy is E=ℏ2/2ma2−e2/a.(2.10) We do not know what a is, but we know that the atom is going to arrange itself to make some kind of compromise so that the energy is as little as possible. In order to minimize E, we differentiate with respect to a, set the derivative equal to zero, and solve for a. The derivative of E is dE/da=−ℏ2/ma3+e2/a2,(2.11) and setting dE/da=0 gives for a the value a0=ℏ2/me2=0.528 angstrom,=0.528×10−10  meter.(2.12) This particular distance is called the Bohr radius,