Chapter III - Waves and Particles: Matter Waves

3.2) A beam of low energy electrons is directed normally at a metal surface, and strong reflection is detected only at an angle of 35 degrees. What is l/D, the ratio of the electron wavelength l to the interatomic spacing D at the surface?

From our treatment of electron diffraction, we know that constructive interference occurs when ml = d sinq. In this case, we have l/d = sinq/m. Since only one diffraction maximum is observed, we have m=1. For q = 35 degrees, l/d = 0.574.

3.11) A beam of electrons, kinetic energy = 54 eV, is directed normally at a nickel surface, and strong reflection is detected only at an angle of 50 degrees. Determine the spacing of nickel atoms on the surface.

The kinetic energy of a non-relativistic electron corresponds to its wavelength by E = p²/2m where p = h/l, or E = h²/(2ml²). This gives l = h/(2mE)^½. When used with our electron diffraction result, ml = d sinq, with m=1 for the diffraction order (not the electron mass), we get d = l/sinq = h/[(2mE)^½ sinq] @ 2.2×10^-10 m.

3.13) In the hydrogen atom, the electron's orbit, not necessarily circular, extends to a distance of about an Angstrom from the proton. If it is to move about as a compact classical particle in the region where it is confined, the electron's wavelength had better always be much smaller than an Angstrom. Is it? How large might be the electron's wavelength - how small might be its speed? If orbiting as a particle, its speed at 1 Angstrom could be no faster than that for circular orbit at that radius. (Why?) Find the corresponding wavelength and compare it to 1 Angstrom. Can the atom be treated classically?

Let's start by assuming that we can treat the electron classically. Then, its speed at an orbit radius of one Angstrom cannot exceed that for a circular orbit or its orbit would need to be non-circular, or extend to orbital distances in excess of one Angstrom. So, taking an orbital radius of 0.1 nm, we set the needed centripetal force equal to the available Coulomb force magnitude:
, or .

The corresponding wavelength would be l = h/p = h/(m_ev) @ 0.3 nm. This wavelength is larger than the 0.1 nm size originally assumed for the electron's radial region, so it cannot be treated classically.

3.26) To how small a region must an electron be confined for borderline relativistic speeds - say, 0.05 c - to become reasonably likely? On the basis of this, would you expect relativistic effects to be prominent for hydrogen's electron, of orbit radius near 10^-10 m? For a lead atom "inner-shell" electron, of orbit radius 10^-12 m?

Our knowledge of the 'state' of an electron in an atom is restricted by the Heisenberg uncertainty relation on position and momentum, DpDx > h/4p. We can estimate a linear positional range from a possible range of speeds for an orbital electron from this. Using the known speed of light and the (rest) mass of an electron, we obtain

This linear dimension is much smaller than the typical orbit radius for an electron in hydrogen, so relativistic effects would not be prominent; but it is comparable in size to the inner-shell radius in a lead atom for which relativistic effects would be expected to be significant.

3.32) A particle is connected to a spring and undergoes one-dimensional motion. (a) Write an expression for the total (kinetic plus potential energy) of the particle in terms of its position x, its mass m, its momentum p, and the force constant k of the spring. (b) Now treat the particle as a wave. Assume that the product of the uncertainties in position and momentum is governed by an uncertainty relation DpDx = h/4p. Also assume that since x is on average zero, the uncertainty Dx is roughly equal to a typical value of |x|. Similarly assume that Dp = |p|. Eliminate p in favor of x in the energy expression. (c) Find the minimum possible energy for the wave.

The total energy of our classical mass-on-a-spring can be written as . If we now treat the particle as a wave, the uncertainty relation applies. The average position and momentum values are zero, since our object oscillates uniformly about its origin. Then, the averages of the squares of these values will be the result from the uncertainty relation. (If they fluctuated about some non-zero value instead, the averages of the squares of the values would be close to the square of the largest value obtained for the parameter.)

Utilizing the values from the uncertainty relation to replace p @ h/(4pDx), we get

The minimum possible energy is then found by differentiating with respect to Dx and obtaining the minimum Dx via

Then, substituting back into our expression for E gives