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Recall from last lecture:

27.4 X-rays

In 1895, Wilhelm Roentgen discovered x-rays, thus marking the beginning of modern physics. X-rays are a form of electromagnetic radiation, like radio waves and light, but having higher frequency and energy than either. The energy of x-rays is higher than the electron binding energies of light atoms. This enables x-rays to easily pass through water and tissues, composed mostly of hydrogen, carbon, nitrogen, and oxygen, while they will tend to scatter off of bones with their high density of calcium.

When originally discovered, it wasn't clear what x-rays where, hence the reason behind the name. It was easily verified that x-rays are not deflected by electric or magnetic fields, and therefore are not charged particles.

It took a demonstration of diffraction of x-rays by a crystal, from which their wavelength could be calculated, and conclude that they are electromagnetic waves.

A common way to produce x-rays is to rapidly decelerate high energy electrons, for instance by having them hit a target made of heavy atoms. An x-ray tube is a device that does just this. The resulting x-rays have a broad spectrum of energies, plus some sharp lines corresponding to atomic energy levels of inner shell electrons.

The x-ray is created when the electron passes near the nucleus of an atom and feels its electric field. This process is like a collision of billiard balls, or the inverse of the photoelectric effect. The maximum x-ray energy, and minimum wavelength results when the electron loses all its energy in a single collision, such that
eDV = hfmax = hc/lmin
or therefore

lmin = hc/eDV

Example: P27.21

What minimum accelerating voltage would be required to produce an x-ray with a wavelength of 0.0300nm?

Since lmin = hc/eDV, we can turn this around to find
DVmin = hc/el = (6.63×10-34Js)(3.0×108m/s)/(1.6×10-19C)(3.00×10-11m) = 4.1×104V = 41kV.

27.5 Diffraction of X-rays by Crystals

Diffraction occurs when light passes through a regularly spaced set of slits. This week, you are using a diffraction grating made by cutting 300 grooves per millimeter in a piece of glass -- grooves work just as well as slits for this work. For a diffraction grating to work well, the spacing of the slits must be comparable to the wavelength of the radiation. Visible light has wavelengths between about 700nm (red) and 400nm (blue). A diffraction grating with 300 grooves per mm has a slit spacing of (1/300)mm = 3mm = 3000nm. This is 4 to 8 times larger than the wavelengths of visible light, sufficient for your laboratory exercise.

The wavelengths of x-rays are about 0.1nm. Your diffraction grating will have little effect on x-rays. But it was realized by Max von Laue that the wavelength of x-rays is comparable to the size of an atom (0.1nm = 1 angstrom), and that a material with a regularly spaced grid of atoms could diffract x-rays, assuming that x-rays are electromagnetic radiation. X-rays are EM radiation, they are diffracted by crystals, and Laue won a Nobel Prize.

Diffraction of x-rays by crystals continues to be an important research tool today. However, the crystals are usually made from proteins, and the technique is called protein crystallography. The diffraction pattern produced is vital in deducing the structure of proteins (folding), enabling researchers to connect the functionality of the protein with the structure. Pharmaceutical researchers are common users of this technique.

27.6 The Compton Effect

The Compton effect is another famous experiment supporting the existence of photons. The Compton effect is the scattering of x-rays from electrons (and for this reason is also called Compton scattering). To understand Compton scattering, we must not only treat EM radiation as quantized photons, we must think of the photons as particles! After all, how can a wave scatter off a point-like particle? Does sound scatter off a pencil? The only way to make sense of Compton scattering, is to treat the photon like a particle not a wave.

Let's do just that and see how this works. Imagine an incoming photon scattering off a stationary electron, like billiard balls colliding on a pool table (drawing, Figure 27.17). After they collide, the photon and electron will move away at some angle to the original photon direction. If we know the energy (momentum) of the incoming photon, and the outgoing angle of the photon, then we can determine the outgoing photon's energy (and the outgoing angle and momentum of the electron) by requiring that energy and momentum be conserved. We treat the photon like a particle with zero mass, and momentum p = E/c. (Note that physicists are more willing to call a wave a particle than to give up energy and momentum conservation.)

It is traditional to express this in terms of the change of the photon's wavelength, since energy and wavelength are related by E = hf = hc/l. The result is:

Dl = l - l0 = (h/mec)(1 - cosq)
where l0 is the wavelength of the incoming x-ray, l is the wavelength of the outgoing x-ray, q is the scattered angle of the outgoing x-ray, and (h/mec) = 0.00243nm is called the Compton wavelength. The Compton wavelength is the amount by which the photon's wavelength changes when it scatters at 90°. The change in wavelegth is smaller if the scattering angle is less than 90°, and is a maximum of twice the Compton wavelength if the photon scatters by 180° (scatters backwards). Since the Compton wavelength is small, the change in wavelength is only perceptible for photons of small wavelength, say less than about 0.1nm, corresponding to x-rays.

Example: P27.29

X-rays with an energy of 300keV undergo Compton scattering from a target. If the scattered rays are deflected at 37° relative to the direction of the incident rays, find (a) the Compton shift at this angle, (b) the energy of the scattered x-ray, and (c) the kinetic energy of the recoiling electron.

(a) The "Compton shift" is the change of wavelength, and for a 37° scattering angle this is:
Dl = (h/mec)(1 - cosq) = (0.00243nm)(1 - cos37°) = 0.00049nm.

(b) We will need the quantity hc several times, so let's calculate it, and convert J to eV.
hc = (6.63×10-34Js)(3.0×108m/s) = 2.0×10-25Jm
hc = (2.0×10-25Jm)/(1.6×10-19J/eV) = 1.24×10-6eV m
The wavelength of the incoming x-ray is
l0 = hc/E0 = (1.24×10-6eV m) / (3.00×105eV) = 4.13×10-12m = 0.00413nm
The outgoing wavelength is
l = l0 + Dl = 0.00413nm + 0.00049nm = 0.00462nm.
Photons of this wavelength have energy:
E = hc/l = (1.24×10-6eV m) / (4.62×10-12m) = 0.268×106eV = 268keV.

We can determine the kinetic energy of the recoiling electron because kinetic energy is conserved. The energy before the scattering is the energy of the incoming x-ray, plus the rest energy of the electron. The energy after the scattering is the energy of the outgoing x-ray, plus the kinetic energy of the recoiling electron, plus the rest energy of the electron. These must be equal, therefore E0 + mec2 = E + KE + mec2, or
KE = E0 - E = 300keV - 268 keV = 32keV.

© Robert Harr 2000