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3.5 The Compton Effect
If light sometimes behaves like a particle, the photon, why not consider the collision of a photon with a free electron from the same point of view as the collision of two billiard balls? Such a collision between a photon and a free electron is called Compton scattering, or the Compton effect, in honor of Arthur Holly Compton (1892-1962). In order to get a massive photon for the collision, X rays are used. (Recall that X rays have a high frequency , and therefore the energy of the X ray, E = h, is large, and thus its mass, m = E/c2, is also large.) In order to get a free
electron, a target made of carbon is used. The outer electrons of the carbon atom are very loosely bound, so compared with the initial energy of the photon, the electron looks like a free electron. Thus, the collision between the photon and the electron can be pictured as shown in figure 3.7. We assume that the electron is initially at
Figure 3.7 Compton scattering.
rest and that the incident photon has an energy (E = h) and a momentum (p = E/c). After the collision, the electron is found to be scattered at an angle from the original direction of the photon. Because the electron has moved after the collision, some energy must have been imparted to it. But where could this energy come from? It must have come from the incident photon. But if that is true, then the scattered photon must have less energy than the incident photon, and therefore, its wavelength should also have changed. Let us call the energy of the scattered photon E’, where
E ’ = h’ and hence, its final momentum is
p’ = E’ = h’ c c
Because momentum is conserved in all collisions, the law of conservation of momentum is applied to the collision of figure 3.7. First however, notice that the collision is two dimensional. Because the vector momentum is conserved, the x- component of the momentum and the y-component of momentum must also be conserved. The law of conservation for the x-component of momentum can be written as
pp + 0 = pp’ cos + pe cos and for the y-component,
0 + 0 = pp’ sin pe sin
where pp is the momentum of the incident photon, pp’ the momentum of the
scattered photon, and pe the momentum of the scattered electron. Substituting the values for the energy and momentum of the photon, these equations become
h = h’ cos + pe cos c c
0 = h’ sin pe sin c
There are more unknowns (’, , , pe) than we can handle at this moment, so let us eliminate from these two equations by rearranging, squaring, and adding them. That is,
pe cos = h h’ cos c c
pe sin = h’ sin c
pe2 cos2 = (h)2 2hh’ cos + (h’)2 cos2
c2 c2 c2
pe2 sin2 = (h’)2 sin2
c2
pe2 (sin2 + cos2) = (h)2 + (h’)2 (sin2 + cos2) 2(h)(h’) cos
c2 c2 c2
But since sin2 + cos2 = 1, we get
pe2 = (h)2 + (h’)2 2(h)(h’) cos (3.22)
c2 c2 c2
The angle has thus been eliminated from the equation. Let us now look for a way to eliminate the term pe, the momentum of the electron. If we square equation 3.19, we can solve for pe2 and obtain
pe2 = Ee 2 E0e2 (3.23)
c2
But the total energy of the electron Ee, given by equation 1.102, is Ee = KEe + E0e
where KEe is the kinetic energy of the electron and E0e is its rest mass. Substituting equation 1.102 back into equation 3.23, gives, for the momentum of the electron, pe2 = (KEe + E0e ) 2 E0e2 c2 = KEe2 + 2E0e KEe + E02 E02 c2 pe2 = KEe2 + 2E0e KEe (3.24) c2
But if the law of conservation of energy is applied to the collision of figure 3.7, we get
E = E ’ + KEe
h = h’ + KEe (3.25) where E is the total energy of the system, E’ is the energy of the scattered photon and KEe is the kinetic energy imparted to the electron during the collision. Thus, the kinetic energy of the electron, found from equation 3.25, is
KEe = h h’ (3.26) Substituting the value of the kinetic energy from equation 3.26 and E0e = m0c2, the
rest energy of the electron, back into equation 3.24, we get, for the momentum of the electron,
pe2 = (h h’)2 + 2m0c2(h h’)
c2
pe2 = (h)2 + (h’)2 2hh’ + 2m0(h h’) (3.27)
c2 c2 c2
Since we now have two separate equations for the momentum of the electron, equations 3.22 and 3.27, we can equate them to eliminate pe. Therefore,
(h)2 + (h’)2 2hh’ + 2m0(h h’) = (h)2 + (h’)2 2(h)(h’) cos c2 c2 c2 c2 c2 c2 Simplifying, 2m0(h h’) = 2hh’ 2(h)(h’) cos c2 c2 h h’ = hh’(1 cos ) m0c2 ' = h (1 cos ) ' m0c2
However, since = c/ this becomes
c/ c/’ = h (1 cos ) (c/)(c/’) m0c2
0 1 1 ' 1 cos ' h m c
0 ( ' ) h 1 cos m c (3.28)Equation 3.28 is called the Compton scattering formula. It gives the change in wavelength of the scattered photon as a function of the scattering angle . The quantity,
h = 2.426 1012 m = 0.002426 nm
m0c
which has the dimensions of a length, is called the Compton wavelength.
Thus, in a collision between an energetic photon and an electron, the scattered light shows a different wavelength than the wavelength of the incident light. In 1923, A. H. Compton confirmed the modified wavelength of the scattered photon and received the Nobel Prize in 1927 for his work.
Example 3.7
Compton scattering. A 90.0-KeV X-ray photon is fired at a carbon target and Compton scattering occurs. Find the wavelength of the incident photon and the wavelength of the scattered photon for scattering angles of (a) 30.00 and (b) 60.00.
The frequency of the incident photon is found from E = h as
3 19 34 90.0 10 eV 1.60 10 J s 6.625 10 J s 1 eV E h = 2.17 1019 Hz
The wavelength of the incident photon is found from 8 19 9 3.00 10 m/s 1 nm 2.17 10 1/s 10 m c =0.0138 nm
The modified wavelength is found from the Compton scattering formula, equation 3.28, as ' = + h (1 cos ) m0c a. ' = 0.0138 nm + (0.002426 nm)(1 cos 30.00) = 0.0141 nm b. ' = 0.0138 nm + (0.002426 nm)(1 cos 60.00) = 0.0150 nm
Solution
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In an actual experiment both the incident and modified wavelengths are found in the scattered photons. The incident wavelength is found in the scattered photons because some of the incident photons are scattered by the atom. In this case, the rest mass of the electron m0 must be replaced in equation 3.28 by the mass
M of the entire atom. Because M is so much greater than m0, the Compton
wavelength h/MC is so small that the change in wavelength for these photons is too small to be observed. Thus, these incident photons are scattered with the same wavelength.