Compton scattering

From Free net encyclopedia

In quantum mechanics, the Compton scattering or Compton effect, observed by Arthur Holly Compton in 1923 that won him the 1927 Nobel Prize in Physics, is the increase in wavelength (decrease in energy) which occurs when X-ray (or gamma ray) photons with energies of around 0.5 MeV to 3.5 MeV interact with electrons in a material. The amount the wavelength increases by is called the Compton shift. Compton's experiment became the ultimate observation that convinced all physicists that light can behave as a stream of particles whose energy is proportional to the frequency.

Because the photons have such high energy, the interaction results in the electron being given enough energy to be completely ejected from its atom, and a photon containing the remaining energy being emitted in a different direction from the original, so that the overall momentum of the system is conserved. (If the photon still has enough energy, the process may be repeated.) Because of the overall reduction in energy of the photon, there is a corresponding increase in its wavelength. Thus overall there is a slight 'reddening' and scattering of the photons as they pass through the material. This scattering is known as Compton Scattering.

In a material where there are free electrons, this effect will occur at all photon energies and hence all wavelengths. In other materials, it is observed only with high-energy photons; photons of visible light, for example, do not have sufficient energy to eject the bound electrons.

The effect is important in scientific terms because it demonstrates that light cannot be explained purely as a wave phenomenon. Thomson scattering, the classical theory of charged particles scattered by an electromagnetic wave, cannot explain any shift in wavelength. Light must behave as if it consists of particles in order to explain the Compton scattering. It is also of prime importance to radiobiology, as it happens to be the most probable interaction of X rays with atomic nuclei in living beings and is applied in radiation therapy.

The Compton scattering has on occasion been proposed as an alternative explanation for the phenomenon of the redshift by opponents of the Big Bang theory, although this is not generally accepted because the influence of the Compton scattering would be noticeable in the spectral lines of distant objects and this is not observed.

1 The equations
2 Derivation
3 See also
4 External links

[edit]

The equations

Image:Compton scattering diagram.png Compton used a combination of three fundamental formulas representing the various aspects of classical and modern physics, combining them to describe the quantum behaviour of light.

Light as a particle, as noted previously in the photoelectric effect.
Relativistic dynamics Special Theory of Relativity
Trigonometry - Law of cosines

The final result gives us the Compton scattering equation:

<math>

  \Delta\lambda = \frac{h}{m_e c}(1-\cos{\theta})

</math>
Here m_e is the electron mass and h/(m_ec) is known as the "Compton wavelength".
θ is the angle by which the photon's heading changes.
m_e, electron mass, is 9.109 x 10^-31 kilogram.
h, the energy associated with photons, is 6.63×10^-34 joule-second or 4.14×10^-15 electronvolt-second.
c, the speed of light, is 3×10⁸ meters per second.

Collectively, the Compton wavelength is 2.43×10^-12 meter.

[edit]

Derivation

We use that:

<math>E_{\gamma} + E_{e} = E_{\gamma'} + E_{e'}\,</math>

(Conservation of energy, where <math>E_{\gamma}</math> is the energy of a photon before the collision and <math>E_e</math> is the energy of an electron before collision — its rest mass). The variables with a prime are used for those after the collision.
And:

<math>\vec p_{\gamma} + \vec{p_{e}} = \vec{p_{\gamma'}} + \vec{p_{e'}}\,</math>

(Conservation of momentum, with the <math>p_e=0</math> because we assume that the electron is at rest.)
We then use <math>E = hf = pc</math>:

<math>\vec{p_{e'}} = \vec{p_{\gamma}} - \vec{p_{\gamma'}}\,</math>

<math>{\vec{p_{e'}}}^2 = {(\vec{p_{\gamma}} - \vec{p_{\gamma'}})}^2</math>

<math>{\vec{p_{e'}}}^2 = \vec{p_{\gamma}}^2 - 2\cdot\vec{p_{\gamma}}\cdot\vec{p_{\gamma'}} + \vec{p_{\gamma'}}^2</math>

<math>\vec{p_{e'}} \cdot \vec{p_{e'}} = \vec{p_{\gamma}} \cdot \vec{p_{\gamma}}- 2\cdot\vec{p_{\gamma}}\cdot\vec{p_{\gamma'}} + \vec{p_{\gamma'}} \cdot \vec{p_{\gamma'}}</math>

<math>{p_{e'}}^2 \cdot \cos(0) = p_{\gamma}^2 \cdot \cos(0) - 2 \cdot p_{\gamma} \cdot p_{\gamma'} \cdot \cos(\theta) + p_{\gamma'}^2\cdot \cos(0)</math>

The <math>\cos(\theta)</math> term appears because the momenta are spacial vectors, all of which lie in a single 2D plane, thus their inner product is the product of their norms multiplied by the cosine of the angle between them.
substituting <math>p_{\gamma}</math> with <math>\frac{hf}{c}</math> and <math>p_{\gamma'}</math> with <math>\frac{hf'}{c}</math>, we derive

<math>p_{e'}^2 = \frac{h^2 f^2}{c^2} + \frac{h^2 f'^2}{c^2} - \frac{2h^2 ff'\cos{\theta}}{c^2}</math>

Now we fill in for the energy part:

<math>E_{\gamma} + E_{e} = E_{\gamma'} + E_{e'}\,</math>

We solve this for p_e':

Then we have two equations for <math>p_{e'}^2</math>, which we equate:

<math>\frac{(hf + mc^2-hf')^2 -m^2c^4}{c^2} = \frac{h^2 f^2}{c^2} + \frac{h^2 f'^2}{c^2} - \frac{2h^2 ff'\cos{\theta}}{c^2}</math>

Now it's just a question of rewriting:

<math>h^2f^2+h^2f'^2-2h^2ff'+2h(f-f')mc^2 = h^2f^2+h^2f'^2-2h^2ff'\cos{\theta}\,</math>

<math>-2h^2ff'+2h(f-f')mc^2 = -2h^2ff'\cos{\theta}\,</math>

<math>hff'-(f-f')mc^2 = hff'\cos{\theta}\,</math>

<math>hff'(1-\cos{\theta}) = (f-f')mc^2\,</math>

<math>h\frac{c}{\lambda'}\frac{c}{\lambda}(1-\cos{\theta}) =\left(\frac{c}{\lambda}-\frac{c}{\lambda'}\right)mc^2</math>

<math>h\frac{c}{\lambda'}\frac{c}{\lambda}(1-\cos{\theta}) = \left(\frac{c\lambda'}{\lambda\lambda'}-\frac{c\lambda}{\lambda'\lambda}\right)mc^2</math>

<math> h(1-\cos{\theta}) = \frac{\lambda'}{c}\frac{\lambda}{c}\left(\frac{c\lambda'}{\lambda'\lambda}-\frac{c\lambda}{\lambda\lambda'}\right)mc^2</math>

<math>h(1-\cos{\theta}) = \left(\frac{\lambda'}{c}-\frac{\lambda}{c}\right)mc^2</math>

<math>\frac{h}{mc}(1-\cos{\theta}) =\lambda'-\lambda</math>

[edit]