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[...]atom bombe[...]

 

Fint det, kanskje du snart skal øse av din uendelige visdom, og vise sammenhengen mellom atombombe, ICBM og månelandinger? :roll:

The Compton shift formula

A photon of wavelength \lambda \, comes in from the left, collides with a target at rest, and a new photon of wavelength \lambda ' \, emerges at an angle \theta \,.

See also: Klein-Nishina formula

 

Compton used a combination of three fundamental formulas representing the various aspects of classical and modern physics, combining them to describe the quantum behavior 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:

 

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

 

where

 

\lambda\, is the wavelength of the photon before scattering,

\lambda'\, is the wavelength of the photon after scattering,

me is the mass of the electron,

\theta\, is the angle by which the photon's heading changes (between 0° and 180°),

h is Planck's constant, and

c is the speed of light.

 

\frac{h}{m_e c} = 2.43 \times 10^{-12}\,\text{m} is known as the Compton wavelength. λ' − λ can be between 0 (for θ = 0°) and two times the Compton wavelength (for θ = 180°).

 

[edit] Derivation

 

Begin with conservation of energy and conservation of momentum:

 

E_\gamma + E_e = E_{\gamma^\prime} + E_{e^\prime} \quad \quad (1) \,

\vec p_\gamma = \vec{p}_{\gamma^\prime} + \vec{p}_{e^\prime} \quad \quad \quad \quad \quad (2) \,

 

where

 

E_\gamma \, and p_\gamma \, are the energy and momentum of the photon and

E_e \, and p_e \, are the energy and momentum of the electron.

 

[edit] Solving (Part 1)

 

Now we fill in for the energy part:

 

E_{\gamma} + E_{e} = E_{\gamma'} + E_{e'}\,

hf + mc^2 = hf' + \sqrt{(p_{e'}c)^2 + (mc^2)^2}\,

 

The square of the second equation gives an equation for pe':

 

p_{e'}^2c^2 = (hf + mc^2-hf')^2-m^2c^4 \quad \quad \quad \quad \quad (3) \,

 

[edit] Solving (Part 2)

 

Rearrange equation (2)

 

\vec{p}_{e'} = \vec{p}_\gamma - \vec{p}_{\gamma'} \,

 

and square it to see

 

p_{e'}^2 = (\vec{p}_\gamma - \vec{p}_{\gamma'}) \cdot (\vec{p}_\gamma - \vec{p}_{\gamma'})

p_{e'}^2 = p_{\gamma}^2 + p_{\gamma'}^2 - 2\vec{p_{\gamma}} \cdot \vec{p_{\gamma'}}

p_{e'}^2 = p_\gamma^2 + p_{\gamma'}^2 - 2|p_{\gamma}||p_{\gamma'}|\cos(\theta) \,

 

Energy and momentum of photons are connected by the relativistic equation p_{\gamma}=\frac{E_{\gamma}}{c}, so E_{\gamma}^2=p_{\gamma}^2c^2=(h f)^2.

 

Therefore, multiplying by c2, we have also

 

p_{e'}^2c^2 = (h f)^2 + (h f')^2 - 2(hf)(h f')\cos{\theta} \quad \quad \quad (4)

 

[edit] Putting it together

 

Now we have the two equations (3 & 4) for p_{e'}^2c^2, which we equate:

 

\left(h f\right)^2 + \left(h f'\right)^2 - 2h^2 ff'\cos{\theta} = (hf + mc^2-hf')^2 -m^2c^4 \,

 

Energies of a photon at 500 keV and an electron after Compton scattering.

 

Next we multiply out the right-hand term (hf + mc2 − hf')2 and cancel square terms on both sides and get:

 

-2h^2ff'\cos{\theta} = -2h^2ff'+2hfmc^2-2hf'mc^2. \,

 

Combining two terms, this becomes:

 

2h^2ff'(1-\cos(\theta)) = 2hfmc^2 - 2hf'mc^2. \,

 

After dividing both sides by \left(2hff'mc^2\right) , we get:

 

\frac{h}{mc^2}\left(1-\cos \theta \right) = \frac{1}{f^\prime} - \frac{1}{f} \,

 

This is equivalent to the Compton scattering equation, but it is usually written in terms of wavelength rather than frequency. To make that switch use

 

f=\frac{c}{\lambda} \,

 

so that finally,

 

\lambda'-\lambda = \frac{h}{mc}(1-\cos{\theta}) \,

] Applications

 

 

 

 

her har du sammenhengen :ohmy:

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