Compton Effect

The Compton effect explains the change in wavelength of photons after colliding with an object (and transferring energy). It further supports the particle nature of Light through means of X-ray radiation.

Definition §

The X-rays are emitted as high-energy photons. When they collide with an electron, they transfer a quantised amount of energy to the electron, and lose that energy, due to the conservation of energy. The angle of scattering is accounted for by the conservation of momentum.

The change in energy (of the photon), also results in a change of wavelength. Specifically, the decrease in energy results in an increase in wavelength/

Compton Effect

$$\Delta \lambda ={\lambda }_f-{\lambda }_i={\lambda }_{compton}(1-\cos (\varphi ))=\frac{h}{m_{e}c}(1-\cos (\varphi ))$$

Note

• $\Delta \lambda$ = Change in Wavelength of the photon (in $\text{m}$)
• ${\lambda }_i$ = Wavelength before collision with electron
• ${\lambda }_f$ = Wavelength after collision
• $h$ = Planck’s Constant $\approx 6.63\times 10^{-34}\text{ Js}$
• $m_e$ = Mass of an Electron (in $\text{kg}$)
• $c$ = Speed of light $\approx 3\times 10^8\text{ m/s}$
• $\varphi$ = Angle of photon deflection (NOT ELECTRON!)

Derivation §

The equation for the Compton Effect can be derived using the conservation of energy the conservation of momentum, and the energy-momentum relation of a relativistic particle.

Begin by considering a isolated system consisting of a single photon and electron. Then, we can apply the conservation of energy:

$$E_{sys}=E_{sys}^{\prime }⟹E_{electron}+E_{photon}=E_{electron}^{\prime }+E_{photon}^{\prime }$$

And the conservation of momentum

$$\vec{p}=(\vec{p}{)}^{\prime }⟹{\vec{p}}_e+{\vec{p}}_{ph}=({\vec{p}}_e{)}^{\prime }+({\vec{p}}_{ph}{)}^{\prime }$$

Before the collision, the electron is considered to have zero momentum*

$${\vec{p}}_{ph}=({\vec{p}}_e{)}^{\prime }+({\vec{p}}_{ph}{)}^{\prime }$$

The initial energy of the electron is given by $E=m{c}^2$, and it’s final energy is given by:

$$m_e{c}^2+E_{photon}=\sqrt{(p_{e}c{)}^2+(m_e{c}^2{)}^2}+E_{photon}^{\prime }$$

Then we can use the Planck-Einstein Equation:

$$m_e{c}^2+hf=\sqrt{(p_{e}c{)}^2+(m_e{c}^2{)}^2}+h{f}^{\prime }$$

* Heisenberg principle excluded#tosee

#todo finish proof