Convincing evidence that light is made up of particles (photons), and that photons have momentum, can be seen when a photon with energy hf collides with a stationary electron. Some of the energy and momentum is transferred to the electron (this is known as the Compton effect), but both energy and momentum are conserved in this elastic collision. After the collision the photon has energy hf^{/} and the electron has acquired a kinetic energy K.

Conservation of energy: hf = hf^{/} + K

Combining this with the momentum conservation equations, it can be shown that the wavelength of the outgoing photon is related to the wavelength of the incident photon by the equation:

Dl = l^{/} - l = (h/m_{e}c) (1 - cosq)

The combination of factors h/m_{e}c = 2.43 x 10^{-12} m, where m_{e} is the mass of the electron, is known as the Compton wavelength. The collision causes the photon wavelength to increase by somewhere between 0 (for a scattering angle of 0°) and twice the Compton wavelength (for a scattering angle of 180°).

A photon with a wavelength of 6.00 x 10^{-12} m collides with an electron. After the collision the photon's wavelength is found to have been changed by exactly one Compton wavelength (2.43 x 10^{-12} m).

(a) What is the photon's wavelength after the collision?

- 3.57 x 10
^{-12}m - 8.43 x 10
^{-12}m - It could be either one of the above

The photon gives up some of its energy to the electron. If its energy goes down its frequency decreases and its wavelength increases.

(b) Through what angle has the photon been deflected in this collision?

- less than 90°
- 90°
- more than 90° but less than 180°
- 180°

Dl = (h/m_{e}c) (1 - cosq)

In this situation we have (1 - cosq) = 1 so cosq = 0 and q must be 90°.

(c) What is the angle for the electron after the collision?

- less than 90°
- 90°
- more than 90° but less than 180°
- 180°

To keep the total momentum the same the electron has both x and y momentum, so its angle must be less than 90°.

(d) What is the electron's kinetic energy, in eV, after the collision?

Energy is conserved in the collision, so: hf = hf^{/} + K

K = hf - hf^{/} = hc/l - hc/l^{/} = hc [ 1/l - 1/l^{/} ]

K = (6.63 x 10^{-34}) (3 x 10^{8}) [ 1/6.00x10^{-12} - 1/8.43x10^{-12}]

K = 9.56 x 10^{-15} J = 59700 eV