The Compton Effect

The Compton Effect (in physics)‏ The scattering of photons by high-energy photons High-energy X-ray photons hitting a metal foil eject electrons and also scatter lower-energy X-ray photons Similar to the photoelectric effect except higher energies APPLETS erials/ModPhys/QM/Compton/compton.html

Energy Analysis Much like the photoelectric effect, Compton also used the particle theory to explain his results Would energy be conserved?YES! Substituting in known values for energy we have:

Momentum Analysis What about momentum? Is it conserved in the collision? How can a 'particle' with no mass have momentum? *Flashback* → E = mc 2 So X-rays with energy E have a mass-equivalence of E/c 2 Since p = mv, we have With v = c for a photon,

Momentum Analysis Our expression for momentum doesn't include mass! Since we know that E = hf for photons, and using the universal wave equation c = f۟λ The magnitude of the momentum of a photon p in kg*m/s is equal to Planck's constant h divided by the wavelength λ in metres Momentum is conserved in X-ray scattering collisions Momentum of a Photon

Implications Dr. Compton's analysis supported the particle theory of light Photons have a discrete energy and value for momentum Compton won the Nobel Prize in 1927 for his work

Example A photon with wavelength λ = 6.00 x 10^-12 [m] collides with an electron. After the collision the photon's wavelength is changed by exactly one Compton wavelength (λ = 2.43 x 10^-12 [m]). Find: a) The photon's wavelength after the collision b) The energy of the photon after the collision expressed in [keV] Solution: a) Because the photon loses energy to the electron, the photon's frequency must decrease (E = hf). Since c = fλ, the wavelength must increase.

Solution b) E = hf = hc/λ E = 2.36 x 10^-14 [J] E = 147 [keV] Therefore the energy of the photon after the collision is E = 147 [keV].

Homework p. 607 Q# {Dr. A.H. Compton}

de Broglie Wavelength Pronounced 'de-Broy', Not 'de Brog-lie'

The Wave Nature of Matter From the Compton Effect, we know that a photon with wavelength λ has momentum However, a particle with momentum p has a wavelength λ As usual, units of mass m are [kg] and speed v is [m/s] In 1923, Louis de Broglie proposed this then ridiculous idea that is now known as the de Broglie wavelength

Implications A wavelength associated with particles of non-zero mass became known as a matter wave Matter waves – the name given to wave properties associated with matter de Broglie's work on the electron discovered that it diffracts revealing wave characteristics de Broglie won the 1929 Nobel Prize in physics for his work on electron analysis [1 st Nobel Prize to be awarded on the basis of a PhD thesis!]

Example If you are moving down the hall at a speed of v = 1.0 m/s, what is your de Broglie wavelength? [If you do not know your mass (in kg), just use m = 70 kg.] Solution: λ = 6.63 x 10^-34 / [1*(70)] λ = 9.47 x 10^-36 [m] Therefore your de Broglie wavelength is λ = 9.47 x 10^-36 [m] It is obvious that the de Broglie wavelength of large objects (humans) is incredibly small compared to that of microscopic particles (accelerated electrons have a dB wavelength of order 10^-10 m)

Homework p. 614 Q# 1-6 Prince Louis-Victor-Pierre-Raymond, 7th duc de Broglie