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

http://www.kcvs.ca/site/projects/physics.html

http://faculty.gvsu.edu/majumdak/public_html/OnlineMaterials/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:

e l ec t r o nR a yXR a yX EEE '

2'

2

1m vh fh f

Momentum Analysis

What about momentum? Is it conserved in the collision?

How can a 'particle' with no mass have momentum?

*Flashback* → E = mc2

So X-rays with energy E have a mass-equivalence of E/c2

Since p = mv, we have

With v = c for a photon,

vcE

p

2

c

Ep

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 = f ۟λ

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

fh f

p h

p 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

ExampleA 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. m1 21 04 3.2

m1 21 04 3.8

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 isE = 147 [keV].

Homework

p. 607 Q# 18-22

{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

h

p

p

h

m v

h

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 [1st 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)

m v

h

Homework

p. 614 Q# 1-6

Prince Louis-Victor-Pierre-Raymond, 7th duc de Broglie