Compton Effect and de Broglie Waves

Compton Effect The photoelectric effect demonstrates that a photon has kinetic energy, even though it has no mass. Einstein predicted that if a photon has kinetic energy, than it must have similar properties to other classical physics such as – momentum – AND HE WAS RIGHT. The tough part is explaining and proving this in a reasonable way, since you don't exactly feel light hammering you into the floor. The momentum that the light photons have must be very small, and not based on the common way of calculating momentum using p = mv (since light/photon has no mass). Instead the formula was based on the wavelength and frequency of the light, just like Planck's formula.

Momentum of a Wave Equation The equation for momentum of a photon (light wave) 𝜌 = ℎ𝑓 𝑐 = ℎ λ 𝜌 = momentum (kg · m/s) h = Planck's Constant (always 6.63e-34 J · s ) λ = wavelength (m) f = frequency (Hz) of (1/s) c = speed of light (m/s)

Compton Effect In 1923 A.H. Compton tested this idea. He started shooting high frequency x-rays at various materials and found that his results seemed to support the idea of photons having momentum. In one setup he shot high frequency x-rays at a piece of graphite (a solid piece of carbon). If light was a wave, we would expect the x-rays would go through the carbon and come out the other side with their wavelength smaller Basically we can explain this as the waves squishing when they hit the graphite, like a ball squishing when it hits the ground.

Compton Effect If light was a wave, it should have behaved like the illustration above.

Compton Effect But this is not what happened. Instead, Compton found that the x-rays scattered after hitting the target, changing the direction they were moving and actually getting a longer wavelength. Remember, longer wavelength means smaller frequency. Since E = hf, the scattered photons had less energy! Somehow, the x-ray photons were losing energy going through the graphite. So where did the energy go? He also found that electrons were being thrown off the target at an angle. Note the work function of carbon is 4.7−4.9 eV. So it will release electrons if the proper wave hits it.

Compton Effect This is what actually happened, and this is what happens when two objects collide, like in momentum, and what happens with energy transfer.

Compton Effect Compton was able to explain all he was seeing (which became known as the Compton Effect)by using principles of classical physics and quantum mechanics: The conservation of energy (the energy the photon lost had to go somewhere). The conservation of momentum (to explain the angles things were shooting off at). The photon theory of light (to figure out the momentum of the photons) Classical Physics Quantum Mechanics

Compton Effect If we looked at this in terms of momentum, we'd need to be careful about using the correct momentum equations for each part. Incident x-rays The original x-rays have a small wavelength, and the formula p=h/λ shows us that this means it has a lot of momentum. Scattered x-rays The x-rays that made it through have a bigger wavelength, so p=h/λ means it has less momentum. Ejected electrons We would calculate the electron's momentum using a classic p=mv calculation.

Compton Effect This image shows all three principles that Compton observed: 1) Law of Conservation of Energy 2) Law of Conservation of Momentum 3) Using light theory - identifying the momentum of the scattered photon – we can figure out the velocity of the electron. TADA light can act both like a wave and a particle - the math works.

Compton Effect recoil electron target electron at rest incident photon scattered photon

De Broglie Waves Now if light can act like both a wave and a particle, then can an object (like a baseball) act like a wave? In 1923 Prince Louis de Broglie proposed this as a new idea in physics. All the stuff discovered so far has shown that what appeared to be waves (EMR) sometimes acted like a particle, so de Broglie just wanted to know if particles could act like waves. Nobody really took de Broglie seriously until Einstein read his paper and agreed with his ideas.

De Broglie Equation de Broglie suggested combining a couple of formulas, one of them a particle formula, the other a wave formula. If p = mv and p = h/ λ  then mv = h/ λ or… λ = h /mv  This is de Broglie equation WHERE λ = wavelength (m) h = Planck's Constant (always 6.63 x 10-34 J · s ) m = mass (kg) v = velocity (m/s) Again we see how light can be both a wave and a particle.

De Broglie Waves Example # 1 Determine the de Broglie wavelength of a 0.200 kg ball moving at 15.0 m/s λ = h /mv λ = (6.63 x 10-34 J · s )/(0.200 kg x 15.0 m/s) λ = 2.21 x 10-34 m

De Broglie Waves Example #2 What is the wavelength of an electron (mass = 9.11 x 10 -31 kg) traveling at 5.31 x 106 m/s? λ = h /mv λ = (6.63 x 10-34 J · s )/ (9.11 x 10 -31 kg)(5.31 x 106 m/s ) λ = 1.37 x 10 -10 m

De Broglie Waves Example #3 What is the wavelength in meters of a proton traveling at 255,000,000 m/s and the mass of the proton is 1.673 x 10 -27 kg. Note this proton is travelling 85% of the speed of light. WOWZA that is fast. λ = h /mv λ = (6.63 x 10-34 J · s )/ (1.673 x 10-27 kg)(2.55 x 108 m/s ) λ = 1.55 x 10 -15 m The faster the object and the smaller the is produces a smaller wavelength. Interesting… hmmm…. That means it is easier to measure it.