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Connections: conservation of momentum

Not only is momentum conserved in all realms of physics, but all types of particles are found to have momentum. We expect particles with mass to have momentum, but now we see that massless particles including photons also carry momentum.

Momentum is conserved in quantum mechanics just as it is in relativity and classical physics. Some of the earliest direct experimental evidence of this came from scattering of x-ray photons by electrons in substances, named Compton scattering after the American physicist Arthur H. Compton (1892–1962). Around 1923, Compton observed that x rays scattered from materials had a decreased energy and correctly analyzed this as being due to the scattering of photons from electrons. This phenomenon could be handled as a collision between two particles—a photon and an electron at rest in the material. Energy and momentum are conserved in the collision. (See [link] ) He won a Nobel Prize in 1929 for the discovery of this scattering, now called the Compton effect    , because it helped prove that photon momentum    is given by

p = h λ , size 12{p = { {h} over {λ} } } {}

where h size 12{h} {} is Planck’s constant and λ size 12{λ} {} is the photon wavelength. (Note that relativistic momentum given as p = γ mu size 12{p=γ ital "mu"} {} is valid only for particles having mass.)

Collision of an electron with a photon of energy E equal to h f is shown. The electron is represented as a spherical ball and the photon as an ellipse enclosing a wave. After collision the energy of the photon becomes E prime equal to h f prime and the final energy of an electron K E sub e is equal to E minus E prime. The direction of electron and photon before and after collision is represented by arrows.
The Compton effect is the name given to the scattering of a photon by an electron. Energy and momentum are conserved, resulting in a reduction of both for the scattered photon. Studying this effect, Compton verified that photons have momentum.

We can see that photon momentum is small, since p = h / λ size 12{p = h/λ} {} and h size 12{h} {} is very small. It is for this reason that we do not ordinarily observe photon momentum. Our mirrors do not recoil when light reflects from them (except perhaps in cartoons). Compton saw the effects of photon momentum because he was observing x rays, which have a small wavelength and a relatively large momentum, interacting with the lightest of particles, the electron.

Electron and photon momentum compared

(a) Calculate the momentum of a visible photon that has a wavelength of 500 nm. (b) Find the velocity of an electron having the same momentum. (c) What is the energy of the electron, and how does it compare with the energy of the photon?

Strategy

Finding the photon momentum is a straightforward application of its definition: p = h λ size 12{p = { {h} over {λ} } } {} . If we find the photon momentum is small, then we can assume that an electron with the same momentum will be nonrelativistic, making it easy to find its velocity and kinetic energy from the classical formulas.

Solution for (a)

Photon momentum is given by the equation:

p = h λ . size 12{p = { {h} over {λ} } } {}

Entering the given photon wavelength yields

p = 6 . 63 × 10 –34 J s 500 × 10 –9 m = 1 . 33 × 10 –27 kg m/s . size 12{p = { {6 "." "63 " times " 10" rSup { size 8{"–34"} } " J " cdot " s"} over {"500 " times " 10" rSup { size 8{"–9"} } " m"} } =" 1" "." "33 " times " 10" rSup { size 8{"–27"} } " kg " cdot " m/s"} {}

Solution for (b)

Since this momentum is indeed small, we will use the classical expression p = mv size 12{p= ital "mv"} {} to find the velocity of an electron with this momentum. Solving for v size 12{v} {} and using the known value for the mass of an electron gives

v = p m = 1 . 33 × 10 –27 kg m/s 9 . 11 × 10 –31 kg = 1460 m/s 1460 m/s . size 12{v = { {p} over {m} } = { {1 "." "33 " times " 10" rSup { size 8{"–27"} } " kg " cdot " m/s"} over {9 "." "11 " times " 10" rSup { size 8{"–31"} } " kg"} } =" 1460 m/s"} {}

Solution for (c)

The electron has kinetic energy, which is classically given by

KE e = 1 2 mv 2 . size 12{"KE" rSub { size 8{e} } = { {1} over {2} } ital "mv" rSup { size 8{2} } } {}

Thus,

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
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Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
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"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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