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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,

KE e = 1 2 ( 9 . 11 × 10 –3 kg ) ( 1455 m/s ) 2 = 9.64 × 10 –25 J .

Converting this to eV by multiplying by ( 1 eV ) / ( 1 . 602 × 10 –19 J ) size 12{ \( "1 eV" \) / \( 1 "." "602" times "10" rSup { size 8{"–19"} } `J \) } {} yields

KE e = 6.02 × 10 –6 eV . size 12{"KE" rSub { size 8{e} } =" 6" "." "06 " times " 10" rSup { size 8{"–6"} } " eV"} {}

The photon energy E is

E = hc λ = 1240 eV nm 500 nm = 2 . 48 eV , size 12{E = { { ital "hc"} over {λ} } = { {" 1240 eV " cdot " nm"} over {"500"" nm"} } = 2 "." "48"" eV"} {}

which is about five orders of magnitude greater.

Discussion

Photon momentum is indeed small. Even if we have huge numbers of them, the total momentum they carry is small. An electron with the same momentum has a 1460 m/s velocity, which is clearly nonrelativistic. A more massive particle with the same momentum would have an even smaller velocity. This is borne out by the fact that it takes far less energy to give an electron the same momentum as a photon. But on a quantum-mechanical scale, especially for high-energy photons interacting with small masses, photon momentum is significant. Even on a large scale, photon momentum can have an effect if there are enough of them and if there is nothing to prevent the slow recoil of matter. Comet tails are one example, but there are also proposals to build space sails that use huge low-mass mirrors (made of aluminized Mylar) to reflect sunlight. In the vacuum of space, the mirrors would gradually recoil and could actually take spacecraft from place to place in the solar system. (See [link] .)

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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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?
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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
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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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