Learning Objectives

Learning Objectives

By the end of this section, you will be able to do the following:

  • Relate the linear momentum of a photon to its energy or wavelength, and apply linear momentum conservation to simple processes involving the emission, absorption, or reflection of photons
  • Account qualitatively for the increase of photon wavelength that is observed, and explain the significance of the Compton wavelength

The information presented in this section supports the following AP® learning objectives and science practices:

  • 5.D.1.6 The student is able to make predictions of the dynamical properties of a system undergoing a collision by application of the principle of linear momentum conservation and the principle of the conservation of energy in situations in which an elastic collision may also be assumed. (S.P. 6.4)
  • 5.D.1.7 The student is able to classify a given collision situation as elastic or inelastic, justify the selection of conservation of linear momentum and restoration of kinetic energy as the appropriate principles for analyzing an elastic collision, solve for missing variables, and calculate their values. (S.P. 2.1, 2.2)

Measuring Photon Momentum

Measuring Photon Momentum

The quantum of EM radiation we call a photon has properties analogous to those of particles we can see, such as grains of sand. A photon interacts as a unit in collisions or when absorbed, rather than as an extensive wave. Massive quanta, like electrons, also act like macroscopic particles—something we expect, because they are the smallest units of matter. Particles carry momentum as well as energy. Despite photons having no mass, there has long been evidence that EM radiation carries momentum. In fact, Maxwell and others who studied EM waves predicted that they would carry momentum. It is now a well-established fact that photons do have momentum. In fact, photon momentum is suggested by the photoelectric effect, where photons knock electrons out of a substance. Figure 12.17 shows macroscopic evidence of photon momentum.

(a) Trajectory of a comet with a nucleus and tail as it passes by the Sun is shown as a partial parabolic path with Sun near the vertex of the parabolic path. (b) The photograph of a moving Hale Bopp comet in space is shown as bright lighted object.
Figure 12.17 The tails of the Hale-Bopp comet point away from the Sun, evidence that light has momentum. Dust emanating from the body of the comet forms this tail. Particles of dust are pushed away from the Sun by light reflecting from them. The blue ionized gas tail is also produced by photons interacting with atoms in the comet material. (Geoff Chester, U.S. Navy, via Wikimedia Commons)

Figure 12.17 shows a comet with two prominent tails. What most people do not know about the tails is that they always point away from the sun rather than trailing behind the comet. Comet tails are composed of gases and dust evaporated from the body of the comet and ionized gas. The dust particles recoil away from the sun when photons scatter from them. Evidently, photons carry momentum in the direction of their motion, away from the sun, and some of this momentum is transferred to dust particles in collisions. Gas atoms and molecules in the blue tail are most affected by other particles of radiation, such as protons and electrons emanating from the sun, rather than by the momentum of photons.

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 Figure 12.18) 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

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

where hh size 12{h} {} is Planck’s constant and λλ size 12{λ} {} is the photon wavelength. Note that relativistic momentum given as p=γmup=γmu 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.
Figure 12.18 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/λp=h/λ size 12{p = h/λ} {} and hh 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.

Example 12.5 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?


Finding the photon momentum is a straightforward application of its definition: p=hλp=hλ 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

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

Entering the given photon wavelength yields

12.24 p=6.63× 10–34 J s500×10–9 m= 1.33 × 10–27 kg m/s.p=6.63× 10–34 J s500×10–9 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=mvp=mv size 12{p= ital "mv"} {} to find the velocity of an electron with this momentum. Solving for vv size 12{v} {} and using the known value for the mass of an electron gives

12.25 v=pm=1.33×10–27 kg m/s9.11×10–31 kg= 1460 m/s 1,460 m/s.v=pm=1.33×10–27 kg m/s9.11×10–31 kg= 1460 m/s 1,460 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"} } =" 1,460 m/s"} {}

Solution for (c)

The electron has kinetic energy, which is classically given by

12.26 KEe=12mv2.KEe=12mv2. size 12{"KE" rSub { size 8{e} } = { {1} over {2} } ital "mv" rSup { size 8{2} } } {}


12.27 KEe=12(9.11×10–3 kg)(1,455 m/s)2= 9.64×10–25 J.KEe=12(9.11×10–3 kg)(1,455 m/s)2= 9.64×10–25 J.

Converting this to eV by multiplying by (1 eV)/(1.602×10–19J)(1 eV)/(1.602×10–19J) size 12{ \( "1 eV" \) / \( 1 "." "602" times "10" rSup { size 8{"–19"} } `J \) } {} yields

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

The photon energy EE is

12.29 E=hcλ= 1,240 eV nm500 nm=2.48 eV,E=hcλ= 1,240 eV nm500 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.


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 1,460 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 made of aluminized polyester resin that use huge low-mass mirrors 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 Figure 12.19.)

(a) A payload having an umbrella-shaped solar sail attached to it is shown. The direction of movement of payload and direction of incident photons are shown using arrows. (b) A photograph of the top view of a silvery space sail.
Figure 12.19 (a) Space sails have been proposed that use the momentum of sunlight reflecting from gigantic low-mass sails to propel spacecraft about the solar system. A Russian test model of this called the Cosmos 1) was launched in 2005, but did not make it into orbit due to a rocket failure. (b) A U.S. version of this, labeled LightSail-1 will have a 40-m2 sail. (Kim Newton/NASA)

Relativistic Photon Momentum

Relativistic Photon Momentum

There is a relationship between photon momentum pp size 12{p} {} and photon energy EE size 12{E} {} that is consistent with the relation given previously for the relativistic total energy of a particle as E2=(pc)2+(mc)2E2=(pc)2+(mc)2 size 12{E rSup { size 8{2} } = \( ital "pc" \) rSup { size 8{2} } + \( ital "mc" \) rSup { size 8{2} } } {}. We know mm size 12{m} {} is zero for a photon, but pp size 12{p} {} is not, so that E2=(pc)2+(mc)2E2=(pc)2+(mc)2 size 12{E rSup { size 8{2} } = \( ital "pc" \) rSup { size 8{2} } + \( ital "mc" \) rSup { size 8{2} } } {} becomes

12.30 E=pc,E=pc, size 12{E = ital "pc"} {}


12.31 p=Ec(photons).p=Ec(photons). size 12{p = { {E} over {c} } } {}

To check the validity of this relation, note that E=hc/λE=hc/λ size 12{E = ital "hc"/λ} {} for a photon. Substituting this into p=E/cp=E/c size 12{p = E"/c"} {} yields

12.32 p=hc/λ/c=hλ,p=hc/λ/c=hλ, size 12{p = left ( ital "hc"/λ right )/c = { {h} over {λ} } } {}

as determined experimentally and discussed above. Thus, p=E/cp=E/c size 12{p = E"/c"} {} is equivalent to Compton’s result p=h/λp=h/λ size 12{p = h/λ} {}. For a further verification of the relationship between photon energy and momentum, see Example 12.6.

Photon Detectors

Almost all detection systems talked about thus far—eyes, photographic plates, photomultiplier tubes in microscopes, and CCD cameras—rely on particle-like properties of photons interacting with a sensitive area. A change is caused and either the change is cascaded or zillions of points are recorded to form an image we detect. These detectors are used in biomedical imaging systems, and there is ongoing research into improving the efficiency of receiving photons, particularly by cooling detection systems and reducing thermal effects.

Example 12.6 Photon Energy and Momentum

Show that p=E/cp=E/c size 12{p = E"/c"} {} for the photon considered in the Example 12.5.


We will take the energy EE size 12{E} {} found in Example 12.5, divide it by the speed of light, and see if the same momentum is obtained as before.


Given that the energy of the photon is 2.48 eV and converting this to joules, we get

12.33 p=Ec=(2.48 eV)(1.60 × 10–19 J/eV)3.00 × 108 m/s= 1.33 × 10–27 kg m/s.p=Ec=(2.48 eV)(1.60 × 10–19 J/eV)3.00 × 108 m/s= 1.33 × 10–27 kg m/s. size 12{p = { {E} over {c} } = { { \( 2 "." "48 eV" \) \( 1 "." "60 " times " 10" rSup { size 8{"–19"} } " J/eV" \) } over {3 "." "00 " times " 10" rSup { size 8{8} } " m/s"} } =" 1" "." "33 " times " 10" rSup { size 8{"–27"} } " kg " cdot " m/s"} {}


This value for momentum is the same as found before (note that unrounded values are used in all calculations to avoid even small rounding errors), an expected verification of the relationship p=E/cp=E/c size 12{p = E"/c"} {}. This also means the relationship between energy, momentum, and mass given by E2=(pc)2+(mc)2E2=(pc)2+(mc)2 size 12{E rSup { size 8{2} } = \( ital "pc" \) rSup { size 8{2} } + \( ital "mc" \) rSup { size 8{2} } } {} applies to both matter and photons. Once again, note that pp size 12{p} {} is not zero, even when mm size 12{m} {} is.

Problem-Solving Suggestion

Note that the forms of the constants h= 4.14 × 10–15 eV sh= 4.14 × 10–15 eV s size 12{h =" 4" "." "14 " times " 10" rSup { size 8{"–15"} } " eV " cdot " s"} {} and hc= 1240 eV nmhc= 1240 eV nm size 12{ ital "hc" =" 1240 eV " cdot " nm"} {} may be particularly useful for this section’s Problems and Exercises.