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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] .)

(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.
(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 (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, is scheduled for trial launches in the first part of this decade. It will have a 40-m 2 sail. (credit: Kim Newton/NASA)

Relativistic photon momentum

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

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

or

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

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

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

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

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.

Photon energy and momentum

Show that p = E / c size 12{p = E"/c"} {} for the photon considered in the [link] .

Strategy

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

Solution

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

p = E c = ( 2.48 eV ) ( 1 . 60 × 10 –19 J/eV ) 3 . 00 × 10 8 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"} {}

Discussion

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 / c size 12{p = E"/c"} {} . This also means the relationship between energy, momentum, and mass given by E 2 = ( 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 p size 12{p} {} is not zero, even when m size 12{m} {} is.

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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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