Relativistic Photon Momentum

There is a relationship between photon momentum \(p\) and photon energy \(E\) that is consistent with the relation given previously for the relativistic total energy of a particle as \({E}^{2}=(\text{pc}{)}^{2}+(\text{mc}{)}^{2}\). We know \(m\) is zero for a photon, but \(p\) is not, so that \({E}^{2}=(\text{pc}{)}^{2}+(\text{mc}{)}^{2}\) becomes

\(E=\text{pc},\)

or

\(p=\cfrac{E}{c}(photons).\)

To check the validity of this relation, note that \(E=\text{hc}/\lambda \) for a photon. Substituting this into \(p=E/c\) yields

\(p=(\text{hc}/\lambda )/c=\cfrac{h}{\lambda },\)

as determined experimentally and discussed above. Thus, \(p=E/c\) is equivalent to Compton’s result \(p=h/\lambda \). For a further verification of the relationship between photon energy and momentum, see this example.

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: Photon Energy and Momentum

Show that \(p=E/c\) for the photon considered in the this example.

Strategy

We will take the energy \(E\) found in this example, 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

\(p=\cfrac{E}{c}=\cfrac{(2.48 eV)(1\text{.}\text{60}×{\text{10}}^{\text{–19}}\phantom{\rule{0.25em}{0ex}}\text{J/eV})}{3\text{.}\text{00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}}=\text{1}\text{.}\text{33}×{\text{10}}^{\text{–27}}\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot \text{m/s}.\)

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 \(p=E/c\). This also means the relationship between energy, momentum, and mass given by \({E}^{2}=(\text{pc}{)}^{2}+(\text{mc}{)}^{2}\) applies to both matter and photons. Once again, note that \(p\) is not zero, even when \(m\) is.

Problem-Solving Suggestion

Note that the forms of the constants \(h=\text{4}\text{.}\text{14}×{\text{10}}^{\text{–15}}\phantom{\rule{0.25em}{0ex}}\text{eV}\cdot \text{s}\) and \(\text{hc}=\text{1240 eV}\cdot \text{nm}\) may be particularly useful for this section’s Problems and Exercises.

This lesson is part of:

Introduction to Quantum Physics

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