Photon Energy and Momentum · Physics Playground

Strategy

We will take the energy $E$ found in Example 29.5, 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 = \frac{E}{c} = \frac{(2.48 eV) (1 . 60 \times 10^{–19} J/eV)}{3 . 00 \times 10^{8} m/s} = 1 . 33 \times 10^{–27} kg \cdot m/s .

29.33

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 $E^{2} = (pc)^{2} + (mc)^{2}$ applies to both matter and photons. Once again, note that $p$ is not zero, even when $m$ is.