where again p is the photon momentum, E is the photon energy and c is the speed of light. When you divide the photon energy found in #6 by the photon momentum found in #4, do you get the speed of light?

If not, check your work for questions #4 through #6.

Yes

No

1 answer

To answer your question, we can use the relationship between the energy \(E\) of a photon, its momentum \(p\), and the speed of light \(c\).

The energy of a photon is given by the formula: \[ E = h \nu \] where \(h\) is Planck's constant and \(\nu\) is the frequency of the photon.

The momentum of a photon is given by: \[ p = \frac{E}{c} \]

If you divide the photon energy \(E\) by the photon momentum \(p\), you get: \[ \frac{E}{p} = \frac{E}{\frac{E}{c}} = c \]

Thus, when you divide the photon energy by the photon momentum, you indeed get the speed of light \(c\).

Therefore, the answer to your question is Yes.