Comparing momentum of electron and photon and the correct usage of electron volt

Question

I was trying to solve this question where we had a photon and an electron with the same energy $E$, and we were asked to compare their momentum.

I know, for photons we have $E=p_{\gamma}c$ and for electron we have $E=\frac{p_e^2}{2m}$

Thus, $p_{\gamma} = \frac{E}{c}$ and $p_e = \sqrt{2mE}$

Now my problem is, if the energy is given in electron volts, the electron has a higher momentum than the photon. For example, if the given energy is $E=10\space eV,$ then :

$p_{\gamma} = \frac{E_0 \times1.6\times10^{-19}}{3\times10^8}$ = $5.33 \times 10 \times10^{-28}$

$p_e = \sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-19}\times E_0}$ = $5.4 \times10^{-25} \sqrt{10}$

As we can see, $p_e \gt p_\gamma$.

However, what if the energy is given in joules. In that case, will the momentum of the electron still be greater than the photon ? According to my calculations, it will be the exact opposite.

Is this correct ? There are some materials that state, the momentum of electrons would always be greater than that of photon for the same energy. However, according to my calculations, this is only true if the energy is given in electron volts.

At around $1.023 MeV$, I've found that both of them have the same momentum. I'm inclined to say, that if the energy is below this, then we have an electron with a larger momentum. However, if the energy is more than this then the photon should have the larger momentum. Am I correct in saying this ? Are the materials that claim, electron always have a larger momentum wrong ?

Another question : While using the formula $p=\sqrt{2mE},$ who do we have to convert the energy into joules ? If the energy is given in electron volts, why do we convert it into joules by multiplying with the charge of an electron, after plugging it into the above formula ? I know if I didn't convert it into joules, I'd get wrong results, if the energy is given in electron volts. However, I don't know the theoretical reason behind why that is done.

Does it have something to do with the unit of mass being in $kg$ ? Maybe if we choose to keep the energy in $eV$ we must also convert the mass and the momentum in these natural units ? I'm not sure.

For example, in the example, where the energy was $10$ $eV$, if I didn't convert it into joules, I'd have got the exact opposite results. Can someone please point out the inconsistency, as I know my final answer should not depend on units? Then why must I convert into joules, for the formula to work ?

Answer 1

you should use Joule since otherwise the formula for massless particles or electrons do not give a momentum in $\frac{kg \cdot m}{s}$ For an electron it makes good sense tu gib the energy in eV the energy it gets by falling through 1V calculating this for light divided by c is not so meaningfull. since in the car of slow e you take the squarroot of E in the case of the massless particle you divide E by c. so you would get different dimensions.

Answer 2

First, if you compare the momentums for energies of about 1 MeV, you should use the relativistic formula for the energy of the electron. Second, use everywhere the same system of units, and you should not have a problem.