# Momentum of neutron, proton, electron

1. Jan 27, 2013

### wahaj

If a proton, a neutron and an electron have the same total energy than rank the the particles in terms of most momentum to least momentum

This is the first problem I attempted after learning about special relativity. I used the equation
$$E^2 = \rho^2 c^2 + (m c^2 )^2$$

Since the energy is constant I gave it a value of 1 so
$$1 = \rho^2 c^2 + (m c^2 )^2$$
rearranging the above equation I get
$$\rho = \sqrt {\frac{1}{c^2} - m^2 c^2 }$$

putting in values for m I get the energy for all 3 particles to be 3.3 nJ.
I have two questions
1) did I do this right? if not then where did I go wrong?
2) If I did this right then why is the energy the same for all particles? Since the mass is different for all three particles wouldn't the momentum also be different. I can understand the proton and neutron having the same energy because their masses are almost the same but the electron's mass is significantly different than a proton's and neutron's.

**So far I have only had an intro to special relativity, I know the transformation equations for space, time, velocity, linear momentum and energy. keep that in mind when answering my question**

2. Jan 27, 2013

### Staff: Mentor

First, a quick note: you appear to be writing $\rho$ for momentum; the usual symbol is $p$. It's a minor point, but it will make it easier for others to understand what you are writing (and also easier for you to compare your notes with the literature).

But you already said the energy was 1. Where does the 3.3 nJ come from?

Energy and momentum are not the same thing. The energy is the same because you made it be the same; it's 1 for all the particles (see above). Your formula for $\rho$ (which should really be written as $p$, see above) gives the momentum; the only variable in it is $m$, and your formula makes it evident that as $m$ gets bigger, the momentum $\rho$ gets smaller.

3. Jan 27, 2013

### wahaj

in my book the p was written fancy so I thought it might be rho. guess not. and I messed up the units its not supposed to be joules is supposed to be kg·m/s. I always mix up units for some reason.
Like you said as my mass gets bigger momentum gets smaller meaning that the momentum for the three particles should be different. So my solution is wrong. Can you tell me where I went wrong?

4. Jan 27, 2013

### Staff: Mentor

Your equation for $\rho$ is correct, and if you plug in different values for $m$, you will get different values for $\rho$. Have you done that?

5. Jan 27, 2013

### wahaj

I have but my calculator is giving me the same 3.3 * 10-9 for all masses

6. Jan 27, 2013

### Staff: Mentor

Masses, or momentum values? I assume you mean momentum, but the fact that you have confused your terms a couple of times now (energy vs. momentum, and now mass vs. momentum) is a sign that you may want to step back and take things a bit slower.

If your calculator can't give distinguishable answers for this problem, then your calculator is not accurate enough. Anyway, this calculation is simple enough that you should be able to do it by hand.

Also, you assigned an energy of "1" to all three particles; but "1" in what units? 1 Joule? That would seem the simplest since you appear to be using SI units; but that is a very, very large energy for a single particle, so you will have to go to many, many decimal places to see a difference in momentum.

Last edited: Jan 27, 2013
7. Jan 27, 2013

### wahaj

I guess. I am using a ti 84 I was't expecting it to run out of memory so fast.
thanks for helping me out. at least now I know that I should do calculations for this class by hand.

8. Jan 27, 2013

### Staff: Mentor

Not necessarily; did you try it on the calculator with a much smaller value than "1" for the energy? For example, try it with an energy around 10^-10.

Edit: I should clarify that I am not saying you shouldn't try to do calculations by hand whenever possible. Before you do any calculation with a calculator you should already know what you expect the answer to be, at least to a rough order of magnitude, which means you should at least know how to do rough order of magnitude calculations by hand.

Last edited: Jan 27, 2013
9. Jan 27, 2013

### wahaj

nope its still the same. I've checked several times and I don't think I am entering the numbers in wrong. I tried doing this by hand but 9E-16 - 2.5178E-38 is kinda impossible to do. compared to the left number the right number is virtually 0. taking the square root of 9E-16 gives me 3.3E-9 which is the answer I've been getting all along. So I understand what the calculator is doing. instead of 1 I did put in 10E-10 but that gave me the same answer.

10. Jan 27, 2013

### Staff: Mentor

You don't actually need to calculate the momentum of any of them, just rank them. Looking at the equation you can see that the more the mass the less the momentum. Therefore, simply rank them in reverse order of mass.

11. Jan 27, 2013

### wahaj

yes that would be the way to go. thanks for help everyone

12. Jan 27, 2013

### Staff: Mentor

Not if you're just trying to figure a rough order of magnitude. This is telling you that, as I said, 1 Joule is an enormous energy for a single particle, much larger than the energy equivalent of its mass for either a proton (or neutron) or an electron. So practically all of its energy is kinetic energy; you have to go to the 22nd decimal place to see the rest energy (mc^2) for a proton, and even more decimal places to see it for an electron.

You put in 10^E-10 for what? Remember that the "1" in your original formula represents E^2, not E. If E is around 10^-10, then E^2 is around 10^-20, so that's what needs to go in place of the "1".

But again, before you do anything with the calculator you should first work through the rough order of magnitude calculation by hand. That tells you that the m^2 c^2 term is something like 10^-38 for the neutron and proton, and something like 10^-44 for the electron. So whatever the E^2/c^2 term is, you're subtracting a much smaller number from it for the electron than for the proton, therefore the electron's momentum will always end up higher. It may only be higher when you go to a lot of decimal places, but it will always be higher. So you can put the momentum values in order without ever having to use your calculator at all.

Also, once you know the m^2 c^2 terms, you can see what E^2 / c^2 has to be to be of the same general order of magnitude. That tells you what sort of E values you can usefully punch into the calculator to see the difference in momentum values show up before you've gone to too many decimal places.

13. Jan 27, 2013

### wahaj

ok i got this thanks