# How to convert Gev/c into m/s

1. Apr 19, 2012

### iamBevan

As the title said :)

I'm trying to find the velocity of a particle with a momentum of between 23 and 150 GeV/c. I found that 1 GeV/c = 5.36 x 10^-19 kg-m/s, and tried to divide by the mass of the particle - this just game me values between 7m/s and some crazy numbers.

What am I doing wrong :(

2. Apr 19, 2012

### Bob S

One way is to use the relation
$$E^2=\left(mc^2 \right)^2+\left(pc \right)^2$$
where $pc$ is 23 to 150 GeV (momentum in energy units), and $mc^2$ is the particle's rest mass (proton is 0.938 GeV). Then use $\beta =pc/E$ to get $\beta$, and $v=\beta c$.

Last edited: Apr 19, 2012
3. Apr 21, 2012

### iamBevan

Thank you for your help Bob S - I still can't manage to get the answer though.

When I set m to 5.5208x10^27kg, and p to 25GeV/c I end up getting a value that is faster than c when I solve for v. Can anyone help with this?

4. Apr 21, 2012

### iamBevan

If 1 GeV/c = 5.36 x 10-19 kg-m/s though, why can't I do 25(5.36x10^-19)/particle's mass?

5. Apr 21, 2012

### Bob S

Using the relation
$$E^2=\left(mc^2 \right)^2+\left(pc \right)^2$$
where $pc=$ 50 GeV and $mc^2=$ 0.938 GeV, E = 50.008798 GeV.
So β= 50/ 50.008798= 0.99982 and βc = 2.9974 x 1010 cm/sec

6. Apr 21, 2012

### Staff: Mentor

Because p ≠ mv, if you're using the particle's "rest mass" in kg. The correct equation is

$$p = \frac{mv}{\sqrt{1 - v^2/c^2}}$$