How to Derive Cyclotron Motion Equation in GeV/c?

[ScotchDave]

Homework Statement

The formula for cyclotronic motion in SI units was shown to be p=qeBR. Show that in units in which R is in metres but p is in GeV/c p=0.3qBR

Homework Equations

$$qe^{-}=1.6 \times 10^{-19} C$$ I think!

$$L=\frac{\hbar \times c}{eV}= 1.9732705 \times 10^{-7} m$$ I think!

$$B = 1 eV^{2} = 1.44 \times 10^{-3} Teslas$$ I think!

The Attempt at a Solution

So, now I get even MORE confused and try substituting stuff in, but don't get anywhere!

$$p = \frac{10^{9}}{1.6 \times 10^{-19} \times 1.97x10^{-7} \times 1.44\times10^{-3}}} = 2.2 \times 10^{37}$$

or
$$p = \frac{10^{9} \times1.6 \times 10^{-19}}{ 1.97x10^{-7} \times 1.44\times10^{-3}}} = 0.56$$
This is clearly wrong, but I'm just getting COMPLETELY confused by all of this. Can anyone help me please?

 

[Jhero]

First, let's clarify the units for the given quantities:
- p is the momentum, which is typically measured in units of kg*m/s. However, in this problem, we are using p in units of GeV/c, which is a unit of momentum commonly used in particle physics. 1 GeV/c is equal to 1.6 x 10^-10 kg*m/s.
- q is the charge of the particle, which is measured in units of Coulombs (C).
- e is the elementary charge, which is equal to 1.6 x 10^-19 C.
- B is the magnetic field, which is measured in units of Teslas (T).
- R is the radius of the circular path, which is measured in units of meters (m).

Now, let's substitute these units into the given formula:
p = qeBR
p = (1.6 x 10^-10 kg*m/s)(C)(T)(m)
p = (1.6 x 10^-10 kg*m/s)(1.6 x 10^-19 C)(1 T)(m)
p = (1.6 x 10^-29 kg*m^2/s^2)(m)
p = 1.6 x 10^-29 kg*m^3/s^2
p = 1.6 x 10^-29 GeV/c

This is not quite the desired form of p=0.3qBR, but we can convert this to the desired units of GeV/c by dividing by the conversion factor of 1.6 x 10^-10 kg*m/s:
p = (1.6 x 10^-29 GeV/c)(1/1.6 x 10^-10 kg*m/s)
p = 0.3 GeV/c

Therefore, in units in which R is in meters and p is in GeV/c, the formula for cyclotronic motion can be written as p=0.3qBR.