# Relativity -- Momentum and energy

1. Sep 13, 2016

### erisedk

1. The problem statement, all variables and given/known data
(a) A proton at rest has energy Eproton = mprotonc2 ≈ 938 MeV. (Its momentum is zero). The protons which circulated inside the Fermilab Bevatron had energies close to 1000 GeV (1GeV = 1000 MeV). What value of γ did a Tevatron proton have?

(b) Somehow a Bevatron proton captures an electron, becoming a fast-moving hydrogen atom without changing its speed. An electron at rest has Eelectron = mec2 ≈ 0.511 MeV. What is the electron's energy in the rest frame of Fermilab's sedentary buffalo herd?

2. Relevant equations
$\gamma = \dfrac{1}{1-\beta^2}$
where $β = \frac{v}{c}$
KE = TE - BE (TE is total energy, BE is rest energy or binding energy)

3. The attempt at a solution
I just want to check if my solution is correct or not.
(a) p (momentum) = $\sqrt{2mK}$
K = 106 MeV - 938 MeV
So, mv = $\sqrt{2mK}$
v = $\sqrt{\frac{2K}{m}}$
Substitute for K = 1.6 × 10-13 J and m = 1.67 × 10-27 kg and get v in m/s
Substitute v in the equation for $\gamma$

(b) Energy = KE + BE
= $\frac{m_e v^2}{2}$ + 0.511 × 106 × 1.6 × 10-19
Substitute in v from above equation, and sub in mass of electron.

Are these calculations right?

2. Sep 13, 2016

### PeroK

What about using $E = \gamma mc^2$

3. Sep 13, 2016

### erisedk

For what part? The second?

4. Sep 13, 2016

### PeroK

Why not both?

5. Sep 13, 2016

### erisedk

So you mean,
(a) $\gamma = \dfrac{10^3}{938}$

and
(b) $E = \dfrac{10^3}{938} × 0.511 MeV$

Why? Also, is my answer wrong?

6. Sep 13, 2016

### PeroK

Those should be $10^6$, otherwise yes.

Note that, for a particle:

$\gamma = E/E_0$

So if you have the energies, there is no need for all those intermediate calculations, which I didn't check.

7. Sep 13, 2016

### erisedk

Oh, okay I get it!
And yeah, that was a careless error, it is 106
Thank you!