When Is Kinetic Energy Equal to Rest Energy for a Proton?

mattst88
Messages
28
Reaction score
0

Homework Statement



What is the speed of a proton when its kinetic energy is equal to its rest energy?

Homework Equations



K = mc^2(\gamma - 1)
E_0 = mc^2
\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}

The Attempt at a Solution



K = E_0
mc^2(\gamma - 1) = mc^2
\gamma = 2
\frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} = 2
0.5 = \sqrt{1 - \frac{v^2}{c^2}}
0.5^2 = 1 - \frac{v^2}{c^2}
c^2 \sqrt{1 - 0.25} = v^2
v = c \sqrt{0.75} = 0.866 c

Am I right to use relativistic energy? Have I come to the correct answer? Please advise.
 
Physics news on Phys.org
correct (for any massive particle, not just the proton)
 
It's given a gas of particles all identical which has T fixed and spin S. Let's ##g(\epsilon)## the density of orbital states and ##g(\epsilon) = g_0## for ##\forall \epsilon \in [\epsilon_0, \epsilon_1]##, zero otherwise. How to compute the number of accessible quantum states of one particle? This is my attempt, and I suspect that is not good. Let S=0 and then bosons in a system. Simply, if we have the density of orbitals we have to integrate ##g(\epsilon)## and we have...
Back
Top