Final answer:
The momentum of a proton whose kinetic energy is half of its rest energy can be calculated using relativistic energy and momentum relationships, resulting in a momentum of 1049 MeV/c.
Step-by-step explanation:
The question pertains to the concept of relativistic kinetic energy and momentum in physics. Given that the kinetic energy (KE) of a proton is half of its rest energy (mc²), we know from special relativity that the total energy E is the sum of its rest energy and kinetic energy: E = mc² + KE. Since KE is (1/2)mc², the total energy E is (3/2)mc². Using the rest mass energy of a proton which is approximately 938.3 MeV, the total energy is 938.3 MeV * 3/2 = 1407.45 MeV. We can now employ the relativistic relationship between energy, mass, and momentum: E² = (pc)² + (mc²)², where p is the momentum and c is the speed of light. Solving for p, we get:
- p = √[(E/c)² - (m0c)²]
- p = √[((1407.45 MeV/c)/c)² - (938.3 MeV/c)²]
- p = √[(1407.45² - 938.3²) MeV²/c²]
- p = 1049 MeV/c
Therefore, the momentum of the proton expressed in MeV/c is 1049.