Question
[ans: 0.25 kg] 2. A proton moves with a speed of 0 .800c through a particle accelerator In the accelerator's frame of reference,calculate (a) the total and (b) the kinetic energies of the proton in megaelectron-volts TA A [ans: (a) 1.57times 10^3 MeV; (b) 6.26times 10^2MeV]
Solution
3.7
(298 Votes)
Brynlee
Master · Tutor for 5 years
Answer
Here's the solution:**Understanding the Problem**We're dealing with relativistic energies here because the proton's speed is a significant fraction of the speed of light (c). We'll need to use the relativistic formulas for total energy and kinetic energy.**Given:*** v = 0.800c (proton's speed)* m₀ (rest mass of a proton) = 938.3 MeV/c² (This is a standard value you should know or have access to)**Formulas:*** **Total Relativistic Energy (E):** E = γm₀c²* **Relativistic Kinetic Energy (K):** K = (γ - 1)m₀c²* **Lorentz Factor (γ):** γ = 1 / sqrt(1 - (v²/c²))**Calculations:****(a) Total Energy (E):**1. **Calculate γ:** γ = 1 / sqrt(1 - (0.800c)²/c²) γ = 1 / sqrt(1 - 0.64) γ = 1 / sqrt(0.36) γ = 1 / 0.6 γ = 1.667 (approximately)2. **Calculate E:** E = γm₀c² E = (1.667) * (938.3 MeV/c²) * c² E = 1564.5 MeV ≈ 1.57 x 10³ MeV**(b) Kinetic Energy (K):**1. **Use the calculated γ:** K = (γ - 1)m₀c² K = (1.667 - 1) * (938.3 MeV/c²) * c² K = (0.667) * 938.3 MeV K = 625.9 MeV ≈ 6.26 x 10² MeV**Final Answers:*** (a) Total Energy: 1.57 x 10³ MeV* (b) Kinetic Energy: 6.26 x 10² MeVTherefore, the provided answers in the original question are correct.