What is the initial kinetic energy of the proton (in eV) given it stops from 7.0×10^5 m/s?

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Study for the UCF PHY2054 General Physics Exam. Use flashcards and multiple-choice questions complete with hints and explanations. Boost your understanding and get exam-ready!

Multiple Choice

What is the initial kinetic energy of the proton (in eV) given it stops from 7.0×10^5 m/s?

Explanation:
The initial kinetic energy is found from KE = 1/2 m v^2. Since a proton’s speed here is much less than the speed of light, we can use the nonrelativistic formula. With m ≈ 1.673×10^-27 kg and v = 7.0×10^5 m/s, v^2 = 4.9×10^11, so KE ≈ 0.5 × 1.673×10^-27 × 4.9×10^11 ≈ 4.10×10^-16 J. Converting to eV using 1 eV = 1.602×10^-19 J gives KE ≈ (4.10×10^-16)/(1.602×10^-19) ≈ 2.56×10^3 eV, i.e., about 2560 eV.

The initial kinetic energy is found from KE = 1/2 m v^2. Since a proton’s speed here is much less than the speed of light, we can use the nonrelativistic formula. With m ≈ 1.673×10^-27 kg and v = 7.0×10^5 m/s, v^2 = 4.9×10^11, so KE ≈ 0.5 × 1.673×10^-27 × 4.9×10^11 ≈ 4.10×10^-16 J. Converting to eV using 1 eV = 1.602×10^-19 J gives KE ≈ (4.10×10^-16)/(1.602×10^-19) ≈ 2.56×10^3 eV, i.e., about 2560 eV.

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