An electron with an initial speed of 460,000 m/s is brought to rest by an electric field. What is the electron’s initial kinetic energy in electron volts?

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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

An electron with an initial speed of 460,000 m/s is brought to rest by an electric field. What is the electron’s initial kinetic energy in electron volts?

Explanation:
Kinetic energy is given by KE = 1/2 m v^2, and we use the electron mass m ≈ 9.11×10^-31 kg with the speed v = 4.60×10^5 m/s. Compute KE in joules: 0.5 × 9.11×10^-31 × (4.60×10^5)^2 ≈ 9.6×10^-20 J. Convert to electron volts using 1 eV = 1.602×10^-19 J: KE ≈ (9.6×10^-20) / (1.602×10^-19) ≈ 0.60 eV. The electron’s initial kinetic energy is about 0.60 eV.

Kinetic energy is given by KE = 1/2 m v^2, and we use the electron mass m ≈ 9.11×10^-31 kg with the speed v = 4.60×10^5 m/s. Compute KE in joules: 0.5 × 9.11×10^-31 × (4.60×10^5)^2 ≈ 9.6×10^-20 J. Convert to electron volts using 1 eV = 1.602×10^-19 J: KE ≈ (9.6×10^-20) / (1.602×10^-19) ≈ 0.60 eV. The electron’s initial kinetic energy is about 0.60 eV.

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