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Chapter 17: Problem 47

An alpha particle (charge \(+2 e\) ) moves through a potential difference \(\Delta V=-0.50 \mathrm{kV} .\) Its initial kinetic energy is $1.20 \times 10^{-16} \mathrm{J} .$ What is its final kinetic energy?

Short Answer

Expert verified
Answer: The final kinetic energy of the alpha particle is -4.0 × 10^-18 J.

Step by step solution

01

Identify the relevant equations

We will use the following equation to find the change in potential energy that results from the particle moving through the potential difference: Change in Potential Energy (ΔE) = Charge (q) × Potential Difference (ΔV)
02

Calculate the charge of the alpha particle

An alpha particle has a charge of +2e, where e is the elementary charge. The elementary charge, e, is approximately 1.6 × 10^-19 Coulombs (C). Therefore, the charge of the alpha particle (q) is: q = (+2)(1.6 × 10^-19 C) = 3.2 × 10^-19 C
03

Calculate the change in potential energy

Using the equation from step 1, we can now calculate the change in potential energy as the particle moves through the potential difference ΔV: ΔE = q × ΔV = (3.2 × 10^-19 C)(-0.50 × 10^3 V) = -1.6 × 10^-16 J
04

Calculate the final kinetic energy

The final kinetic energy of the alpha particle is the sum of its initial kinetic energy and the change in potential energy: Final Kinetic Energy = Initial Kinetic Energy + ΔE Final Kinetic Energy = (1.20 × 10^-16 J) + (-1.60 × 10^-16 J) = -4.0 × 10^-18 J So, the final kinetic energy of the alpha particle is -4.0 × 10^-18 J.

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