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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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Most popular questions from this chapter

A parallel plate capacitor is connected to a \(12-\mathrm{V}\) battery. While the battery remains connected, the plates are pushed together so the spacing is decreased. What is the effect on (a) the potential difference between the plates? (b) the electric field between the plates? (c) the magnitude of charge on the plates?
The potential difference across a cell membrane is \(-90 \mathrm{mV} .\) If the membrane's thickness is \(10 \mathrm{nm},\) what is the magnitude of the electric field in the membrane? Assume the field is uniform.
As an electron moves through a region of space, its speed decreases from $8.50 \times 10^{6} \mathrm{m} / \mathrm{s}\( to \)2.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ The electric force is the only force acting on the electron. (a) Did the electron move to a higher potential or a lower potential? (b) Across what potential difference did the electron travel?
The capacitor of Problem 79 is initially charged to a \(150-\mathrm{V}\) potential difference. The plates are then physically separated by another \(0.750 \mathrm{mm}\) in such a way that none of the charge can leak off the plates. Find (a) the new capacitance and (b) the new energy stored in the capacitor. Explain the result using conservation of energy.
A certain capacitor stores \(450 \mathrm{J}\) of energy when it holds $8.0 \times 10^{-2} \mathrm{C}$ of charge. What is (a) the capacitance of this capacitor and (b) the potential difference across the plates?
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