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Chapter 17: Problem 106

An alpha particle (helium nucleus, charge \(+2 e\) ) starts from rest and travels a distance of \(1.0 \mathrm{cm}\) under the influence of a uniform electric field of magnitude \(10.0 \mathrm{kV} / \mathrm{m}\) What is the final kinetic energy of the alpha particle?

Short Answer

Expert verified
Answer: The final kinetic energy of the alpha particle is 3.2 × 10^{-17} J.

Step by step solution

01

Find the force on the alpha particle due to the electric field

The force on a charged particle in an electric field is given by the equation: F = qE where F is the force, q is the charge of the particle and E is the electric field strength. Since the charge of an alpha particle is +2e and the electric field strength is 10.0 kV/m, we have: F = (2e)(10.0 kV/m) = (2e)(10,000 N/C)
02

Find the work done by the force to move the alpha particle

The work done by a constant force on an object is given by the equation: W = Fd where W is the work done, F is the force and d is the distance moved. Using the force obtained in step 1 and the given distance of 1.0 cm, we calculate the work done: W = (2e)(10,000 N/C)(0.01 m) = (2e)(100 J/C)
03

Use the work-energy theorem to find the final kinetic energy

The work-energy theorem states that the work done on an object is equal to the change in kinetic energy: W = ΔK Since the alpha particle starts from rest, its initial kinetic energy is zero. Therefore, the work done on the alpha particle is equal to its final kinetic energy: ΔK = (2e)(100 J/C)
04

Calculate the final kinetic energy of the alpha particle

Now we will plug in the value of the fundamental charge e to find the final kinetic energy of the alpha particle: ΔK = (2)(1.6 × 10^{-19} C)(100 J/C) = 3.2 × 10^{-17} J So, the final kinetic energy of the alpha particle is 3.2 × 10^{-17} J.

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Most popular questions from this chapter

A defibrillator is used to restart a person's heart after it stops beating. Energy is delivered to the heart by discharging a capacitor through the body tissues near the heart. If the capacitance of the defibrillator is $9 \mu \mathrm{F}\( and the energy delivered is to be \)300 \mathrm{J},$ to what potential difference must the capacitor be charged?
A parallel plate capacitor has a capacitance of \(1.20 \mathrm{nF}\) and is connected to a \(12-\mathrm{V}\) battery. (a) What is the magnitude of the charge on each plate? (b) If the plate separation is doubled while the plates remain connected to the battery, what happens to the charge on each plate and the electric field between the plates?
A van de Graaff generator has a metal sphere of radius \(15 \mathrm{cm} .\) To what potential can it be charged before the electric field at its surface exceeds \(3.0 \times 10^{6} \mathrm{N} / \mathrm{C}\) (which is sufficient to break down dry air and initiate a spark)?
If an electron moves from one point at a potential of \(-100.0 \mathrm{V}\) to another point at a potential of \(+100.0 \mathrm{V}\) how much work is done by the electric field?
A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\). The parallel plates are separated by 0.40 mm of air. What energy is stored in this capacitor?
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