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Chapter 22: Problem 21

A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a \(100-\mathrm{V}\) potential difference. How much energy does each gain?

Short Answer

Expert verified
A proton gains \(1.602176634 \times 10^{-17} J\), an alpha particle gains \(3.204353268 \times 10^{-17} J\), and a singly ionized helium atom gains \(1.602176634 \times 10^{-17} J\) of kinetic energy.

Step by step solution

01

Calculate the Energy Gain for a Proton

Use the equation \(Energy = Charge \times Potential\). Replacing the Charge as the charge of a proton (1.602176634 \times 10^{-19} C) and the Potential as 100V \(Energy = 1.602176634 \times 10^{-19} C \times 100V = 1.602176634 \times 10^{-17} J\).
02

Calculate the Energy Gain for an Alpha Particle

Use the same formula as in step 1, but replace the charge with the charge of an alpha particle (2 \times 1.602176634 \times 10^{-19} C). So, \(Energy = 2 \times 1.602176634 \times 10^{-19} C \times 100V = 3.204353268 \times 10^{-17} J\).
03

Calculate the Energy Gain for a Singly Ionized Helium Atom

Use the same formula as in steps 1 and 2, but replace the charge with the charge of a singly ionized helium atom (1.602176634 \times 10^{-19} C). So, \(Energy = 1.602176634 \times 10^{-19} C \times 100V = 1.602176634 \times 10^{-17} J\).

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Most popular questions from this chapter

A uranium nucleus (mass 238 u, charge \(92 e\) ) decays, emitting an alpha particle (mass 4 u, charge 2e) and leaving a thorium nucleus (mass 234 u, charge \(90 e\) ). At the instant the alpha particle leaves the nucleus, the centers of the two are 7.4 fim apart and essentially at rest. Treating each particle as a spherical charge distribution, find their speeds when they're a great distance apart.

Two identical charges \(q\) lie on the \(x\) -axis at \(\pm a\). (a) Find an ex \(-\) pression for the potential at all points in the \(x-y\) plane. (b) Show that your result reduces to the potential of a point charge for dis tances large compared with \(a\)

The potential at the surface of a 10 -cm-radius sphere is \(4.8 \mathrm{kV}\) What's the sphere's total charge, assuming charge is distributed in a spherically symmetric way?

Proton-beam therapy can be preferable to X rays for cancer treatment (although much more expensive) because protons deliver most of their energy to the tumor, with less damage to healthy tissue. A cyclotron used to accelerate protons for cancer treatment repeatedly passes the protons through a 15 -kV potential difference. (a) How many passes are needed to bring the protons' kinetic energy to \(1.2 \times 10^{-11} \mathrm{J} ?\) (b) What's that energy in eV?

You're sizing a new electric transmission line, and you can save money with thinner wire. The potential difference between the line and the ground, \(60 \mathrm{m}\) below, is \(115 \mathrm{kV}\). The field at the wire surface cannot exceed \(25 \%\) of the 3 -MV/m breakdown field in air. Neglecting charges in the ground itself, what minimum wire diameter do you specify? (Hint: You'll have to do a numerical calculation.)
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