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Chapter 25: Problem 13

What would happen to the drift velocity of electrons in a wire if the resistance due to collisions between the electrons and the atoms in the crystal lattice of the metal disappeared?

Short Answer

Expert verified
Answer: When the resistance due to collisions between electrons and atoms in the crystal lattice is removed, the drift velocity of the electrons in the wire would increase as a result of fewer collisions and more efficient movement of electrons through the wire.

Step by step solution

01

Understand the drift velocity of electrons in a wire

Drift velocity is the average velocity of electrons in a conductive material like a wire, due to an applied electric field. This drift velocity is quite slow as the electrons are constantly colliding with the atoms in the metal, which cause resistance.
02

Relationship between drift velocity and resistance

The drift velocity (v_d) of electrons in a wire can be calculated using the formula: v_d = I / (n * e * A) where I is the current through the wire, n is the number of electrons per unit volume, e is the charge of an electron, and A is the cross-sectional area of the wire. The resistance (R) of the wire is calculated using Ohm's law: R = V / I where V is the voltage across the wire and I is the current through the wire. In a metal, the resistance is mainly due to collisions between the electrons and the atoms in the lattice. These collisions slow down the electrons and decrease their overall drift velocity.
03

Effect of removing resistance due to collisions

If the resistance due to collisions between electrons and atoms in the crystal lattice were to disappear, the electrons would no longer be slowed down in their path. This means that they would experience fewer collisions and would be able to move more freely through the metal. As a result, the drift velocity of electrons in the wire would increase. In other words, without the resistance due to collisions, the electrons would move more efficiently through the wire, causing a decrease in the time it takes for them to travel through the conductive material. Consequently, the drift velocity of the electrons would increase.
04

Conclusion

In conclusion, if the resistance due to collisions between electrons and atoms in the crystal lattice of a metal were to disappear, the drift velocity of the electrons would increase as a result of fewer collisions and more efficient movement of electrons through the wire.

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Most popular questions from this chapter

A charged-particle beam is used to inject a charge, \(Q_{0}\), into a small, irregularly shaped region (not a cavity, just some region within the solid block) in the interior of a block of ohmic material with conductivity \(\sigma\) and permittivity \(\epsilon\) at time \(t=0\). Eventually, all the injected charge will move to the outer surface of the block, but how quickly? a) Derive a differential equation for the charge, \(Q(t)\), in the injection region as a function of time. b) Solve the equation from part (a) to find \(Q(t)\) for all \(t \geq 0\). c) For copper, a good conductor, and for quartz (crystalline \(\mathrm{SiO}_{2}\) ), an insulator, calculate the time for the charge in the injection region to decrease by half. Look up the necessary values. Assume that the effective "dielectric constant" of copper is \(1.00000 .\)

The resistivity of a conductor is \(\rho=1.00 \cdot 10^{-5} \Omega \mathrm{m}\). If a cylindrical wire is made of this conductor, with a crosssectional area of \(1.00 \cdot 10^{-6} \mathrm{~m}^{2},\) what should the length of the wire be for its resistance to be \(10.0 \Omega ?\)

A battery has a potential difference of \(14.50 \mathrm{~V}\) when it is not connected in a circuit. When a \(17.91-\Omega\) resistor is connected across the battery, the potential difference of the battery drops to \(12.68 \mathrm{~V}\). What is the internal resistance of the battery?

Two cylindrical wires, 1 and \(2,\) made of the same material, have the same resistance. If the length of wire 2 is twice that of wire 1 , what is the ratio of their cross-sectional areas, \(A_{1}\) and \(A_{2} ?\) a) \(A_{1} / A_{2}=2\) c) \(\mathrm{A}_{1} / \mathrm{A}_{2}=0.5\) b) \(A_{1} / A_{2}=4\) d) \(A_{1} / A_{2}=0.25\)

A copper wire has radius \(r=0.0250 \mathrm{~cm},\) is \(3.00 \mathrm{~m}\) long, has resistivity \(\rho=1.72 \cdot 10^{-8} \Omega \mathrm{m},\) and carries a current of \(0.400 \mathrm{~A}\). The wire has density of charge carriers of \(8.50 \cdot 10^{28}\) electrons \(/ \mathrm{m}^{3}\) a) What is the resistance, \(R,\) of the wire? b) What is the electric potential difference, \(\Delta V\), across the wire? c) What is the electric field, \(E\), in the wire?
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