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Chapter 35: Problem 74

An electron is accelerated from rest through a potential of \(1.0 \cdot 10^{6} \mathrm{~V}\). What is its final speed?

Short Answer

Expert verified
Answer: The final speed of the electron is approximately \(1.874 \cdot 10^{7} \mathrm{\frac{m}{s}}\).

Step by step solution

01

Identify given values and constants

In this problem, we have the following data: Potential, V = \(1.0 \cdot 10^{6} \mathrm{~V}\) (Voltage through which the electron is accelerated) Electron charge, e = \(-1.6 \cdot 10^{-19} \mathrm{~C}\) Electron mass, m = \(9.11 \cdot 10^{-31} \mathrm{~kg}\)
02

Calculate the work done on the electron

The work (W) done on the electron as it accelerates through the electric potential (V) is given by: \(W = e \times V\) Plug in the values for e and V: \(W = -1.6 \cdot 10^{-19} \mathrm{~C} \times 1.0 \cdot 10^{6} \mathrm{~V}\) \(W = -1.6 \cdot 10^{-13} \mathrm{~J}\)
03

Use the work-energy theorem to relate the work done to the final kinetic energy

According to the work-energy theorem, the work done on the electron equals the change in its kinetic energy: \(W = KE_{final} - KE_{initial}\) Since the electron starts from rest, its initial kinetic energy is 0. Thus, \(W = KE_{final}\)
04

Calculate the final kinetic energy of the electron

From step 2 and step 3, we have: \( KE_{final} = -1.6 \cdot 10^{-13} \mathrm{~J}\)
05

Calculate the final speed of the electron using the kinetic energy formula

The formula for kinetic energy is given by: \(KE = \frac{1}{2}mv^2\) Where m is the mass of the electron and v is its final speed. We can now solve for the final speed (v) by rearranging the formula: \(v = \sqrt{\frac{2 \times KE}{m}}\) Plug in the values for KE and m: \(v = \sqrt{\frac{2 \times -1.6 \cdot 10^{-13} \mathrm{~J}}{9.11 \cdot 10^{-31} \mathrm{~kg}}}\) \(v \approx 1.874 \cdot 10^{7} \mathrm{\frac{m}{s}}\) So, the final speed of the electron is approximately \(1.874 \cdot 10^{7} \mathrm{\frac{m}{s}}\).

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Most popular questions from this chapter

Consider a positively charged particle moving at constant speed parallel to a current-carrying wire, in the direction of the current. As you know (after studying Chapters 27 and 28), the particle is attracted to the wire by the magnetic force due to the current. Now suppose another observer moves along with the particle, so according to him the particle is at rest. Of course, a particle at rest feels no magnetic force. Does that observer see the particle attracted to the wire or not? How can that be? (Either answer seems to lead to a contradiction: If the particle is attracted, it must be by an electric force because there is no magnetic force, but there is no electric field from a neutral wire; if the particle is not attracted, you see that the particle is, in fact, moving toward the wire.)

Consider a one-dimensional collision at relativistic speeds between two particles with masses \(m_{1}\) and \(m_{2}\). Particle 1 is initially moving with a speed of \(0.700 c\) and collides with particle \(2,\) which is initially at rest. After the collision, particle 1 recoils with speed \(0.500 c\), while particle 2 starts moving with a speed of \(0.200 c\). What is the ratio \(m_{2} / m_{1} ?\)

Use light cones and world lines to help solve the following problem. Eddie and Martin are throwing water balloons very fast at a target. At \(t=-13 \mu s,\) the target is at \(x=0,\) Eddie is at \(x=-2 \mathrm{~km},\) and Martin is at \(x=5 \mathrm{~km},\) and all three remain in these positions for all time. The target is hit at \(t=0 .\) Who made the successful shot? Prove this using the light cone for the target. When the target is hit, it sends out a radio signal. When does Martin know the target has been hit? When does Eddie know the target has been hit? Use the world lines to show this. Before starting to draw your diagrams, consider: If your \(x\) position is measured in \(\mathrm{km}\) and you are plotting \(t\) versus \(x / c,\) what units must \(t\) be in, to the first significant figure?

The Relativistic Heavy Ion Collider (RHIC) can produce colliding beams of gold nuclei with beam kinetic energy of \(A \cdot 100 .\) GeV each in the center-of- mass frame, where \(A\) is the number of nucleons in gold (197). You can approximate the mass energy of a nucleon as approximately \(1.00 \mathrm{GeV}\). What is the equivalent fixed-target beam energy in this case?

The explosive yield of the atomic bomb dropped on Hiroshima near the end of World War II was approximately 15.0 kilotons of TNT. One kiloton is about \(4.18 \cdot 10^{12} \mathrm{~J}\) of energy. Find the amount of mass that was converted into energy in this bomb.
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