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Chapter 23: Problem 27

Before the advent of solid-state electronics, vacuum tubes were widely used in radios and other devices. A simple type of vacuum tube known as a diode consists essentially of two electrodes within a highly evacuated enclosure. One electrode, the cathode, is maintained at a high temperature and emits electrons from its surface. A potential difference of a few hundred volts is maintained between the cathode and the other electrode, known as the anode, with the anode at the higher potential. Suppose that in a particular vacuum tube the potential of the anode is 295 \(\mathrm{V}\) higher than that of the cathode. An electron leaves the surface of the cathode with zero initial speed. Find its speed when it strikes the anode.

Short Answer

Expert verified
The electron's speed when it strikes the anode is approximately 3.216 million m/s.

Step by step solution

01

Understanding the Problem

We are given a vacuum tube with a potential difference of 295 V between the anode and the cathode. An electron is emitted from the cathode with zero initial speed and we need to find the speed of the electron when it reaches the anode.
02

Using Energy Conservation

In this situation, we can apply the principle of energy conservation. The electron gains kinetic energy equal to the loss of electric potential energy as it moves from the cathode to the anode. The equation is \[ \Delta KE + \Delta PE = 0 \]which simplifies to \[ \frac{1}{2}mv^2 = e\Delta V \] where \( e \) is the elementary charge (\(1.6 \times 10^{-19} \text{C}\)), \( \Delta V = 295 \text{V} \), \( m \) is the electron mass (\(9.11 \times 10^{-31} \text{kg}\)), and \( v \) is the speed to find.
03

Solving for Final Speed

Rearrange the equation from step 2 to solve for \( v \): \[ v = \sqrt{\frac{2e\Delta V}{m}} \] Substitute the known values: \[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 295}{9.11 \times 10^{-31}}} \] Calculate to find \( v \).
04

Calculating the Result

Substituting the values into the equation gives: \[ v = \sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 295}{9.11 \times 10^{-31}}} \approx \sqrt{1.033 \times 10^{13}} \approx 3.216 \times 10^6 \text{ m/s} \] Thus, the speed of the electron when it strikes the anode is approximately \(3.216 \times 10^6 \text{ m/s}.\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Motion
In a vacuum tube diode, electron motion is a crucial element. Inside the tube, electrons move from the cathode towards the anode. This happens because of the potential difference maintained between the two electrodes. The cathode emits electrons when heated, a process known as thermionic emission. The emitted electrons start with virtually no speed.

As electrons travel towards the anode, they accelerate due to the force exerted by the electric field within the tube. This field is created by the potential difference, or voltage, between the cathode and anode. An electron, being negatively charged, moves towards the positively charged anode. This motion of electrons is a fundamental action that allows vacuum tubes to control electrical currents.
  • Electrons are emitted from a heated cathode.
  • They accelerate towards the anode due to the electric field.
  • The motion is directed by the potential difference in the tube.
Energy Conservation
The principle of energy conservation plays a pivotal role in understanding how vacuum tubes work. As electrons travel from the cathode to the anode, they convert potential energy into kinetic energy. This energy transformation is a classic example of energy conservation laws at work.

In a vacuum tube diode, each electron starts with zero kinetic energy when emitted from the cathode. As it moves towards the anode, the potential energy of the electron decreases, while its kinetic energy increases. This is due to the work done on the electron by the electric field.
  • The total energy of the electron remains constant.
  • Potential energy lost is converted to kinetic energy.
  • Understanding this conversion helps calculate the electron's speed.
Potential Difference
Potential difference, often referred to as voltage, is the driving force behind electron motion in a vacuum tube diode. This difference is maintained between the cathode and anode, with the anode being at a higher potential.

The potential difference creates an electric field inside the tube, which propels electrons towards the anode. The magnitude of this potential difference impacts the speed at which electrons move. In the given problem, a potential difference of 295 V is applied, facilitating the acceleration of the electron as it makes its journey between the electrodes.
  • Voltage is the key motivator of electron movement.
  • The anode is at a higher potential compared to the cathode.
  • A higher potential difference results in a greater electron velocity.

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Most popular questions from this chapter

A metal sphere with radius \(R_{1}\) has a charge \(Q_{1}\) . Take the electric potential to be zero at an infinite distance from the sphere. (a) What are the electric field and electric potential at the surface of the sphere? This sphere is now connected by a long, thin conducting wire to another sphere of radius \(R_{2}\) that is several meters from the first sphere. Before the connection is made, this second sphere is uncharged. After electrostatic equilibrium has been reached, what are (b) the total charge on each sphere; (c) the electric potential at the surface of each sphere (d) the electric field at the surface of each sphere? Assume that the amount of charge on the wire is much less than the charge on each sphere.

Two metal spheres of different sizes are charged such that the electric potential is the same at the surface of each. Sphere \(A\) has a radius three times that of sphere \(B\) . Let \(Q_{A}\) and \(Q_{B}\) be the charges on the two spheres, and let \(E_{A}\) and \(E_{B}\) be the electric-field magnitudes at the surfaces of the two spheres. What are (a) the ratio \(Q_{B} / Q_{A}\) and (b) the ratio \(E_{B} / E_{A} ?\)

A charge of 28.0 \(\mathrm{nC}\) is placed in a uniform electric field that is directed vertically upward and has a magnitude of \(4.00 \times 10^{4} \mathrm{V} / \mathrm{m}\) . What work is done by the electric force when the charge moves (a) 0.450 \(\mathrm{m}\) to the right; (b) 0.670 \(\mathrm{m}\) upward; (c) 2.60 \(\mathrm{m}\) at an angle of \(45.0^{\circ}\) downward from the horizontal?

Two charges of equal magnitude \(Q\) are beld a distance \(d\) apart. Consider only points on the line passing through both charges. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential (relative to infinity) is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?). and (b) Repeat part (a) for two charges having opposite signs.

Two positive point charges, each of magnitude \(q,\) are fixed on the \(y\) -axis at the points \(y=+a\) and \(y=-a\) . Take the potential to be zero at an infinite distance from the charges. (a) Show the positions of the charges in a diagram. (b) What is the potenntial \(V_{0}\) at the origin? (c) Show that the potential at any point on the \(x\) -axis is $$ V=\frac{1}{4 \pi \epsilon_{0}} \frac{2 q}{\sqrt{a^{2}+x^{2}}} $$ (d) Graph the potential on the \(x\) -axis as a function of \(x\) over the range from \(x=-4 a\) to \(x=+4 a\) . (e) What is the potential when \(x \gg a ?\) Explain why this result is obtained.
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