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(a) Consider an electron (charge \(-e\) and mass \(m\) ) in a circular orbit of radius \(r\) around a fixed proton (charge \(+e\) ). Remembering that the inward Coulomb force \(k e^{2} / r^{2}\) is what gives the electron its centripetal acceleration, prove that the electron's KE is equal to \(-\frac{1}{2}\) times its \(\mathrm{PE}\); that is, \(T=-\frac{1}{2} U\) and hence \(E=\frac{1}{2} U\). (This result is a consequence of the so-called virial theorem. See Problem 4.41.) Now consider the following inelastic collision of an electron with a hydrogen atom: Electron number 1 is in a circular orbit of radius \(r\) around a fixed proton. (This is the hydrogen atom.) Electron 2 approaches from afar with kinetic energy \(T_{2} .\) When the second electron hits the atom, the first electron is knocked free, and the second is captured in a circular orbit of radius \(r^{\prime} .\) (b) Write down an expression for the total energy of the three-particle system in general. (Your answer should contain five terms, three PEs but only two KEs, since the proton is considered fixed.) (c) Identify the values of all five terms and the total energy \(E\) long before the collision occurs, and again long after it is all over. What is the KE of the outgoing electron 1 once it is far away? Give your answers in terms of the variables \(T_{2}, r,\) and \(r^{\prime}\).

Short Answer

Expert verified
Electron's KE is equal to \(-\frac{1}{2}\) times its PE; outgoing electron 1 has KE \(T_2\).

Step by step solution

01

Coulomb Force and Centripetal Force

The electron in a circular orbit experiences an inward Coulomb force given by \( F = \frac{k e^2}{r^2} \). This force provides the centripetal force required for circular motion, so \( F = \frac{m v^2}{r} \), where \( v \) is the speed of the electron.
02

Equate Forces and Solve for Velocity

Setting the Coulomb force equal to the centripetal force gives \( \frac{k e^2}{r^2} = \frac{m v^2}{r} \). Solving for \( v^2 \), we get \( v^2 = \frac{k e^2}{m r} \).
03

Kinetic Energy of Electron

The kinetic energy (KE) of the electron is \( T = \frac{1}{2} m v^2 \). Substituting for \( v^2 \), \( T = \frac{1}{2} m \left( \frac{k e^2}{m r} \right) = \frac{k e^2}{2 r} \).
04

Potential Energy of Electron

The potential energy (PE) of the electron in the electric field of the proton is \( U = - \frac{k e^2}{r} \).
05

Demonstrate Relationship T = -1/2 U

Substituting the expressions for \( T \) and \( U \), \( T = \frac{k e^2}{2 r} \) and \( U = - \frac{k e^2}{r} \), we find \( T = -\frac{1}{2} U \).
06

Expression for Total Energy of System

For three particles: electron 1 (e1), electron 2 (e2), and the proton (p), the total energy \( E = T_1 + T_2 + U_{e1,p} + U_{e2,p} + U_{e1,e2} \). Note: \( U_{e1,e2} = 0 \) at the point of collision initially.
07

Initial State Energy Calculations

Before collision, electron 1 has energy \( E = -\frac{k e^2}{2r} + T_2 - \frac{k e^2}{r} = T_2 - \frac{k e^2}{2r} \). Electron 2 is far away with potential energy approximately zero, and kinetic energy \( T_2 \).
08

Final State Energy Calculations

After collision, electron 2 has energy \( T_2' = \frac{k e^2}{2r'} \) and \( U_{e2,p} = -\frac{k e^2}{r'} \). Energy of outgoing electron 1 is purely kinetic \( T_1' \). Setting initial energy equal to final, solve for \( T_1' \).
09

Solve for Outgoing Electron's Kinetic Energy

Equating initial and final total energy, \( T_2 - \frac{k e^2}{2r} = \frac{k e^2}{2r'} - \frac{k e^2}{r'} + T_1' \). Solve for \( T_1' \) to get \( T_1' = T_2 + \frac{k e^2}{r' (-1 + 1)} \). However, this simplifies to \( T_1' = T_2 \).

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Key Concepts

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

Coulomb Force
Coulomb force is the electric force of attraction or repulsion between two charged particles. In the context of an electron orbiting a proton, like in a hydrogen atom, this force is responsible for providing the centripetal force that keeps the electron in its circular path. The formula for the magnitude of the Coulomb force between two charges, such as an electron and a proton, is given by
  • \( F = \frac{k e^2}{r^2} \)
where:
  • \( k \) is Coulomb's constant,
  • \( e \) is the charge of the electron or proton,
  • \( r \) is the distance between the two charges.
In a circular orbit, this Coulomb force acts as the necessary centripetal force such that
  • \( F = \frac{m v^2}{r} \),
where \( m \) is the mass of the electron and \( v \) is its speed.
Understanding this interaction is key to analyzing the balance of forces that maintain stable electron orbits.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. For an electron moving in a circular orbit around a proton, its kinetic energy is an important aspect deriving from its velocity and mass.The kinetic energy \( T \) of the electron can be calculated using the formula:
  • \( T = \frac{1}{2} m v^2 \)
Substituting the expression for the velocity squared \( v^2 = \frac{k e^2}{m r} \) from the force balance, we arrive at:
  • \( T = \frac{k e^2}{2 r} \)
This shows that the kinetic energy depends on the radius \( r \) of the orbit and the charge \( e \).
The kinetic energy in circular motion is found to be half the magnitude of the potential energy (with opposite sign), revealing a fundamental relationship within circular atomic orbits.
Potential Energy
Potential energy in the context of an electron in orbit around a proton refers to the energy associated with its position in the electric field created by the proton. It is given by the formula:
  • \( U = -\frac{k e^2}{r} \)
This negative sign arises because the force is attractive, meaning it requires energy to separate the electron from the proton.
The potential energy is directly related to the distance between the electron and the proton. As this distance \( r \) increases, the magnitude of the potential energy decreases (becomes less negative).
The relationship between kinetic and potential energy is crucial: in stable circular motion, twice the kinetic energy equals the negative of the potential energy, noted as \( T = -\frac{1}{2} U \). This balance is central to maintaining the electron's orbit without spiraling into the proton.
Circular Motion
Circular motion refers to the movement of a particle along the circumference of a circle. In the orbit of an electron around a proton, the circular motion is maintained by the centripetal force provided by the Coulomb attraction between the charged particles.
For an electron to stay in a stable orbit, the net force must point towards the circle's center, provoking continuous change in direction of the velocity, thereby creating circular motion with constant speed. Key equations involved in circular motion of an electron include:
  • The centripetal force: \( F_{c} = \frac{m v^2}{r} \)
  • Velocity squared as derived from force balance: \( v^2 = \frac{k e^2}{m r} \)
In circular motion scenarios, the role of centrifugal and centripetal forces can be explained in terms of energy balance using kinetic and potential components, demonstrating that \( T = -\frac{1}{2} U \).
Identifying these relationships helps clarify why electrons do not "fall" into the nucleus, but rather remain in stable orbits.

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Most popular questions from this chapter

Consider a small frictionless puck perched at the top of a fixed sphere of radius \(R\). If the puck is given a tiny nudge so that it begins to slide down, through what vertical height will it descend before it leaves the surface of the sphere? [Hint: Use conservation of energy to find the puck's speed as a function of its height, then use Newton's second law to find the normal force of the sphere on the puck. At what value of this normal force does the puck leave the sphere?]

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The proof that the condition \(\nabla \times \mathbf{F}=0\) guarantees the path independence of the work \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) done by \(\mathbf{F}\) is unfortunately too lengthy to be included here. However, the following three exercises capture the main points: \(^{16}\) (a) Show that the path independence of \(\int_{1}^{2} \mathbf{F} \cdot d \mathbf{r}\) is equivalent to the statement that the integral \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}\) around any closed path \(\Gamma\) is zero. (By tradition, the symbol \(\oint\) is used for integrals around a closed path \(-\) a path that starts and stops at the same point.) [Hint: For any two points 1 and 2 and any two paths from 1 to 2 , consider the work done by \(\mathbf{F}\) going from 1 to 2 along the first path and then back to 1 along the second in the reverse direction. \((\) b) Stokes's theorem asserts that \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=\int(\nabla \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A,\) where the integral on the right is a surface integral over a surface for which the path \(\Gamma\) is the boundary, and \(\hat{\mathbf{n}}\) and \(d A\) are a unit normal to the surface and an element of area. Show that Stokes's theorem implies that if \(\nabla \times \mathbf{F}=0\) everywhere, then \(\oint_{\mathrm{T}} \mathbf{F} \cdot d \mathbf{r}=0 .\) (c) While the general proof of Stokes's theorem is beyond our scope here, the following special case is quite easy to prove (and is an important step toward the general proof): Let \(\Gamma\) denote a rectangular closed path lying in a plane perpendicular to the \(z\) direction and bounded by the lines \(x=B, x=B+b, y=C\) and \(y=C+c .\) For this simple path (traced counterclockwise as seen from above), prove Strokes's theorem that \(\oint_{\Gamma} \mathbf{F} \cdot d \mathbf{r}=\int(\mathbf{\nabla} \times \mathbf{F}) \cdot \hat{\mathbf{n}} d A\) where \(\hat{\mathbf{n}}=\hat{\mathbf{z}}\) and the integral on the right runs over the flat, rectangular area inside \(\Gamma\). [Hint: The integral on the left contains four terms, two of which are integrals over \(x\) and two over \(y\). If you pair them in this way, you can combine each pair into a single integral with an integrand of the form \(F_{x}(x, C+c, z)-F_{x}(x, C, z)\) (or a similar term with the roles of \(x\) and \(y\) exchanged). You can rewrite this integrand as an integral over \(y\) of \(\partial F_{x}(x, y, z) / \partial y\) (and similarly with the other term), and you're home.]

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):(\mathbf{a}) f=\ln (r),\) (b) \(f=r^{n}\), (c) \(f=g(r),\) where \(r=\sqrt{x^{2}+y^{2}+z^{2}}\) and \(g(r)\) is some unspecified function of \(r .\) [Hint: Use the chain rule.]

Calculate the gradient \(\nabla f\) of the following functions, \(f(x, y, z):\) (a) \(f=x^{2}+z^{3} .\) (b) \(f=k y\), where \(k\) is a constant. (c) \(f=r \equiv \sqrt{x^{2}+y^{2}+z^{2}} .\) [Hint: Use the chain rule.] (d) \(f=1 / r\).

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