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Chapter 27: Problem 49

Show that the magnetic dipole moment of an electron orbiting in a hydrogen atom is proportional to its angular momentum, \(L: \mu=-e L / 2 m,\) where \(-e\) is the charge of the electron and \(m\) is its mass.

Short Answer

Expert verified
Question: Show that the magnetic dipole moment of an electron orbiting in a hydrogen atom is proportional to its angular momentum with the relation given by \(\mu = -\frac{eL}{2m}\). Answer: By deriving expressions for the magnetic dipole moment and angular momentum of an electron in a hydrogen atom, we have shown that they are proportional with the relation \(\mu = -\frac{eL}{2m}\).

Step by step solution

01

Define the variables and given parameters

In this problem, we have the following variables and parameters: - \(L\): Angular momentum of the electron - \(\mu\): Magnetic dipole moment of the electron - \(e\): Charge of the electron (\(e < 0\)) - \(m\): Mass of the electron
02

Derive expression for magnetic dipole moment

We know that the magnetic dipole moment of a charged particle moving in a circle is given by: $$\mu = IA = I\pi r^2$$ where \(I\) is the current produced by the moving electron and \(A\) is the area of the circular orbit with radius \(r\). The current \(I\) can be defined as the charge \(e\) divided by the time \(T\) it takes for the electron to complete one orbit around the nucleus: $$I = \frac{e}{T}$$
03

Angular frequency and period of the orbit

To calculate the period \(T\) of the circular orbit, we first need to determine the velocity \(v\) of the electron and the radius \(r\) of the orbit. For this, we'll use the angular frequency of the orbit, denoted by \(\omega\). The angular frequency is given by the following equation that relates angular frequency, velocity, and radius of the orbit: $$\omega = \frac{v}{r}$$ Furthermore, the period \(T\) is related to the angular frequency as follows: $$T = \frac{2\pi}{\omega}$$
04

Determine the centripetal force

The centripetal force on the electron is given by the expression \(F_c = \frac{mv^2}{r}\). The only force acting on the electron in the hydrogen atom is the electrostatic force between the electron and the nucleus (proton). This force is given by Coulomb's Law: $$F_e = \frac{ke^2}{r^2}$$ where \(k\) is the Coulomb's constant. As the centripetal force and electrostatic force are equal, we can write: $$F_c = F_e \Rightarrow \frac{mv^2}{r} = \frac{ke^2}{r^2}$$
05

Find expressions for velocity and radius

By solving the equation from Step-4, we can find expressions for the velocity \(v\) and the radius \(r\) of the electron's orbit, which are given by: $$v = \frac{ke}{mr}$$ $$r = \frac{ke^2}{mv^2}$$
06

Calculate the angular momentum

Now, we can calculate the angular momentum of the electron (L) using the following equation: $$L = mvr$$ Plugging in the expressions for \(v\) and \(r\), obtained in Step 5, we get: $$L = m \left(\frac{ke}{mr}\right) \left(\frac{ke^2}{mv^2}\right) = \frac{ke^2}{v}$$
07

Calculate the magnetic dipole moment

With the expressions for \(I\) and \(A\), we can now calculate the magnetic dipole moment \(\mu\). Using the expression for \(T\) from Step 3, plug it into the definition of the current \(I\): $$I = \frac{e}{T} = \frac{e\omega}{2\pi}$$ Now, using the expression for \(r\) from Step 5 in the expression \(A = \pi r^2\), we get: $$A = \pi \left(\frac{ke^2}{mv^2}\right)^2$$ Finally, let's calculate the magnetic dipole moment \(\mu\) using the expressions for \(I\) and \(A\): $$\mu = IA = \frac{e\omega}{2\pi} \cdot \pi \left(\frac{ke^2}{mv^2}\right)^2 = \frac{- e \omega ke^2}{2mv^2}$$
08

Show proportionality between magnetic dipole moment and angular momentum

To show the proportionality between the magnetic dipole moment and the angular momentum, let's substitute the expression for \(\omega\) from Step 3 and \(L\) from Step 6 into the expression for \(\mu\): $$\mu = \frac{- e \omega ke^2}{2mv^2} = \frac{-e}{2m} \cdot \frac{v}{r} \cdot \frac{ke^2}{v} = -\frac{eL}{2m}$$ The above equation proves that the magnetic dipole moment is indeed proportional to the angular momentum with the relation \(\mu = -\frac{eL}{2m}\).

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Key Concepts

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

Electron Orbit
The concept of an electron orbit is a fundamental one in classical physics and serves as a simplistic model to understand more complex quantum behavior. An electron orbit refers to the path that an electron takes around the nucleus of an atom. In early atomic models, such as the Bohr model, this path was envisioned as similar to the orbit of planets around the sun, though we now know that electrons exhibit wave-like properties and exist in regions of probability called orbitals.

Angular Momentum
The concept of angular momentum plays a crucial role when analyzing the movement of an electron in its orbit. Angular momentum, denoted by the symbol L, is a measure of the amount of rotational motion an object has. For an electron orbiting the nucleus, it can be represented mathematically as the product of the electron's mass (m), its velocity (v), and the radius (r) of its orbit:

Coulomb's Law
Coulomb's Law describes the force of attraction or repulsion between two charged particles. In the context of an electron orbiting a nucleus, Coulomb's Law is used to describe the electrostatic force acting between the negatively charged electron and the positively charged nucleus. The force is proportional to the product of the charges and inversely proportional to the square of the distance between them, expressed as:

Centripetal Force
The term centripetal force refers to the net force required to keep an object moving in a circular path. In the case of an electron orbiting a nucleus, centripetal force is what keeps the electron in its circular orbit. This force is directed towards the center of the orbit and is necessary to counteract the electron's tendency to move in a straight line due to its inertia. Centripetal force is provided by the electrostatic force described by Coulomb's law, ensuring the electron stays in its prescribed orbit around the nucleus.

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Most popular questions from this chapter

A rectangular coil with 20 windings carries a current of 2.00 mA flowing in the counterclockwise direction. It has two sides that are parallel to the \(y\) -axis and have length \(8.00 \mathrm{~cm}\) and two sides that are parallel to the \(x\) -axis and have length \(6.00 \mathrm{~cm} .\) A uniform magnetic field of \(50.0 \mu \mathrm{T}\) acts in the positive \(x\) -direction. What torque must be applied to the loop to hold it steady?

A magnetic field is oriented in a certain direction in a horizontal plane. An electron moves in a certain direction in the horizontal plane. For this situation, there a) is one possible direction for the magnetic force on the electron. b) are two possible directions for the magnetic force on the electron. c) are infinite possible directions for the magnetic force on the electron.

An alpha particle \(\left(m=6.6 \cdot 10^{-27} \mathrm{~kg}, q=+2 e\right)\) is accelerated by a potential difference of \(2700 \mathrm{~V}\) and moves in a plane perpendicular to a constant magnetic field of magnitude \(0.340 \mathrm{~T}\), which curves the trajectory of the alpha particle. Determine the radius of curvature and the period of revolution.

The velocity selector described in Solved Problem 27.2 is used in a variety of devices to produce a beam of charged particles of uniform velocity. Suppose the fields in such a selector are given by \(\vec{E}=\left(1.00 \cdot 10^{4} \mathrm{~V} / \mathrm{m}\right) \hat{x}\) and \(\vec{B}=(50.0 \mathrm{mT}) \hat{y} .\) Find the velocity in the \(z\) -direction with which a charged particle can travel through the selector without being deflected.

A copper wire with density \(\rho=8960 \mathrm{~kg} / \mathrm{m}^{3}\) is formed into a circular loop of radius \(50.0 \mathrm{~cm} .\) The cross-sectional area of the wire is \(1.00 \cdot 10^{-5} \mathrm{~m}^{2},\) and a potential difference of \(0.012 \mathrm{~V}\) is applied to the wire. What is the maximum angular acceleration of the loop when it is placed in a magnetic field of magnitude \(0.25 \mathrm{~T}\) ? The loop rotates about an axis through a diameter.
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