Problem 21 - University Physics with Modern Physics

Question

CP Classical Electron Spin. (a) If you treat an electron as a classical spherical object with a radius of \(1.0 \times 10^{-17} \mathrm{m}\) , what angular speed is necessary to produce a spin angular momentum of magnitude \(\sqrt{\frac{3}{4}} \hbar ?\) (b) Use \(v=r \omega\) and the result of part (a) to calculate the speed \(v\) of a point at the electron's equator. What does your result suggest about the validity of this model?

Step-by-Step Solution

Answer

The classical model suggests non-physical results, supporting quantum mechanics.

Step 1: Understanding the Electron's Spin Angular Momentum

The spin angular momentum of an electron is given as \( \sqrt{\frac{3}{4}} \hbar \), where \( \hbar \) is the reduced Planck's constant, \( \hbar = 1.0545718 \times 10^{-34} \, \text{J} \cdot \text{s} \). Our goal is to find the angular velocity using the relation for classical angular momentum \( L = I \omega \), where \( I \) is the moment of inertia of a sphere.

Step 2: Calculate the Moment of Inertia for a Sphere

The moment of inertia \( I \) for a solid sphere about an axis through its center is \( I = \frac{2}{5} m r^2 \), where \( m \) is the mass of the electron \( \approx 9.10938356 \times 10^{-31} \text{kg} \) and \( r \) is the given radius \(1.0 \times 10^{-17} \text{m}\). Compute \( I \): \[ I = \frac{2}{5} \times 9.10938356 \times 10^{-31} \times (1.0 \times 10^{-17})^2 \text{kg m}^2 \].

Step 3: Set up the Equation for Angular Velocity

Using the relation \( L = I \omega \) and substituting \( L = \sqrt{\frac{3}{4}} \hbar \), we have: \[ \sqrt{\frac{3}{4}} \hbar = \left( \frac{2}{5} m r^2 \right) \omega \].

Step 4: Solve for Angular Velocity \(\omega\)

Reorganize the equation to solve for \( \omega \):\[ \omega = \frac{\sqrt{\frac{3}{4}} \hbar}{\frac{2}{5} m r^2} \].Substitute the known values and calculate \( \omega \).

Step 5: Determine the Linear Speed at the Equator

Using the relationship \( v = r \omega \) where \( r = 1.0 \times 10^{-17} \text{m} \) and \( \omega \) is computed from Step 4, calculate the linear speed \( v \).

Step 6: Interpret the Results

Consider the calculated speed \( v \). If \( v \) exceeds the speed of light \( c = 3.0 \times 10^8 \text{m/s} \), the model of the electron as a classical spinning object is invalid. This supports the quantum mechanical view of electron spin.

Key Concepts

Classical Mechanics

The framework of classical mechanics forms the bedrock for understanding motion on everyday scales, exploring the physics of macroscopic objects. It encompasses the laws that describe how objects behave when subject to forces, specifically the principles laid out by Newton.
Consider an electron modeled as a classical spherical object. This approach attempts to explain phenomena like spin using familiar laws of motion.
The classical treatment involves concepts like moment of inertia and angular momentum, imagining the electron to spin like a tiny top. While helpful, classical mechanics alone cannot fully capture the nature of subatomic particles like electrons, pardoning its limitations when a quantum mechanical perspective is taken.

Angular Momentum

Angular momentum is a measure of how much rotation an object has, taking into account its moment of inertia and angular velocity. In classical mechanics, it's calculated using the formula:

• \( L = I \omega \)

where \( L \) is the angular momentum, \( I \) is the moment of inertia, and \( \omega \) is the angular velocity.
In the context of the problem, the spin angular momentum of an electron is given as \( \sqrt{\frac{3}{4}} \hbar \), a quantum mechanical value. Using classical equations to derive \( \omega \), we illustrate a tangible mental model to visualize electron behavior, albeit not fully accurate. Yet, envisioning electrons akin to spinning spheres can be an intriguing exercise bridging the classical to the quantum realm.

Moment of Inertia

The moment of inertia quantifies an object's resistance to changes in its rotational motion, much like mass does for linear motion. For a solid sphere, the equation is:

• \( I = \frac{2}{5} m r^2 \)

where \( m \) is the mass and \( r \) is the radius of the sphere.
Within our exercise, we calculate the moment of inertia for a supposed classical electron. Using an assumed radius and the known mass of the electron, this aids in determining its angular velocity.
The concept, while straightforward in mechanics, becomes murky when applied to subatomic particles, prompting a shift towards quantum mechanics for a more accurate depiction.

Quantum Mechanics

Quantum mechanics revolutionizes our understanding of particles at tiny scales, where classical mechanics falls short. Unlike the deterministic nature of classical physics, quantum mechanics embraces probabilities and wave-particle duality.
In the quantum view, electron "spin" isn't akin to a spinning sphere. Instead, it is intrinsic angular momentum with no classical counterpart. The electron's spin of \( \sqrt{\frac{3}{4}} \hbar \) emerges naturally only when quantum principles are applied, providing clarity and explaining behavior impossible under classical paradigms.
By exploring these concepts, one appreciates quantum mechanics' power, resolving puzzles classical models, like spinning electrons, cannot solve. The speed calculations exceeding light beyond the classical model's limits reaffirm quantum mechanical dominance in accurately depicting reality.