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Chapter 41: Problem 23

\(\textbf{Classical Electron Spin}\). (a) If you treat an electron as a classical spherical object with a radius of 1.0 \(\times\) 10\(^{-17}\) m, what angular speed is necessary to produce a spin angular momentum of magnitude \(\sqrt{3\over4}\hslash\) ? (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?

Short Answer

Expert verified
The required angular speed is unrealistic, suggesting the classical model is flawed.

Step by step solution

01

Identify the given values and formula for angular momentum

We have the radius of the electron, \( r = 1.0 \times 10^{-17} \text{ m} \), and the desired spin angular momentum \( L = \sqrt{\frac{3}{4}}\hslash \). The angular momentum \( L \) for a solid sphere is given by \( L = I\omega \), where \( I = \frac{2}{5}mr^2 \) is the moment of inertia, \( m \) is the mass of the electron, and \( \omega \) is angular speed.
02

Calculate the mass of the electron

The mass of an electron is \( m = 9.11 \times 10^{-31} \text{ kg} \). We'll use this value in calculating the moment of inertia.
03

Solve for angular speed \( \omega \)

Substitute \( I = \frac{2}{5}mr^2 \) and \( L = I\omega \) into the equation \( L = \sqrt{\frac{3}{4}}\hslash \). We solve for \( \omega \):\[ \omega = \frac{L}{I} = \frac{\sqrt{\frac{3}{4}}\hslash}{\frac{2}{5}mr^2} \]. Substitute \( \hslash = 1.05 \times 10^{-34} \text{ Js} \), \( m = 9.11 \times 10^{-31} \text{ kg} \), and \( r = 1.0 \times 10^{-17} \text{ m} \) to find \( \omega \).
04

Compute the speed \( v \) using \( v = r\omega \)

Using \( \omega \) from Step 3, calculate \( v = r\omega \, \) where \( r \) is the radius of the electron. This will give the speed at the electron's equator.
05

Analyze the speed result

Compare this calculated speed \( v \) with the speed of light. Because the calculated speed will far exceed the speed of light, this result suggests that treating an electron as a classical spinning sphere is not valid.

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Key Concepts

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

Angular Momentum
Angular momentum is an essential concept in physics, particularly when analyzing rotating objects, such as a classic spinning electron. Much like linear momentum in a straight line, angular momentum represents the momentum of an object in rotation. It accounts for how difficult it is to stop the object from spinning.
For a solid sphere, such as our hypothetical electron, angular momentum \( L \) is calculated using the formula \( L = I\omega \), where \( I \) is the moment of inertia and \( \omega \) is the angular speed.
Key points to remember:
  • It is conserved in closed systems, meaning that the angular momentum will remain constant unless acted upon by an external force.
  • Measured in \( \text{kg} \cdot \text{m}^2/\text{s} \), indicating a complete set of rotational motion units.
  • For quantized systems, like electrons, the classical analogy may break down and require quantum mechanical models.
Moment of Inertia
The moment of inertia \( I \) represents how mass is distributed in an object concerning its axis of rotation. For an electron treated as a solid sphere, this quantifies its rotational inertia and directly influences how easily the electron can achieve angular acceleration under a torque.
The formula for the moment of inertia of a solid sphere is \( I = \frac{2}{5}mr^2 \), where \( m \) is the mass and \( r \) is the radius.
This concept translates to:
  • Larger moments of inertia mean more difficulty in changing the object's rotating speed, akin to mass in linear motion.
  • A crucial factor in calculating angular momentum and predicting rotational behavior.
Understanding the moment of inertia helps in grasping why classical models of electron spin, like the one in the exercise, can face limitations.
Classical Mechanics
Classical mechanics is the framework used to describe the motion of macroscopic objects at everyday speeds and scales. Though profoundly successful for a wide range of phenomena, it sometimes falls short when applied to subatomic particles like electrons. In this exercise, we model the electron's spin as that of a classical spinning sphere.
Key assumptions in classical mechanics include:
  • Continuity of energy and momentum, whereas quantum mechanics introduces discretized or quantized values.
  • Applicability to large-scale objects where relativity effects are negligible.
  • Deterministic approach, predicting exact system evolutions opposed to probabilistic quantum mechanics.
Using classical mechanics for an electron spin highlights these limits, as the calculated speed exceeds feasible physical parameters.
Angular Speed
Angular speed \( \omega \) is a measure of how fast an object rotates or spins around a central point, expressed in radians per second (rad/s). In our exercise's context, angular speed is pivotal in connecting the spin鈥檚 angular momentum to its physical geometry.
To find angular speed from angular momentum, use \( \omega = \frac{L}{I} \).
If the model treats electrons classically, then:
  • Angular speed provides valuable insights into rotational dynamics, setting the stage for calculating surface speed at the electron's equator using \( v = r\omega \).
  • Exceeding the speed of light, when calculated, suggests inadequacies in classical assumptions for subatomic particles.
Always compare results within the theoretical boundaries, and transition to quantum mechanics when classical mechanics fails.

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Most popular questions from this chapter

A particle is in the three-dimensional cubical box of Section 41.2. (a) Consider the cubical volume defined by \(0 \leq x \leq L/4, 0 \leq y \leq L/4\), and \(0 \leq z \leq L/4\). What fraction of the total volume of the box is this cubical volume? (b) If the particle is in the ground state \((n_X = 1, n_Y = 1, n_Z = 1)\), calculate the probability that the particle will be found in the cubical volume defined in part (a). (c) Repeat the calculation of part (b) when the particle is in the state \(n_X = 2, n_Y = 1, n_Z = 1\).

(a) Write out the ground-state electron configuration (\(1s^2, 2s^2,\dots\)) for the beryllium atom. (b) What element of nextlarger \(Z\) has chemical properties similar to those of beryllium? Give the ground-state electron configuration of this element. (c) Use the procedure of part (b) to predict what element of nextlarger \(Z\) than in (b) will have chemical properties similar to those of the element you found in part (b), and give its ground-state electron configuration.

(a) Write out the ground-state electron configuration (\(1s^2, 2s^2,\dots\)) for the carbon atom. (b) What element of nextlarger \(Z\) has chemical properties similar to those of carbon? Give the ground-state electron configuration for this element.

For germanium (\(Ge, Z = 32\)), make a list of the number of electrons in each subshell (\(1s, 2s, 2p,\dots\)). Use the allowed values of the quantum numbers along with the exclusion principle; do not refer to Table 41.3.

A hydrogen atom initially in an \(n\) = \(3,\) \(l\) = 1 state makes a transition to the \(n\) = \(2\), \(l\) = \(0\), \(j\) = \\(\frac{1}{2}\\) state. Find the difference in wavelength between the following two photons: one emitted in a transition that starts in the \(n\) = \(3\), \(l\) = \(1\), \(j\) = \\(\frac{3}{2}\\) state and one that starts instead in the \(n\) = \(3\), \(l\) = \(1\), \(j\) = \\(\frac{1}{2}\\) state. Which photon has the longer wavelength?
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