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Chapter 8: Problem 94

In a simple model of the hydrogen atom, an electron orbits a proton at \(2.18 \times 10^{6} \mathrm{~m} / \mathrm{s}\) in a circle of radius \(5.29 \times 10^{-11} .\) Find the magnitude and direction of the electron's angular momentum.

Short Answer

Expert verified
The angular momentum is \(1.056 \times 10^{-34} \text{ kg m}^2/ ext{s}\), pointing upwards.

Step by step solution

01

Recall the formula for angular momentum

The angular momentum \( L \) of a particle moving in a circular orbit is given by the formula \( L = mvr \), where \( m \) is the mass of the particle, \( v \) is the tangential speed, and \( r \) is the radius of the circle.
02

Identify the given values

In the problem, we are given the electron's speed \( v = 2.18 \times 10^6 \text{ m/s} \) and the radius of the circle \( r = 5.29 \times 10^{-11} \text{ m} \). The mass of an electron \( m \) is \( 9.11 \times 10^{-31} \text{ kg} \).
03

Calculate the angular momentum

Substitute the given values into the angular momentum formula \( L = mvr \): \[L = (9.11 \times 10^{-31} \text{ kg})(2.18 \times 10^6 \text{ m/s})(5.29 \times 10^{-11} \text{ m})\]Solve this to find \( L \).
04

Execute the calculation

Perform the multiplication: \[ L = 9.11 \times 2.18 \times 5.29 \times 10^{-31+6-11} \text{ kg m}^2/ ext{s} \]First calculate the coefficients:\[ 9.11 \times 2.18 \times 5.29 \approx 1.056 \]Now calculate the powers of 10:\[ 10^{-31+6-11} = 10^{-36} \]Thus, the angular momentum \( L = 1.056 \times 10^{-34} \text{ kg m}^2/ ext{s} \).
05

Determine the direction of angular momentum

The direction of angular momentum is perpendicular to the plane of the orbit, following the right-hand rule. If the electron is orbiting counterclockwise when viewed from above, the angular momentum vector points upwards.

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Key Concepts

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

Hydrogen Atom Model
The hydrogen atom, one of the simplest atoms, consists of only a single electron orbiting around a single proton. This simplicity makes it an ideal subject for understanding fundamental atomic structures.

In many educational models, the electron is depicted in a circular orbit around the proton, much like a planet around the sun, although in reality, electrons occupy more complex quantum mechanical orbitals.
  • This model helps in visualizing atomic behavior with respect to classical motion concepts like angular momentum.
  • It introduces students to foundational physics principles, aiding in understanding more complex atoms.
The hydrogen atom model is crucial for building a bridge between classical physics and the quantum world, helping students visualize atomic structure.
Electron Motion
Electron motion in the hydrogen atom model focuses on an electron's circular path around a proton. Electrons move with a specific tangential speed, contributing to the calculation of angular momentum.

The constant speed indicates uniform circular motion, which means:
  • The magnitude of the velocity is constant, while the direction changes continuously.
  • This results in acceleration towards the center of the circle, known as centripetal acceleration.
Understanding electron motion is fundamental to comprehend how angular momentum arises from circular movement, using the relationship between speed, mass, and radius.
Circular Orbit
The concept of a circular orbit is central to understanding how angular momentum functions within atomic systems like the hydrogen atom. A circular orbit means the path of the electron is perfectly circular and has a constant radius.

This assumption simplifies calculations and helps illustrate key points:
  • The constant radius (r = 5.29 596 ext{ m}) simplifies the application of formulas like angular momentum (L = mvr).
  • Despite the reality being more complex due to quantum mechanics, this model is a great starting point for learning essential physics principles.
A circular orbit also intuitively helps understand how altering any of the parameters like mass, speed, or radius affects angular momentum.
Right-Hand Rule
The right-hand rule is a handy tool used to determine the direction of angular momentum vector. When dealing with circular motion, understanding the direction of vectors is crucial.

To apply the right-hand rule:
  • Point your right thumb in the direction of the angular momentum vector, which is perpendicular to the plane of motion.
  • Rotate your fingers in the direction of the electron's motion. If electron motion is counterclockwise, your thumb will point upwards, indicating the direction of the angular momentum.
This visual tool simplifies vector direction understanding, an otherwise abstract concept, helping students comprehend how angular quantities relate to physical orientation.

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Most popular questions from this chapter

A figure skater spins with arms extended horizontally. Explain what happens, and why, when she brings her arms tight to her body.

A drag racer has 76.2 -cm-diameter wheels. Haw fast are they turning when the drag racer is doing \(140 \mathrm{~km} / \mathrm{h} ?\)

A bicycle wheel rotates with increasing angular velocity. Compare the tangential acceleration of a point on the rim with that of a point midway along one spoke.

Explain why the rotational inertia of a hollow ball is greater than that of a uniform solid ball with the same mass and radius.

In a car's drive train, a flywheel with rotational inertia \(26.0 \mathrm{~kg} \cdot \mathrm{m}^{2}\) rotates at \(310 \mathrm{rad} / \mathrm{s}\). The clutch engages, pressing the clutch plate \(-\) a disk with rotational inertia half that of the flywheel \(-\) against the flywheel, so the two rotate as one. Assuming both are otherwise free from torque, what's the rotational speed of the combined system? (a) \(155 \mathrm{rad} / \mathrm{s} ;\) (b) \(206 \mathrm{rad} / \mathrm{s} ;\) (c) \(267 \mathrm{rad} / \mathrm{s}\) (d) \(310 \mathrm{rad} / \mathrm{s}\).
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