(a) If you treat an electron as a classical spherical object with a radius of 1.0×10−171.0\$$times10^{-17} m$$, what angular speed is necessary to produce a spin angular momentum of magnitude 34h\sqrt{\frac34}h?(b) Use v=rωv=r\omega and the result of part (a) to calculate the speed vv of a point at the electron's equator. What does your result suggest about the validity of this model?

Ch 41: Quantum Mechanics II: Atomic Structure

Young & Freedman Calc - University Physics 14th Edition

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Young & Freedman Calc 14th Edition

Ch 41: Quantum Mechanics II: Atomic Structure

Problem 23

Chapter 41, Problem 23

(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 $\sqrt{\frac{3}{4}} h$ ?
(b) Use $v = r ω$ 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?

Verified step by step guidance

Step 1: Understand the problem. The problem involves calculating the angular speed (ω) of an electron modeled as a classical sphere to produce a given spin angular momentum (S). Then, using the relationship v = rω, calculate the linear speed (v) of a point at the equator of the electron and analyze the result.

Step 2: Recall the formula for spin angular momentum of a rigid body: S = Iω, where I is the moment of inertia and ω is the angular speed. For a solid sphere, the moment of inertia is given by I = (2/5)mr², where m is the mass of the sphere and r is its radius.

Step 3: Substitute the given values into the formula. The radius of the electron is r = 1.0 × 10⁻¹⁷ m, and the spin angular momentum is given as S = (3/4)ħ, where ħ is the reduced Planck's constant (ħ ≈ 1.054 × 10⁻³⁴ J·s). Solve for ω using the equation ω = S / I.

Step 4: Use the relationship v = rω to calculate the linear speed of a point at the equator of the electron. Substitute the radius r and the angular speed ω obtained in the previous step into this formula.

Step 5: Analyze the result. Compare the calculated linear speed (v) to the speed of light (c ≈ 3.0 × 10⁸ m/s). If v exceeds c, this suggests that the classical model of the electron as a rigid sphere is not valid, as it violates the principles of special relativity.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Angular Momentum

Angular momentum is a measure of the rotational motion of an object and is defined as the product of the moment of inertia and the angular velocity. For a spherical object, the moment of inertia can be calculated using its mass and radius. In the context of an electron, the angular momentum is quantized, and understanding how to calculate it is essential for determining the required angular speed to achieve a specific angular momentum.

Classical vs Quantum Models

The classical model treats particles like electrons as solid objects with defined properties such as radius and mass, which can lead to contradictions when applied to quantum-scale phenomena. Electrons exhibit wave-particle duality and are better described by quantum mechanics, where properties like position and momentum are inherently uncertain. This distinction is crucial when evaluating the validity of classical assumptions in the context of subatomic particles.

Linear Velocity and Circular Motion

In circular motion, the linear velocity (v) of a point on the circumference is related to the angular velocity (ω) and the radius (r) of the circle by the equation v = rω. This relationship is fundamental in analyzing the motion of objects in circular paths, such as the hypothetical electron in this problem. Understanding this concept allows for the calculation of the linear speed at the electron's equator based on the angular speed derived from the angular momentum.

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Textbook Question

The hyperfine interaction in a hydrogen atom between the magnetic dipole moment of the proton and the spin magnetic dipole moment of the electron splits the ground level into two levels separated by $5.9 \times 10^{- 6}$ eV. Calculate the wavelength and frequency of the photon emitted when the atom makes a transition between these states, and compare your answer to the value given at the end of Section $41.5$ . In what part of the electromagnetic spectrum does this lie? Such photons are emitted by cold hydrogen clouds in interstellar space; by detecting these photons, astronomers can learn about the number and density of such clouds.

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Textbook Question

A hydrogen atom undergoes a transition from a $2 p$ state to the $1 s$ ground state. In the absence of a magnetic field, the energy of the photon emitted is $122$ nm. The atom is then placed in a strong magnetic field in the $z$ -direction. Ignore spin effects; consider only the interaction of the magnetic field with the atom's orbital magnetic moment. How many different photon wavelengths are observed for the $2 p right arrow 1 s$ transition? What are the $m_{l}$ values for the initial and final states for the transition that leads to each photon wavelength?

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Textbook Question

A hydrogen atom in the $5 g$ state is placed in a magnetic field of $0.600$ T that is in the $z$ -direction. Into how many levels is this state split by the interaction of the atom's orbital magnetic dipole moment with the magnetic field?

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Textbook Question

A hydrogen atom in a particular orbital angular momentum state is found to have $j$ quantum numbers $\frac{7}{2}$ and $\frac{9}{2}$ . If $n equals 5$ , what is the energy difference between the $j = \frac{7}{2}$ and $j = \frac{9}{2}$ levels?

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Textbook Question

A hydrogen atom in a $3 p$ state is placed in a uniform external magnetic field $\vec{B}$ . Consider the interaction of the magnetic field with the atom's orbital magnetic dipole moment. What field magnitude $B$ is required to split the $3 p$ state into multiple levels with an energy difference of $2.71 \times 10^{- 5}$ eV between adjacent levels?

Calculate the energy difference between the $m_{s} = \frac{1}{2}$ ('spin up') and $m_{s} = - \frac{1}{2}$ ('spin down') levels of a hydrogen atom in the $1 s$ state when it is placed in a $1.45$ T magnetic field in the negative $z$ -direction. Which level, $m_{s} = \frac{1}{2}$ or $m_{s} = - \frac{1}{2}$ , has the lower energy?

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