/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 Consider states with angular-mom... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider states with angular-momentum quantum number \(l=2\) (a) In units of \(\hbar\) , what is the largest possible value of \(L_{z}\) ? (b) In units of \(\hbar\) , what is the value of \(L ?\) Which is larger: \(L\) or the maximum possible \(L_{z} ?\) (c) For each allowed value of \(L_{v}\) what angle does the vector \(\vec{L}\) make with the \(+z\) -axis? How does the minimum angle for \(l=2\) compare to the minimum angle for \(l=3\) calculated in Example 41.2\(?\)

Short Answer

Expert verified
(a) Largest possible \(L_z = 2\hbar\). (b) \(L = \sqrt{6}\hbar\), which is larger than \(L_z = 2\hbar\). (c) Angle depends on \(\cos(\theta) = \frac{m_l}{\sqrt{6}}\). For \(l=3\), minimum \(\theta\) is smaller.

Step by step solution

01

Understanding the quantum number

The angular momentum quantum number is given as \(l=2\). This number means that the magnetic quantum number, \(m_l\), can take integer values from \(-l\) to \(+l\), including zero. Therefore, \(m_l = -2, -1, 0, 1, 2\).
02

Finding the largest possible value of Lz

The component \(L_z\) represents the angular momentum along the z-axis, given by \(L_z = m_l\hbar\). To find the largest possible value of \(L_z\), we choose the maximum \(m_l\), which is 2. Thus, the largest possible value of \(L_z\) is \(L_z = 2\hbar\).
03

Calculating the total angular momentum L

The magnitude of the total angular momentum \(L\) is given by the equation \(L = \sqrt{l(l+1)}\hbar\). Plugging in \(l = 2\), we get \[L = \sqrt{2(2+1)}\hbar = \sqrt{6}\hbar\].
04

Comparing L and Lz

We have \(L = \sqrt{6}\hbar\) and the maximum possible \(L_z = 2\hbar\). Calculating, \(\sqrt{6} \approx 2.45\), thus \(\sqrt{6}\hbar\) is larger than \(2\hbar\).
05

Calculating the angle with the z-axis

For a given \(m_l\), the angle \(\theta\) between the total angular momentum \(\vec{L}\) and the \(+z\)-axis can be found using \(\cos(\theta) = \frac{L_z}{L}\).
06

Finding the angles for each allowed Lz value

We compute \(\cos(\theta)\) for each \(m_l\).- For \(m_l = 2\), \(\cos(\theta) = \frac{2\hbar}{\sqrt{6}\hbar} = \frac{2}{\sqrt{6}} = \sqrt{\frac{2}{3}}\).- For \(m_l = 1\), \(\cos(\theta) = \frac{1}{\sqrt{6}}\).- For \(m_l = 0\), \(\cos(\theta) = 0\).- Similarly, negative \(m_l\) values provide symmetric angles due to squaring when calculating angles.
07

Comparing with the case for l=3

For \(l=3\), the smallest angle corresponds to the largest \(m_l\), which is 3. The angle would be steeper because \(\cos(\theta) = \frac{3}{\sqrt{12}}\). Compare \(\theta\) values for smallest angles using these calculations.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Angular Momentum Components
In quantum mechanics, angular momentum components describe how the angular momentum of a particle is directed in space. Specifically, the component along the z-axis is denoted by \(L_z\), which is quantized and takes on values based on specific quantum numbers.
The key here is the magnetic quantum number \(m_l\), which determines the possible values of \(L_z\) in units of \(\hbar\). For a given angular momentum quantum number \(l\), \(L_z\) can have values of \(-l\hbar, -(l-1)\hbar, \ldots, 0, \ldots, (l-1)\hbar, l\hbar\).
  • The Angular momentum component \(L_z\) represents the projection of angular momentum along the z-axis.
  • With \(l = 2\), \(L_z\) can be \(-2\hbar, -1\hbar, 0, 1\hbar, 2\hbar\).
Because \(L_z\) changes discretely, understanding these components helps do complex calculations concerning orientation of particles like electrons around nuclei.
Quantum Mechanics
Quantum mechanics is the branch of physics that deals with the mathematical description of the behavior of particles at the atomic and subatomic levels. It introduces concepts like quantization of energy levels and angular momentum, which differ significantly from classical physics.
In this exercise, we deal with quantized angular momentum, where certain values are allowed for the magnitude and direction components of momentum. These allowable values are described by angular momentum quantum numbers.
  • Angular momentum is not continuous in quantum mechanics; it takes discrete values.
  • The quantization of such properties is pivotal in explaining atomic stability and electron configurations.
By breaking down problems into such quantized values, quantum mechanics allows for the accurate prediction of particle behavior in ways that classical physics cannot.
Magnetic Quantum Number
The magnetic quantum number, denoted as \(m_l\), is central to determining the orientation of angular momentum in space. For a specific angular momentum quantum number \(l\), \(m_l\) can range from \(-l\) to \(+l\), impacting allowed values of \(L_z\).
This number can be seen as a representation of the number of possible orientations that a particle's orbit can have concerning an external magnetic field.
  • For \(l = 2\), the possible \(m_l\) values are \(-2, -1, 0, 1, 2\).
  • Each \(m_l\) corresponds to a different orientation parallel or antiparallel to an applied magnetic field.
Understanding \(m_l\) sheds light on how angular momentum behaves in applied fields and why particles have specific alignments in space.
Total Angular Momentum
The total angular momentum \(L\) of a particle is a vector quantity representing the magnitude of angular motion. Its magnitude is quantized and calculated using \(L = \sqrt{l(l+1)}\hbar\).
This total angular momentum is different from its component \(L_z\); while \(L_z\) is directional along the z-axis, \(L\) is the overall magnitude of the momentum.
  • Using \(l = 2\), calculate \(L = \sqrt{6}\hbar\).
  • It is always larger than the maximum \(L_z = 2\hbar\), which is along one axis.
This quantum relationship highlights how our measure of a particle's momentum must consider different axes and why these vectors are crucial in quantum systems.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the energy difference between the \(m_{s}=\frac{1}{2}\left(^{u} \text { spin }\right.\) \(\mathrm{up}^{\prime \prime}\) 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?

Estimate the minimum and maximum wavelengths of the characteristic \(x\) rays emitted by (a) vanadium \((Z=23)\) and \((b)\) rhenium \((Z=45)\) . Discuss any approximations that you make.

(a) If the intrinsic spin angular momentum \(S\) of the earth had the same limitations as that of the electron, what would be the angular velocity of our planet's spin on its axis? To get a reasonable answer but simplify the calculations, assume that the earth is uniform throughout. (b) Could we, in principle, use the method of part (a) to determine the angular velocity of the electron's spin? Why?

Weird Universe. In another universe, the electron is a \(\operatorname{spin} \frac{3}{2}\) rather than a spin- \(\frac{1}{2}\) particle, but all other physics are the same as in our universe. In this universe, (a) what are the atomic numbers of the lightest two inert gases? (b) What is the ground-state electron configuration of sodium?

An electron in a hydrogen atom is in an \(s\) level, and the atom is in a magnetic field \(\vec{B}=B \hat{k} .\) Explain why the "spin up" state \(\left(m_{s}=+\frac{1}{2}\right)\) has a higher energy than the "spin down" state \(\left(m_{s}=-\frac{1}{2}\right)\)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.