/*! 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 50 (a) Show all the distinct states... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 41: Problem 50

(a) Show all the distinct states for an electron in the \(N\) shell of hydrogen. Include all four quantum numbers. (b) For an \(f\) electron in the \(N\) shell, what is the largest possible orbital angular momentum and the greatest positive value for the component of this angular momentum along any chosen direction (the z-axis)? What is the magnitude of its spin angular momentum? Express these quantities in units of \(\hbar\) . (c) For an electron in the \(d\) state of the \(N\) shell, what are the maximum and minimum angles between its angular momentum vector and any chosen direction (the \(z\) -axis)? (d) What is the largest value of the orbital angular momentum for an \(f\) electron in the \(M\) shell?

Short Answer

Expert verified
(a) N shell: \( L \) values from 0 to 3. (b) Largest \( L \): \( 2\sqrt{3} \hbar \), spin \( \frac{\sqrt{3}}{2} \hbar \). (c) Angles: \( \theta_{\min} \approx 35.26^\circ \), \( \theta_{\max} \approx 144.74^\circ \). (d) \( 2\sqrt{3} \hbar \).

Step by step solution

01

Understand the Electron States in the N Shell

For the hydrogen atom, the principal quantum number \( n \) determines the shell. The N shell corresponds to \( n = 4 \). The azimuthal quantum number \( l \) can take integer values from \( 0 \) to \( n-1 \). The magnetic quantum number \( m_l \) ranges from \( -l \) to \( l \), and the spin quantum number \( m_s \) is either \( +\frac{1}{2} \) or \( -\frac{1}{2} \). To find all distinct electron states, combine these quantum numbers for each possible \( l \) value.
02

Enumerate the Quantum Numbers for the N Shell

For \( n = 4 \), the possible \( l \) values are 0, 1, 2, and 3. For each \( l \), enumerate the \( m_l \) values, and for each \( m_l \), enumerate the two possible \( m_s \). This yields:- \( l = 0 \): \( m_l = 0 \), \( m_s = \pm\frac{1}{2} \)- \( l = 1 \): \( m_l = -1, 0, 1 \); \( m_s = \pm\frac{1}{2} \)- \( l = 2 \): \( m_l = -2, -1, 0, 1, 2 \); \( m_s = \pm\frac{1}{2} \)- \( l = 3 \): \( m_l = -3, -2, -1, 0, 1, 2, 3 \); \( m_s = \pm\frac{1}{2} \)
03

Calculate Largest Possible Angular Momentum for an f Electron

For \( l = 3 \) (since \( f \) corresponds to \( l = 3 \)), the largest orbital angular momentum \( L \) is given by \( L = \sqrt{l(l+1)} \hbar \). For \( l = 3 \),\( L = \sqrt{3(3+1)} \hbar = \sqrt{12} \hbar = 2\sqrt{3} \hbar \). The largest \( m_l \) is 3, so the component along the z-axis is: \( L_z = m_l \hbar = 3\hbar \). An electron's spin angular momentum magnitude is always \( S = \sqrt{\frac{3}{4}} \hbar = \frac{\sqrt{3}}{2} \hbar \).
04

Determine Angles for Angular Momentum in d State

For a \( d \) electron, \( l = 2 \). The total angular momentum magnitude \( L = \sqrt{2(2+1)} \hbar = \sqrt{6} \hbar \). The maximum value of \( m_l \) is 2, making \( L_z = 2\hbar \). The cosine of the angle \( \theta \) with the z-axis is\( \cos \theta = \frac{L_z}{L} \). Hence, the minimum angle (maximum value of \( \cos \theta \)) is:\( \theta_{\min} = \cos^{-1}\left(\frac{2}{\sqrt{6}}\right) \),and the maximum angle (minimum value of \( \cos \theta \)) is:\( \theta_{\max} = \cos^{-1}\left(\frac{-2}{\sqrt{6}}\right) \).
05

Find Largest Orbital Angular Momentum for f Electron in M Shell

For an f electron in the M shell, \( n = 3 \), and \( l = 3 \). The orbital angular momentum \( L = \sqrt{l(l+1)} \hbar = \sqrt{3(3+1)} \hbar = 2\sqrt{3} \hbar \), same as in the N shell.

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 91影视!

Key Concepts

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

Electron States
Understandably, electrons have unique states depending on their position within an atom. This is determined by four main quantum numbers: **the principal quantum number (\( n \)), the azimuthal quantum number (\( l \)), the magnetic quantum number (\( m_l \)), and the spin quantum number (\( m_s \))**.
- **Principal Quantum Number (\( n \)):** Identifies the shell or energy level. For instance, in the N shell of a hydrogen atom, \( n = 4 \).
- **Azimuthal Quantum Number (\( l \)):** Corresponds to the subshell or shape of the electron orbital. It varies from 0 up to \( n-1 \). For each shell, electrons can have values of \( l = 0, 1, 2, \) or 3.
- **Magnetic Quantum Number (\( m_l \)):** Describes the orientation of the orbital鈥檚 angular momentum with values ranging from \( -l \) to \( +l \). For an electron in the N shell, \( l = 3 \) implies \( m_l = -3, -2, -1, 0, 1, 2, 3 \).
- **Spin Quantum Number (\( m_s \)):** The only two possible values are \( +\frac{1}{2} \) or \( -\frac{1}{2} \), indicating the electron's intrinsic spin.
Each electron state is unique and results from combinations of these quantum numbers. Understanding how these states are organized helps us delve deeper into concepts like bonding and reactions happening on an atomic level.
Orbital Angular Momentum
Every electron has an associated orbital angular momentum. This refers to the amount of rotation or angular motion an electron has in its orbital.
- **Magnitude of Orbital Angular Momentum:** Calculated using the formula \( L = \sqrt{l(l+1)} \hbar \). This represents how much angular motion is involved for a given \( l \). In our context, an \( f \) electron (\( l = 3 \)) in any shell will have an orbital angular momentum of \( 2\sqrt{3} \hbar \).
- **Component Along a Direction:** The maximum component of orbital angular momentum along any chosen direction, such as the z-axis, amounts to \( L_z = m_l \hbar \), where \( m_l \) takes values between \( -l \) and \( +l \). An \( f \) electron in the N shell can have a maximum \( L_z \) of \( 3\hbar \).
Understanding orbital angular momentum allows insights into the behavior of electrons around atoms, including how they contribute to the magnetic properties and spectra of atoms.
Spin Angular Momentum
Beyond just orbitals, electrons have an intrinsic form of angular momentum known as spin. Unlike orbital angular momentum, spin is a fundamental quantum property and does not depend on how the electron is orbiting a nucleus.
- **Magnitude of Spin Angular Momentum:** Each electron inherently has the same magnitude of spin angular momentum given by \( S = \sqrt{\frac{3}{4}} \hbar \). This magnitude stays consistent across all electrons when considering their intrinsic nature.
- **Spin Quantum Number (\( m_s \)):** The spin angular momentum vector can point in one of two directions, indicating either \( +\frac{1}{2} \) or \( -\frac{1}{2} \). This is essential for forming electron pairs in atomic orbitals, which is a key element of the Pauli exclusion principle.
Spin angular momentum is crucial for understanding the magnetic properties of materials and how electrons pair with each other in chemical bonds or quantum mechanical contexts.

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

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 \(L_{z}\) , what angle does the vector \(\vec{\boldsymbol{L}}\) make with the \(+z\) -axis? How does the minimum angle for \(l=2\) compare to the minimum angle for \(l=3\) calculated in Example 41.3\(?\)

For magnesium, the first ionization potential is 7.6 \(\mathrm{eV}\) The second ionization potential (additional energy required to remove a second electron is almost twice this, \(15 \mathrm{eV},\) and the third ionization potential is much larger, about 80 \(\mathrm{eV} .\) How can these numbers be understood?

For a particle in a three-dimensional box, what is the degeneracy (number of different quantum states with the same energy) of the following energy levels: (a) 3\(\pi^{2} \hbar^{2} / 2 m L^{2}\) and \((b)\) 9\(\pi^{2} \hbar^{2} / 2 m L^{2 n} ?\)

(a) Write out the ground-state electron configuration \(\left(1 s^{2},\right.\) \(2 s^{2}, \ldots .\) for the beryllium atom. (b) What element of next-larger \(Z\) 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 next-larger \(Z\) 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 hydrogen atom in the 5\(g\) state is placed in a magnetic field of 0.600 T that is in the \(z\) -direction. (a) Into how many levels is this state split by the interaction of the atom's orbital magnetic dipole moment with the magnetic field? (b) What is the energy separation between adjacent levels? (c) What is the energy separation between the level of lowest energy and the level of highest energy?
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Measurements

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Particle Model of Matter

Read Explanation

Wave Optics

Read Explanation

Electricity

Read Explanation
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.