/*! 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 21 A hydrogen atom in a particular ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A hydrogen atom in a particular orbital angular momentum state is found to have \(j\) quantum numbers \(\frac{7}{2}\) and \(\frac{9}{2} .\) What is the letter that labels the value of \(l\) for the state?

Short Answer

Expert verified
The letter for \(l = 4\) is \(g\).

Step by step solution

01

Understanding Quantum Numbers

In the quantum mechanical model of an atom, angular momentum quantum numbers are used to describe the electron's state. The quantum number \(j\) represents the total angular momentum, which is the vector sum of the orbital angular momentum \(l\) and the spin angular momentum \(s\). The possible values of \(j\) are given by \(j = l + s\) or \(j = l - s\).
02

Identifying Spin Quantum Number

Electrons have a spin quantum number \(s\) of \(\frac{1}{2}\). This is because electrons are fermions, and the spin quantum number for such particles is usually \(\pm\frac{1}{2}\).
03

Finding Possible Values of \(l\) for Each \(j\)

Given \(j = \frac{7}{2}\) and \(j = \frac{9}{2}\), we need to find possible \(l\) values. For each \(j\), try calculating \(l = j \pm \frac{1}{2}\).
04

Case 1: Calculating \(l\) for \(j = \frac{7}{2}\)

\(l\) can be calculated as \(l = \frac{7}{2} - \frac{1}{2} = 3\) or \(l = \frac{7}{2} + \frac{1}{2} = 4\). Therefore, possible \(l\) values are 3 or 4 for this \(j\).
05

Case 2: Calculating \(l\) for \(j = \frac{9}{2}\)

\(l\) can be calculated as \(l = \frac{9}{2} - \frac{1}{2} = 4\) or \(l = \frac{9}{2} + \frac{1}{2} = 5\). Therefore, possible \(l\) values are 4 or 5 for this \(j\).
06

Identifying Common \(l\) Value

Comparing possible \(l\) values from both \(j\), a common value is \(l = 4\). This value is consistent with both \(j = \frac{7}{2}\) and \(j = \frac{9}{2}\).
07

Matching \(l\) to Orbital Letter

The letter that corresponds to \(l = 4\) is \(g\), where the letters for \(l\) are assigned as \(s\) for 0, \(p\) for 1, \(d\) for 2, \(f\) for 3, and \(g\) for 4.

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.

Orbital Angular Momentum
In quantum mechanics, the concept of orbital angular momentum is crucial for understanding how electrons behave within an atom. Orbital angular momentum, denoted by the quantum number \( l \), defines the shape of the electron's orbital and plays a key role in determining the atom's structure and chemical behavior.
  • The quantum number \( l \) can take on any integer value from 0 up to \( n-1 \), where \( n \) is the principal quantum number, determining the electron's energy level.
  • These values of \( l \) correlate to specific letters that label the orbitals: \( s \) for 0, \( p \) for 1, \( d \) for 2, \( f \) for 3, and \( g \) for 4, continuing alphabetically from there.

The orbital angular momentum is quantized, meaning it can only take on specific discrete values. This is a fundamental property of electrons in atoms: unlike classical particles which can have a continuum of possible angular momentum values, quantum particles like electrons are restricted by these discrete quantum numbers.
This quantization explains why only certain electron transitions are observed, giving rise to distinctive spectral lines.
Spin Quantum Number
The spin quantum number, represented by \( s \), is a unique quantum number that arises due to the intrinsic spin of particles such as electrons. Unlike orbital angular momentum, which depends on how an electron moves through space, the spin quantum number describes an inherent geometric property of the electron itself. This property is akin to a top spinning around its axis.
  • The possible values for an electron's spin quantum number are \( +\frac{1}{2} \) and \( -\frac{1}{2} \). This binary choice reflects the two allowed spin states of the electron, sometimes referred to as "spin up" or "spin down".
  • Spin is a crucial element in the Pauli exclusion principle, which states that no two electrons in an atom can have the exact same set of quantum numbers.

Understanding electron spin was a defining moment in the development of quantum mechanics, and it helps explain many physical phenomena such as magnetism.The combination of orbital and spin angular momenta leads to the total angular momentum quantum number \( j \). Knowing \( s \) allows us to calculate \( j \) values and solve problems involving electron configurations in atoms.
Hydrogen Atom
The hydrogen atom is the simplest atom and serves as the cornerstone for understanding atomic physics and quantum mechanics. It consists of a single electron orbiting a single proton, which makes it the perfect model to apply quantum mechanics.
  • The key quantum numbers that describe the hydrogen atom are: the principal quantum number \( n \), the orbital angular momentum quantum number \( l \), the magnetic quantum number \( m_l \), and the spin quantum number \( s \).
  • Since the hydrogen atom is isolated from interactions that may alter its electron states, it provides clear spectral lines, making it ideal for comparing theoretical predictions of quantum mechanics against experimental data.

The importance of the hydrogen atom in physics cannot be overemphasized—it laid the foundation for the quantum theory of atoms, leading to a deeper understanding of more complex atoms and molecules.Modern research into multi-electron systems still relies heavily on principles first laid out by studying the hydrogen atom.In problems involving quantum numbers, like the given exercise on orbital angular momentum states, the hydrogen atom's simplicity makes it a go-to example for illustrating complex quantum principles in a clear and understandable manner.

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 the transition from a 3\(d\) to a 2\(p\) state of hydrogen in an external magnetic field. Assume that the effects of electron spin can be ignored (which is not actually the case) so that the magnetic field interacts only with the orbital angular momentum. Identify each allowed transition by the \(m_{I}\) values of the initial and final states. For each of these allowed transitions, determine the shift of the transition energy from the zero-field value and show that there are three different transition energies.

A hydrogen atom is in a \(d\) state. In the absence of an external magnetic field the states with different \(m_{I}\) values have (approximately) the same energy. Consider the interaction of the magnetic field with the atom's orbital magnetic dipole moment. (a) Calculate the splitting (in electron volts) of the \(m_{l}\) levels when the atom is put in a 0.400 - T magnetic field that is in the \(+\) z-direction. (b) Which \(m_{l}\) level will have the lowest energy? (c) Draw an energy-level diagram that shows the \(d\) levels with and without the external magnetic field.

Calculate, in units of \(\hbar\) , the magnitude of the maximum orbital angular momentum for an electron in a hydrogen atom for states with a principal quantum number of \(2,20,\) and 200 . Compare each with the value of \(n\) h postulated in the Bohr model. What trend do you see?

For an ion with nuclear charge \(Z\) and a single electron, the electric potential energy is \(-\mathrm{Ze}^{2} / 4 \pi \epsilon_{0} r\) and the expression for the energies of the states and for the normalized wave functions are obtained from those for hydrogen by replacing \(e^{2}\) by \(\mathrm{Ze}^{2}\) . Consider the \(\mathrm{N}^{6+}\) ion, with seven protons and one electron. (a) What is the wavelength of the photon emitted when the \(\mathrm{N}^{6+}\) ion makes a transition from the \(n=2\) state to the \(n=1\) ground state?

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

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.