Problem 11 - Physics for Scientists and Engineers with Modern Physics

Question

(II) The magnitude of the orbital angular momentum in an excited state of hydrogen is \(6.84 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\) and the \(z\) component is \(2.11 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s} .\) What are all the possible values of \(n, \ell,\) and \(m_{\ell}\) for this state?

Step-by-Step Solution

Answer

\(n = 7, \ell = 6, m_\ell = 2\).

Step 1: Identify the relationship for angular momenta

The magnitude of the orbital angular momentum is given by the formula \( L = \sqrt{\ell(\ell+1)} \hbar \), where \( \hbar = 1.0545718 \times 10^{-34} \, \mathrm{J \cdot s} \). The \( z \) component of the angular momentum is \( L_z = m_\ell \hbar \). Given the values, we need to find \( \ell \) and \( m_\ell \).

Step 2: Calculate \( \ell \)

Substitute the given magnitude of the angular momentum into the formula: \[ L = \sqrt{\ell(\ell+1)} \hbar = 6.84 \times 10^{-34} \, \mathrm{J \cdot s} \]Solve for \( \ell(\ell+1) \):\[ \sqrt{\ell(\ell+1)} = \frac{6.84 \times 10^{-34}}{1.0545718 \times 10^{-34}} \approx 6.49 \]\[ \ell(\ell+1) \approx 42.13 \]Solve the quadratic equation \( \ell^2 + \ell - 42.13 = 0 \) to find \( \ell \).

Step 3: Determine integer \( \ell \)

Solve the quadratic equation approximately by factoring or using the quadratic formula \( \ell = \frac{-1 \pm \sqrt{1 + 4 \times 42.13}}{2} \). This simplifies to \( \ell \approx 6 \), as \( \ell \) must be an integer.

Step 4: Calculate \( m_\ell \)

Using the formula for the \( z \) component:\[ L_z = m_\ell \hbar = 2.11 \times 10^{-34} \, \mathrm{J \cdot s} \]Solve for \( m_\ell \):\[ m_\ell = \frac{2.11 \times 10^{-34}}{1.0545718 \times 10^{-34}} \approx 2 \]\( m_\ell \) must be an integer, so \( m_\ell = 2 \).

Step 5: Determine \( n \) based on \( \ell \) value

In a hydrogen atom, \( n \) must be greater than \( \ell \). Since \( \ell = 6 \), the smallest possible principal quantum number is \( n = 7 \).

Key Concepts

Orbital Angular Momentum

Orbital angular momentum is a fundamental concept in quantum mechanics that describes the angular momentum of particles in an atom, particularly in an electron's orbit.
In essence, it is a measure of how much rotation an electron has as it moves around the nucleus. This concept is crucial for understanding the structure of atoms and how electrons behave on a quantum level.

Mathematically, the magnitude of orbital angular momentum \( L \) in a quantum state can be described using the formula:

• \( L = \sqrt{\ell(\ell+1)} \hbar \)

Here, \( \ell \) is the azimuthal quantum number, also known as the angular momentum quantum number, and \( \hbar \) is the reduced Planck's constant. This formula highlights that the angular momentum is quantized, meaning it can only take on specific discrete values.
This quantization relates to the broader principle in quantum physics where observables such as energy and angular momentum exist in discrete packets rather than continuous ranges.

Quantum Numbers

Quantum numbers are essential in the quantum mechanical model of the atom, where they describe different properties of atomic orbitals and the electrons in those orbitals.
There are four quantum numbers: principal \( n \), azimuthal (angular) \( \ell \), magnetic \( m_\ell \), and spin \( m_s \). Each one provides different information on the electron's position and energy.

**Quantum Number Details**

• **Principal Quantum Number \( n \):** Indicates the energy level of the electron, effectively telling us the size and energy of the orbital.
• **Azimuthal Quantum Number \( \ell \):** Relates to the shape of the orbital. Its values range from 0 to \( n-1 \).
• **Magnetic Quantum Number \( m_\ell \):** Specifies the orientation of the orbital in space, ranging from \( -\ell \) to \( \ell \).
• **Spin Quantum Number \( m_s \):** Represents the electron's spin orientation, typically \( +\frac{1}{2} \) or \( -\frac{1}{2} \).

In the exercise, the orbital angular momentum's magnitude was used to find \( \ell \), and the \( z \)-component helped in determining \( m_\ell \). The quantum numbers together provide a detailed description of the electron's state.

Hydrogen Atom

The hydrogen atom is the simplest atom and has a single proton in its nucleus with one electron orbiting around it. This simplicity has made it a crucial subject of study in quantum mechanics.
Understanding the electron's behavior in a hydrogen atom laid the groundwork for quantum theory, influencing how we understand heavier atoms and even molecules.

In these atoms, electrons are found in specific energy levels or shells, denoted by the principal quantum number \( n \). As \( n \) increases, the electron is found further from the nucleus, which corresponds to higher energy states.
For a hydrogen atom, the relationship between the energy level and the quantum numbers \( n \), \( \ell \), and \( m_\ell \) becomes evident:

• The lowest energy state (ground state) is when \( n = 1 \).
• For any state where \( n>1 \), the atom is in an excited state, meaning the electron is in a higher energy orbit.

In the exercise, a particular excited state was analyzed using its orbital angular momentum, illustrating how the principles of quantum mechanics provide a precise depiction of electron behavior in atoms.