Problem 39 - College Physics

Question

(a) What is the lowest possible energy (in electron volts) of an electron in hydrogen if its orbital angular momentum is \(\sqrt { 12 } \hbar ?\) (b) What are the largest and smallest values of the \(z\) component of the orbital angular momentum (in terms of \(\hat { h }\) ) for the electron in part (a)? (c) What are the largest and smallest values of the spin angular momentum (in terms of \(\hbar\) ) for the electron in part (a)?(d) What are the largest and smallest values of the orbital angular momentum (in terms of \(\hbar\) ) for an electron in the \(M\) shell of hydrogen?

Step-by-Step Solution

Answer

(a) -0.85 eV; (b) Largest: \(3\hbar\), Smallest: \(-3\hbar\); (c) Largest: \(+\frac{1}{2}\hbar\), Smallest: \(-\frac{1}{2}\hbar\); (d) Largest: \(\sqrt{6}\hbar\), Smallest: 0.

Step 1: Determine Orbital Angular Momentum Quantum Number

The orbital angular momentum is given by the formula \( L = \sqrt{l(l+1)}\hbar \). Based on the problem, \( L = \sqrt{12}\hbar \). Thus, we set up the equation \( \sqrt{l(l+1)}\hbar = \sqrt{12}\hbar \). Solving \( l(l+1) = 12 \) gives \( l = 3 \).

Step 2: Calculate Principal Quantum Number

The principal quantum number \( n \) is related to \( l \) by the condition \( n > l \). Therefore, the smallest possible \( n \) is \( n = 4 \).

Step 3: Find Energy of Hydrogen Atom

The energy of an electron in a hydrogen atom is given by the formula \( E_n = -13.6 \frac{1}{n^2} \) eV. Insert \( n = 4 \) to find \( E = -13.6 \frac{1}{4^2} = -0.85 \) eV.

Step 4: Largest and Smallest z-Component of Orbital Angular Momentum

For a given \( l \), the possible values for \( m_l \) are integers from \( -l \) to \( l \). Thus, for \( l = 3 \), \( m_l = -3, -2, -1, 0, 1, 2, 3 \). The largest value of \( L_z = m_l\hbar = 3\hbar \), and the smallest value is \( -3\hbar \).

Step 5: Spin Angular Momentum of Electron

The spin of an electron is always \( s = \frac{1}{2} \). Thus, the total spin angular momentum is \( S = \sqrt{\frac{1}{2}(\frac{1}{2}+1)}\hbar = \sqrt{\frac{3}{4}}\hbar \). The largest \( S_z = +\frac{1}{2}\hbar \) and smallest \( S_z = -\frac{1}{2}\hbar \).

Step 6: Orbital Angular Momentum in M-shell

The \( M \) shell corresponds to \( n = 3 \). For this shell, \( l = 0, 1, 2 \). The largest \( L = \sqrt{2(2+1)}\hbar = \sqrt{6}\hbar \) for \( l = 2 \) and the smallest \( L = \sqrt{0} = 0 \) for \( l = 0 \).

Key Concepts

Orbital Angular Momentum

Orbital angular momentum is a concept that describes the motion of electrons around the nucleus. In quantum mechanics, it is quantized, which means it can only take on specific values. For an electron in an atom, the orbital angular momentum is not just about moving in a circle; it also involves the three-dimensional space around the nucleus. It is denoted by the quantum number \( l \), which can be any integer from 0 up to \( n-1 \), where \( n \) is the principal quantum number.

• Formula: The magnitude of the orbital angular momentum is given by \( L = \sqrt{l(l+1)}\hbar \).
• The orbital angular momentum is important because it helps define the shape of the electron's path around the nucleus. For example, when \( l = 3 \), the path is more complex than when \( l = 0 \).

Learning about orbital angular momentum helps us understand how electrons occupy atomic orbitals and the limits of their behavior due to quantum restrictions.

Principal Quantum Number

The principal quantum number, denoted as \( n \), is a key idea in quantum mechanics for describing electrons in atoms. It indicates the energy level or "shell" that an electron occupies. The principal quantum number is crucial because: - It determines the size of the electron's orbit around the nucleus—the larger \( n \), the bigger the orbit.- It dictates the total energy of the electron within an atom.

• The smallest possible value of \( n \) is 1. As we go higher in \( n \), the electron occupies higher energy levels.
• In our problem, the smallest possible \( n \) for given orbital angular momentum is 4, according to the constraint \( n > l \).

Understanding \( n \) is essential for exploring how atoms function and the energies involved in electronic transitions, which affect properties like color and chemical reactivity.

Hydrogen Atom Energy

The energy levels of a hydrogen atom are perhaps one of the simplest to calculate due to only having a single electron. The energy is quantized and described by the equation:\[E_n = -13.6 \frac{1}{n^2} \text{ eV}\]This equation tells us that the energy an electron has depends on the principal quantum number \( n \) it occupies. For example, when \( n = 4 \), the energy level is calculated as:\[E_4 = -13.6 \frac{1}{16} = -0.85 \text{ eV}\]

• This negative energy value shows the electron is bound to the atom.
• The hydrogen energy levels point to why emitting or absorbing specific energy quanta alters the electron’s position within the atom.

These principles are fundamental for understanding phenomena like atomic spectra and the basis of quantum mechanics in hydrogen.

Spin Angular Momentum

Spin angular momentum is an intrinsic form of angular momentum carried by particles like electrons, independent of any orbital motion. Although electrons are point particles and do not literally "spin," they possess this innate form of angular momentum, characterized by the quantum number \( s \), which is always \( \frac{1}{2} \) for electrons.

• The total spin angular momentum is calculated as \( S = \sqrt{\frac{1}{2}(\frac{1}{2}+1)}\hbar = \sqrt{\frac{3}{4}}\hbar \).
• The component of this spin along any axis, typically the \( z \)-axis, has only two possible values: \( S_z = \pm\frac{1}{2}\hbar \).

Spin is a critical quantum number because it affects how electrons pair up in atomic orbitals, influencing chemical bonds and material properties.