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An electron is in the hydrogen atom with \(n = 3 .\) (a) Find the possible values of \(L\) and \(L _ { z }\) for this electron, in units of \(\hbar .\) (b) For each value of \(L ,\) find all the possible angles between \(L\) and the \(z\) axis.

Short Answer

Expert verified
Possible values of L: 0, \(\sqrt{2}\hbar\), \(\sqrt{6} \hbar\); Lz: values range from \(-l\hbar\) to \(+l\hbar\); angles depend on \(m_l\).

Step by step solution

01

Understanding the Quantum Numbers

The quantum number \( n \) for hydrogen is 3. The possible values for the angular momentum quantum number \( l \) range from 0 to \( n-1 \). So, \( l \) can be 0, 1, or 2.
02

Calculating L for Each l Value

Angular momentum \( L \) is given by \( L = \sqrt{l(l+1)}\ \hbar \). Calculating for each possible \( l \):- For \( l = 0 \), \( L = \sqrt{0(0+1)}\ \hbar = 0 \).- For \( l = 1 \), \( L = \sqrt{1(1+1)}\ \hbar = \sqrt{2}\ \hbar \).- For \( l = 2 \), \( L = \sqrt{2(2+1)}\ \hbar = \sqrt{6}\ \hbar \).
03

Finding Possible Values of Lz

The magnetic quantum number \( m_l \) determines \( L_z \), and its values are \(-l, -(l-1), ..., 0, ..., +(l-1), +l\). Thus:- For \( l = 0 \), \( L_z = 0 \).- For \( l = 1 \), \( L_z = - \hbar, 0, + \hbar \).- For \( l = 2 \), \( L_z = -2\hbar, -\hbar, 0, +\hbar, +2\hbar \).
04

Finding the Angle Between L and the z-Axis

The angle \( \theta \) between \( L \) and the z-axis is determined by the equation \( \cos \theta = \frac{L_z}{L} \). For each \( L \), calculate \( \theta \):- For \( l = 0 \): - \( L_z = 0 \), \( L = 0 \), so the angle is undefined as \( L_z/L \) is indeterminate.- For \( l = 1 \): - \( \cos \theta = 0/\sqrt{2}, \theta = 90^\circ \). - \( \cos \theta = \pm1/\sqrt{2}, \theta \approx 45^\circ, 135^\circ \).- For \( l = 2 \): - \( \cos \theta = 0/\sqrt{6}, \theta = 90^\circ \). - \( \cos \theta = \pm1/\sqrt{6}, \theta \approx 65.5^\circ, 114.5^\circ \). - \( \cos \theta = \pm2/\sqrt{6}, \theta \approx 35.3^\circ, 144.7^\circ \).

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Key Concepts

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

Quantum Numbers
Quantum numbers are essential in quantum mechanics, as they describe the characteristics and behavior of electrons in atoms. For any given electron, you can use four quantum numbers to specify its unique identity:
  • The principal quantum number, denoted as \(n\), indicates the electron's energy level and its relative distance from the nucleus. In our exercise, \(n = 3\) for the hydrogen atom, meaning the electron is in the third energy level.
  • The azimuthal or angular momentum quantum number, \(l\), defines the shape of the electron's orbital. This value can range from \(0\) to \(n-1\) for any specific energy level \(n\). In the situation described, possible values of \(l\) are \(0, 1,\) and \(2\).
  • The magnetic quantum number, \(m_l\), indicates the orientation of the orbital in space. Its possible values range from \(-l\) to \(+l\), including zero.
  • Lastly, the spin quantum number, \(m_s\), describes the electron's intrinsic spin, which can be \(+\frac{1}{2}\) or \(-\frac{1}{2}\).
These numbers collectively describe the electron's probable location and energy, allowing chemists and physicists to understand and predict atomic behavior.
Angular Momentum
Angular momentum in quantum mechanics refers to the rotational counterpart to linear momentum, analogous to how a spinning tire travels along a path. Each electron orbiting an atom has quantized angular momentum, symbolized usually by \(L\).
The magnitude of angular momentum \(L\) is linked to the quantum number \(l\) and is calculated using the formula:\[ L = \sqrt{l(l+1)}\ \hbar \]where \(\hbar\) is the reduced Planck's constant.
When each value of \(l\) is substituted:
  • For \(l = 0\), \(L = 0\ \hbar\)
  • For \(l = 1\), \(L = \sqrt{2}\ \hbar\)
  • For \(l = 2\), \(L = \sqrt{6}\ \hbar\)
This means the electron's possible angular momentum increases with higher \(l\) values. The precise calculation forms the foundation for understanding how electrons occupy different shapes and orientations in space.
Hydrogen Atom
The hydrogen atom is the simplest atom, consisting of one proton and one electron. Its simplicity makes it a great starting point for studying quantum mechanics.
In the hydrogen atom, the electron moves around the proton via a cloud-like probability region, rather than fixed circular orbits as one might initially imagine.
For an electron in energy level \(n = 3\), detailed quantum mechanical models help us predict properties such as energy, positioning, and behavior of the electron. The quantization of angular momentum results in specific shapes (orbitals) with defined energy and spatial orientation, pivotal for understanding complex atoms and chemical reactions.
Electron Configurations
Electron configurations are a systematic way of detailing the arrangement of electrons within an atom. They adhere to the principles of quantum mechanics, ensuring the electrons fill orbitals in a manner that adheres to certain rules:
  • Electrons fill starting from the lowest energy level upwards, known as the Aufbau principle.
  • No two electrons can have identical quantum numbers in the same atom, referred to as the Pauli exclusion principle.
  • Electrons will occupy degenerate orbitals singly, maintaining parallel spins as much as possible, a tendency known as Hund's rule.
For the case of hydrogen, with \(n = 3\), the available orbitals are precisely defined by quantum numbers dictating their specific order. The electron configuration provides a roadmap to understanding chemical behavior and atomic structure, predicting reactivity and bond formation properties.

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Most popular questions from this chapter

The gap between valence and conduction bands in diamond is 5.47 eV. (a) What is the maximum wavelength of a photon that can excite an electron from the top of the valence band into the conduction band? In what region of the electromagnetic spectrum does this photon lie? (b) Explain why pure diamond is transparent and colorless. (Hint: Will photons of visible light that strike a diamond be absorbed or transmitted?) (c) Most gem diamonds have a yellow color. Explain how impurities in the diamond can cause this color.

The dissociation energy of the hydrogen molecule (i.e., the energy required to separate the two atoms) is 4.48 eV. In the gas phase (treated as an ideal gas), at what temperature is the average translational kinetic energy of a molecule equal to this energy?

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 \(n \hbar\) postulated in the Bohr model. What trend do you see?

Write out the electron configuration \(\left( 1 s ^ { 2 } 2 s ^ { 2 } ,\) etc. \right\()\) for Ne, Ar, and Kr. (b) How many electrons does each of these atoms have in its outer shell? (c) Predict the chemical behavior of these three atoms. Explain your reasoning.

(a) Write out the electron configuration \(\left( 1 s ^ { 2 } 2 s ^ { 2 } ,\) etc. \right\()\) for Li and \(\mathrm { Na }\) . (b) How many electrons does each of these atoms have in its outer shell?

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