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Chapter 30: Problem 31

Two of the three electrons in a lithium atom have quantum numbers of \(n=1, \ell=0, m_{c}=0, m_{s}=+\frac{1}{2}\) and \(n=1, \ell=0, m_{c}=0, m_{3}=-\frac{1}{2} .\) What quantum numbers can the third electron have if the atom is in (a) its ground state and (b) its first excited state?

Short Answer

Expert verified
In the ground state, \( n=2, \ell=0, m_\ell=0 \), \( m_s = \pm\frac{1}{2} \). In the excited state, \( n=2, \ell=1, m_\ell \in \{-1,0,1\}\), \( m_s = \pm\frac{1}{2} \).

Step by step solution

01

Understanding the Quantum Numbers

Quantum numbers include the principal quantum number \( n \), azimuthal or angular momentum quantum number \( \ell \), magnetic quantum number \( m_\ell \), and spin quantum number \( m_s \). For lithium's two given electrons, all are in the first shell \( (n=1) \), the \( s \)-orbitals \( (\ell=0) \), and have opposite spins \( m_s = +\frac{1}{2} \) and \( m_s = -\frac{1}{2} \).
02

Explaining the Ground State of Lithium

In its ground state, lithium's third electron will occupy the next available energy level. Since the first shell is full, it moves to the second shell \( (n=2) \). Here, in the \( s \)-subshell \( (\ell=0) \), it can have quantum numbers \( n=2, \ell=0, m_\ell=0 \), and one of \( m_s = +\frac{1}{2} \) or \( m_s = -\frac{1}{2} \).
03

Explaining the Excited State of Lithium

In the first excited state, the third electron moves to an orbital with higher energy, such as the \( 2p \) orbital. Here, we have \( n=2, \ell=1 \). The \( m_\ell \) values can be \(-1, 0, \) or \( 1 \), and \( m_s \) can be \( +\frac{1}{2} \) or \( -\frac{1}{2} \). This gives multiple combinations of quantum numbers.

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Key Concepts

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

Principal Quantum Number
The principal quantum number, denoted as \( n \), is one of the most fundamental quantum numbers. It indicates the main energy level or shell that an electron occupies within an atom. Think of it as the electron's home address regarding its energy. The principal quantum number can take any positive integer value starting from 1.

  • \( n = 1 \) is the closest and lowest energy level, commonly referred to as the ground state.
  • As \( n \) increases, electrons are found in shells that are further from the nucleus and have higher energy levels.
  • For example, in a lithium atom鈥檚 ground state, the two innermost electrons have \( n = 1 \), while the third electron is in the next available energy level \( n = 2 \).
The further away the electron is from the nucleus, the more energy it has, and the higher its principal quantum number. Understanding \( n \) helps in grasping the structure and stability of atoms.
Azimuthal Quantum Number
The azimuthal quantum number, symbolized by \( \ell \), determines the shape of an electron's orbital and is sometimes called the angular momentum quantum number. This quantum number is crucial because it enhances our understanding of chemical bonding and the shapes of molecules.

  • \( \ell \) can take integer values from 0 up to \( n-1 \).
  • When \( \ell = 0 \), the orbital is spherical and is often called an \( s \)-orbital.
  • A value of \( \ell = 1 \) corresponds to \( p \)-orbitals, which have more complex, dumbbell-like shapes.
For instance, the two electrons of lithium in the problem are in \( s \)-orbitals since they have \( \ell = 0 \). The third electron could potentially be in a \( p \)-orbital when the atom reaches an excited state, with \( \ell = 1 \). Understanding \( \ell \) is key to visualizing and predicting electron arrangements.
Spin Quantum Number
The spin quantum number, denoted as \( m_s \), describes the intrinsic angular momentum or "spin" of an electron. Each electron acts as though it is spinning on its axis, and \( m_s \) helps us understand this property.

  • It can take one of two possible values: \( +\frac{1}{2} \) or \( -\frac{1}{2} \).
  • This means an electron can have a "spin-up" or "spin-down" orientation.
  • The differing spins reduce repulsion between electrons, allowing them to coexist within the same orbital.
In the case of a lithium atom, two electrons in the ground state occupy the \( n=1 \) level and have spins \( +\frac{1}{2} \) and \( -\frac{1}{2} \). Only these two electrons with opposite spins can exist in the same orbital, showcasing the importance of the spin quantum number in defining electron distribution.
Lithium Atom Ground State
The ground state configuration describes how electrons are distributed under the lowest possible energy conditions. For lithium, which has three electrons, the simplest configuration can be described by the quantum numbers associated with its electrons.

  • The first two electrons fill the first energy shell, \( n=1 \), in the \( s \)-orbital where \( \ell=0 \).
  • These electrons have spins \( +\frac{1}{2} \) and \( -\frac{1}{2} \), balancing each other.
  • The third electron, in its ground state, takes the next available energy level, \( n=2 \), filling the \( 2s \)-orbital with \( \ell=0, m_\ell=0 \), and either \( m_s=+\frac{1}{2} \) or \( -\frac{1}{2} \).
This configuration ensures that all electrons are in their lowest possible energy state, making the atom as stable as possible.
Excited State Configuration
In an excited state, an electron absorbs energy and moves to a higher energy level or orbital, beyond its original ground state position. For lithium, an excited state does not follow the conventional lowest energy arrangement of electrons.

  • Excited states occur when the third electron in lithium transitions to the \( 2p \)-orbital.
  • Here, \( n=2, \ell=1 \), providing several possible \( m_\ell \) values: \(-1, 0, 1\).
  • The spin quantum number \( m_s \) remains \( +\frac{1}{2} \) or \( -\frac{1}{2} \), offering a variety of electron configurations.
Such transitions allow electrons to interact in new ways, influencing chemical reactions and the absorption and emission of light. Understanding these excited states is critical for applications in spectroscopy and other fields of study where electron behavior is essential.

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Most popular questions from this chapter

A laser peripheral iridotomy is a procedure for treating an eye condition known as narrow-angle glaucoma, in which pressure buildup in the eye can lead to loss of vision. A neodymium YAG laser (wavelength = \(1064 \mathrm{nm}\) ) is used in the procedure to punch a tiny hole in the peripheral iris, thereby relieving the pressure buildup. In one application the laser delivers \(4.1 \times 10^{-3} \mathrm{J}\) of energy to the iris in creating the hole. How many photons does the laser deliver?

In the ground state, the outermost shell \((n=1)\) of helium (He) is filled with electrons, as is the outermost shell \((n=2)\) of neon (Ne). The full outermost shells of these two elements distinguish them as the first two socalled "noble gases." Suppose that the spin quantum number \(m_{s}\) had three possible values, rather than two. If that were the case, which elements would be (a) the first and (b) the second noble gases? Assume that the possible values for the other three quantum numbers are unchanged, and that the Pauli exclusion principle still applies.

When an electron makes a transition between energy levels of an atom, there are no restrictions on the initial and final values of the principal quantum number \(n .\) According to quantum mechanics, however, there is a rule that restricts the initial and final values of the orbital quantum number \(\ell\). This rule is called a selection rule and states that \(\Delta \ell=\pm 1 .\) In other words, when an electron makes a transition between energy levels, the value of \(t\) can only increase or decrease by one. The value of \(\ell\) may not remain the same nor may it increase or decrease by more than one. According to this rule, which of the following energy level transitions are allowed? (a) \(2 \mathrm{s} \rightarrow 1 \mathrm{s}\) (b) \(2 p \rightarrow\) Is \((c) 4 p \rightarrow 2 p\) (d) \(4 \mathrm{s} \rightarrow 2 \mathrm{p}\) (e) \(3 \mathrm{d} \rightarrow 3 \mathrm{s}\)

Doubly ionized lithium \(L i^{2+}(Z=3)\) and triply ionized beryllium \(\mathrm{Be}^{3+}(\mathrm{Z}=4)\) each emit a line spectrum. For a certain series of lines in the lithium spectrum, the shortest wavelength is \(40.5 \mathrm{nm}\). For the same series of lines in the beryllium spectrum, what is the shortest wavelength?

Consider a particle of mass \(m\) that can exist only between \(x=0 \mathrm{m}\) and \(x=+L\) on the \(x\) axis. We could say that this particle is confined to a "box" of length \(L\). In this situation, imagine the standing de Broglie waves that can fit into the box. For example, the drawing shows the first three possibilities. Note in this picture that there are either one, two, or three half-wavelengths that fit into the distance \(L\). Use Equation 29.8 for the de Broglie wavelength of a particle and derive an expression for the allowed energies (only kinetic energy) that the particle can have. This expression involves \(m, L,\) Planck's constant, and a quantum number \(n\) that can have only the values \(1,2,3, \ldots\)
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