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Two of the three electrons in a lithium atom have quantum numbers of \(n=1, \ell=0, m_{\ell}=0, m_{\mathrm{s}}=+\frac{1}{2}\) and \(n=1, \ell=0, m_{e}=0, m_{\mathrm{s}}=-\frac{1}{2}\) 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
(a) Ground state: \(n=2, \ell=0, m_\ell=0, m_s=\pm 1/2\). (b) First excited state: \(n=2, \ell=1, m_\ell=-1, 0, 1\), \(m_s=\pm 1/2\).

Step by step solution

01

Understanding the Question

We have a lithium atom with atomic number 3, which means it has 3 electrons. We are given the quantum numbers for two electrons. The task is to determine the possible quantum numbers for the third electron considering the ground state and the first excited state of the atom.
02

Reviewing Ground State Configuration

In a ground state lithium atom, the electrons fill the lowest energy levels first. The given electrons already occupy the 1s orbital (with quantum numbers). To identify the quantum numbers for the third electron in its ground state, recall that the next lowest energy level is the 2s orbital. This electron would have quantum numbers: \( n=2, \ell=0, m_\ell=0, m_s=\pm \frac{1}{2} \).
03

Determining First Excited State

In the first excited state, one of the electrons will move to a higher energy level than the ground state. The excited electron can move to a 2p orbital as it is the next available level. Hence, the possible quantum numbers for the third electron can be: \( n=2, \ell=1, m_\ell=-1, 0, 1\), and \( m_s=\pm \frac{1}{2} \).
04

Conclusion

In the ground state, the third electron in a lithium atom will have \( n=2, \ell=0, m_\ell=0, m_s=\pm \frac{1}{2} \). In the first excited state, it can take one of the states: \( n=2, \ell=1, m_\ell=-1, 0, 1\), with \( m_s=\pm \frac{1}{2} \).

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Key Concepts

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

Ground State Configuration
Every atom has a specific way that its electrons are organized, especially when it is at its lowest energy level. This arrangement is called the "ground state configuration." Think of it as the most stable way the electrons can settle into the available energy levels or orbitals.
The ground state configuration follows a set order, filling the available orbitals from the lowest energy level to the highest. For lithium, which has three electrons, the first two electrons fit snugly into the 1s orbital with quantum numbers: \( n=1, \ell=0, m_{\ell}=0, m_s=\frac{1}{2} \text{ or } -\frac{1}{2} \).
Now, what about the third electron? To achieve a stable structure, the third electron moves to the next available lower energy level, the 2s orbital. Hence, its quantum numbers are \( n=2, \ell=0, m_{\ell}=0, m_s=\pm \frac{1}{2} \).
Key points to keep in mind include:
  • Electrons fill orbitals starting from the lowest to highest energy levels.
  • The ground state is crucial to maintaining atomic stability.
Excited State of Atom
When an atom gains energy, its electrons can become "excited." This happens when one or more electrons absorb energy and "jump" to a higher energy level, moving further from the nucleus. Thus, the atom transitions from its ground state to an excited state.
For a lithium atom, upon reaching an excited state, the third electron might jump from the 2s orbital to a higher energy orbital, such as the 2p. This shift disrupts the standard electron configuration and offers a different pattern to its quantum numbers.
In the 2p orbital, the possible quantum numbers for the third electron change. These can include \( n=2, \ell=1, m_{\ell}=-1, 0, 1 \), and \( m_s=\pm \frac{1}{2} \).
Remember:
  • Excited states involve electrons moving to higher energy levels.
  • Energy absorption is necessary for these transitions.
  • Quantum numbers expand due to the shift in energy levels which provides more options.
Lithium Atom
Lithium is a fascinating element in chemistry. It holds atomic number 3, revealing it has three electrons under normal conditions. These electrons determine lithium's behavior in both chemical reactions and states of energy, such as in ground and excited states.
The simplicity of lithium makes it an excellent choice to introduce fundamental concepts of quantum mechanics, particularly in understanding electron configurations. By examining the transitions of electrons between ground and excited states, we gain insight into how atoms manage energy changes.
Lithium's significance extends to technology realms such as batteries, showcasing its practical applications beyond chemical curiosity. Consider these facts:
  • Lithium's low number of electrons simplifies its energy state transitions.
  • It helps in studying the distribution and movement of electrons.
  • Its applications stretch from theoretical studies to real-world applications like lithium-ion batteries.
By grasping the concept of lithium's electron behavior, we make great strides in understanding broader atomic structures and quantum number applications.

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Most popular questions from this chapter

Using the Bohr model, determine the ratio of the energy of the \(n\) th orbit of a triply ionized beryllium atom \(\left(\mathrm{Be}^{3+}, Z=4\right)\) to the energy of the \(n\) th orbit of a hydrogen atom (H).

The nucleus of the hydrogen atom has a radius of about \(1 \times 10^{-15} \mathrm{m}\). The electron is normally at a distance of about \(5.3 \times 10^{-11} \mathrm{m}\) from the nucleus. Assuming that the hydrogen atom is a sphere with a radius of \(5.3 \times 10^{-11} \mathrm{m},\) find (a) the volume of the atom, (b) the volume of the nucleus, and (c) the percentage of the volume of the atom that is occupied by the nucleus.

The electron in a hydrogen atom is in the first excited state, when the electron acquires an additional \(2.86 \mathrm{eV}\) of energy. What is the quantum number \(n\) of the state into which the electron moves?

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 \(\ell\) 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) \(2 p \rightarrow 1 s\) (c) \(4 p \rightarrow 2 p(d) 4 s \rightarrow 2 p\) (e) \(3 \mathrm{d} \rightarrow 3 \mathrm{s}\)

It is known that the possible values for the magnetic quantum number \(m_{\ell}\) are \(-4,-3,-2,-1,0,+1,+2,+3,\) and \(+4 .\) Determine the orbital quantum number \(\ell\) and the smallest possible value of the principle quantum number \(n\).

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