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Chapter 40: Problem 27

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

Short Answer

Expert verified
Ground: (2,0,0,卤1/2); Excited: (2,1,-1,卤1/2), (2,1,0,卤1/2), (2,1,1,卤1/2).

Step by step solution

01

Identify known quantum numbers

We are given two sets of quantum numbers for the electrons in a lithium atom: \( (1,0,0,+\frac{1}{2}) \) and \( (1,0,0,-\frac{1}{2}) \). They represent the principal quantum number \( n \), the azimuthal or angular momentum quantum number \( \ell \), the magnetic quantum number \( m_{\ell} \), and the spin magnetic quantum number \( m_{s} \), respectively. These values identify two electrons in the 1s orbital.
02

Third electron in the ground state

In the ground state of the lithium atom, the third electron goes into the next available state. Since electrons fill the lowest energy state available, the next state is \( 2s \). Thus, the third electron has the quantum numbers \( (2,0,0,\pm\frac{1}{2}) \). Here, the principal quantum number \( n = 2 \), and \( \ell = 0 \) for the s orbital.
03

Third electron in the first excited state

For the first excited state, the third electron is promoted to the \( 2p \) orbital. The quantum numbers in this case are \( (2,1,m_{\ell},\pm\frac{1}{2}) \), where \( m_{\ell} \) can be \(-1, 0, 1\). So, the possible sets of quantum numbers are \((2,1,-1,\pm\frac{1}{2})\), \((2,1,0,\pm\frac{1}{2})\), and \((2,1,1,\pm\frac{1}{2})\).
04

List possible quantum numbers for each state

For the ground state, possible quantum numbers for the third electron are \( (2,0,0,+\frac{1}{2}) \) and \( (2,0,0,-\frac{1}{2}) \). For the first excited state, they are \((2,1,-1,\pm\frac{1}{2})\), \((2,1,0,\pm\frac{1}{2})\), and \((2,1,1,\pm\frac{1}{2})\).

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Key Concepts

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

Lithium Atom
A lithium atom is an interesting part of the periodic table. It's the third element, with an atomic number of 3. Simply put, this means lithium has three protons, and in a neutral lithium atom, there are also three electrons.
These electrons are the focus of our study when discussing quantum numbers and electron orbitals. Each electron has its own set of quantum numbers that precisely describe its state and position in the atom.
In the context of lithium, we're mostly concerned with how these electrons are configured when the atom is in its ground or excited states.
Ground State
The ground state of an atom is its lowest energy state. For lithium, this means that all three electrons are arranged in the lowest energy orbitals following the Pauli-exclusion principle and Hund's rule.
In the ground state configuration of a lithium atom, the first two electrons fill the 1s orbital:
  • The first electron has the quantum numbers: \((n = 1, \, \ell = 0, \, m_{\ell} = 0, \, m_{s} = +\frac{1}{2})\).
  • The second electron is: \((n = 1, \, \ell = 0, \, m_{\ell} = 0, \, m_{s} = -\frac{1}{2})\). This pairing minimizes energy and each quantum number follows a rule.
The third electron then goes into the 2s orbital, which is the lowest energy level available after 1s. In its ground state, this electron's quantum numbers are:
  • \((n = 2, \, \ell = 0, \, m_{\ell} = 0, \, m_{s} = \pm\frac{1}{2})\).
Excited State
In an excited state, an atom's electron has absorbed energy and moved to a higher energy level. For a lithium atom, transitions occur when an electron is promoted from the 2s orbital to a higher orbital.
When considering the first excited state for lithium, the third electron is promoted from the 2s to the 2p orbital.
This electron transition changes the quantum numbers describing the electron.
Possible sets of quantum numbers in the first excited state include:
  • \((n = 2, \, \ell = 1, \, m_{\ell} = -1, \, m_{s} = \pm\frac{1}{2})\).
  • \((2, \, 1, \, 0, \, \pm\frac{1}{2})\).
  • \((2, \, 1, \, 1, \, \pm\frac{1}{2})\).
1s Orbital
An orbital is a region of space where the probability of finding an electron is highest. The 1s orbital is the simplest and lowest energy orbital in an atom.
It can hold up to two electrons. In a lithium atom, the 1s orbital is filled by the two lowest energy electrons.
  • The first electron in the 1s orbital can have the quantum numbers: \((n = 1, \, \ell = 0, \, m_{\ell} = 0, \, m_{s} = +\frac{1}{2})\)
  • The second electron pairs with it: \((n = 1, \, \ell = 0, \, m_{\ell} = 0, \, m_{s} = -\frac{1}{2})\).
This filling completes the 1s orbital, giving it stability.
2s Orbital
The 2s orbital is the next higher energy level after the 1s orbital. It can also hold up to two electrons. In a lithium atom, once the 1s orbital is filled, the next electron occupies the 2s orbital in the ground state.
  • The third electron in lithium has quantum numbers: \((n = 2, \, \ell = 0, \, m_{\ell} = 0, \, m_{s} = \pm\frac{1}{2})\).
This simple filling ensures the atom's energy remains at the lowest possible until additional energy boosts the electron to a higher state.
2p Orbital
The 2p orbital comes into play when an electron is excited to a higher level. Unlike s orbitals, p orbitals have an azimuthal quantum number \(\ell = 1\) and are shaped differently, which allows for a larger set of possible orientations.
  • The principal quantum number remains \(n = 2\),
  • \(\ell\) becomes 1 for all 2p orbitals: \(m_{\ell}\) could be -1, 0, or 1.
In a first excited state, the third electron in a lithium atom can have quantum numbers like:
  • \((n = 2, \, \ell = 1, \, m_{\ell} = -1, \, m_{s} = \pm\frac{1}{2})\),
  • \((2, \, 1, \, 0, \, \pm\frac{1}{2})\),
  • \((2, \, 1, \, 1, \, \pm\frac{1}{2})\).
The added orientations compared to s orbitals allow for more flexibility in electron placement during excitation.

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Most popular questions from this chapter

For a helium atom in its ground state, what are quantum numbers \(\left(n, \ell, m_{\ell},\right.\) and \(\left.m_{s}\right)\) for the (a) spin-up electron and (b) spin-down electron?

X rays are produced in an \(x\) -ray tube by electrons accelerated through an electric potential difference of \(50.0 \mathrm{kV}\). Let \(K_{0}\) be the kinetic energy of an electron at the end of the acceleration. The electron collides with a target nucleus (assume the nucleus remains stationary) and then has kinetic energy \(K_{1}=0.500 K_{0}\). (a) What wavelength is associated with the photon that is emitted? The electron collides with another target nucleus (assume it, too, remains stationary) and then has kinetic energy \(K_{2}=0.500 K_{1}\) (b) What wavelength is associated with the photon that is emitted?

Consider the elements selenium \((Z=34),\) bromine \((Z=35)\), and krypton \((Z=36) .\) In their part of the periodic table, the subshells of the electronic states are filled in the sequence \(1 s 2 s 2 p 3 s 3 p 3 d 4 s 4 p \ldots\) What are (a) the highest occupied subshell for selenium and (b) the number of electrons in it, (c) the highest occupied subshell for bromine and (d) the number of electrons in it, and (e) the highest occupied subshell for krypton and (f) the number of electrons in it?

Comet stimulated emission. When a comet approaches the Sun, the increased warmth evaporates water from the ice on the surface of the comet nucleus, producing a thin atmosphere of water vapor around the nucleus. Sunlight can then dissociate \(\mathrm{H}_{2} \mathrm{O}\) molecules in the vapor to \(\mathrm{H}\) atoms and \(\mathrm{OH}\) molecules. The sunlight can also excite the OH molecules to higher energy levels. When the comet is still relatively far from the Sun, the sunlight causes equal excitation to the \(E_{2}\) and \(E_{1}\) levels (Fig. \(40-28 a\) ). Hence, there is no population inversion between the two levels. However, as the comet approaches the Sun, the excitation to the \(E_{1}\) level decreases and population inversion occurs. The reason has to do with one of the many wavelengths - said to be Fraunhofer lines - that are missing in sunlight because, as the light travels outward through the Sun's atmosphere, those particular wavelengths are absorbed by the atmosphere. As a comet approaches the Sun, the Doppler effect due to the comet's speed relative to the Sun shifts the Fraunhofer lines in wavelength, apparently overlapping one of them with the wavelength required for excitation to the \(E_{1}\) level in \(\mathrm{OH}\) molecules. Population inversion then occurs in those molecules, and they radiate stimulated emission (Fig. \(40-28 b\) ). For example, as comet Kouhoutek approached the Sun in December 1973 and January \(1974,\) it radiated stimulated emission at about \(1666 \mathrm{MHz}\) during mid-January. (a) What was the energy difference \(E_{2}-E_{1}\) for that emission? (b) In what region of the electromagnetic spectrum was the emission?

Consider an atom with two closely spaced excited states \(A\) and \(B\). If the atom jumps to ground state from \(A\) or from \(B\), it emits a wavelength of \(500 \mathrm{nm}\) or \(510 \mathrm{nm}\), respectively. What is the energy difference between states \(A\) and \(B ?\)
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