Schaum’s Outline of College Physics

Eugene Hecht

Chapter 44

Multielectron Atoms - all with Video Answers

Educators

Chapter Questions

Problem 1

Estimate the energy required to remove an $n=1$ (i.e., inner-shell) electron from a gold atom $(Z=79)$.
Because an electron in the innermost shell of the atom is not much influenced by distant electrons in outer shells, we can consider it to be the only electron present. Then its energy is given approximately by an appropriately modified version of the energy formula of Chapter 43 that takes into consideration the charge (Ze) of the nucleus. With $n=1$, that formula- which was given on the first page of Chapter 43-is $\mathrm{E}_{n}=-13.6 \mathrm{Z}^{2} / \mathrm{n}^{2}$, whereupon
$$\mathrm{E}_{1}=-13.6(79)^{2}=-84900 \mathrm{eV}=-84.9 \mathrm{keV}$$
To tear the electron loose (i.e., remove it to the $\mathrm{E}_{\infty}=0$ level), we must give it an energy of about $84.9 \mathrm{keV}$.

Suzanne W.

Suzanne W.

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Problem 2

What are the quantum numbers for the electrons in the lithium atom $(Z=3)$ when the atom is in its ground state?
Start with $n=1$ and go up from there until you run out of electrons. Keeping in mind that $\ell=0,1,2, \ldots,(n-1)$ and $m_{\ell}=0$, $\pm 1, \pm 2, \ldots, \pm \ell$ while $m_{s}=\pm \frac{1}{2}$, the Pauli Exclusion Principle tells us that the lithium atom's three electrons can take on the following quantum numbers:
$$\begin{array}{ll}
\text { Electron 1: } & n=1, \quad \ell=0, \quad m_{\ell}=0, \quad m_{s}=+\frac{1}{2} \\
\text { Electron 2: } & n=1, \quad \ell=0, \quad m_{\ell}=0, \quad m_{s}=-\frac{1}{2} \\
\text { Electron 3: } & n=2, \quad \ell=0, \quad m_{\ell}=0, \quad m_{s}=+\frac{1}{2}
\end{array}$$
Notice that, when $n=1, \ell$ must be zero and $m_{\ell}$ must be zero (why?). Then there are only two $n=1$ possibilities, and the third electron has to go into the $n=2$ level. Since it is in the second Bohr orbit, it is more easily removed from the atom than an $n=1$ electron. That is why lithium ionizes easily to $\mathrm{Li}^{+}$.

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Problem 3

Why is sodium $(Z=11)$ the next univalent atom after lithium?
Sodium has a single electron in the $n=3$ shell. To see why this is necessarily so, notice that the Pauli Exclusion Principle allows only two electrons in the $n=1$ shell. The next eight electrons can fit in the $n=2$ shell, as follows:
$$\begin{array}{l}
n=2, \quad \ell=0, \quad m_{\ell}=0, \quad m_{s}=\pm \frac{1}{2} \\
n=2, \quad \ell=1, \quad m_{\ell}=0, \quad m_{s}=\pm \frac{1}{2} \\
n=2, \quad \ell=1, \quad m_{\ell}=1, \quad m_{s}=\pm \frac{1}{2} \\
n=2, \quad \ell=1, \quad m_{\ell}=-1, \quad m_{s}=\pm \frac{1}{2}
\end{array}$$
The eleventh electron must go into the $n=3$ shell, from which it is easily removed to yield $\mathrm{Na}^{+}$.

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Problem 4

(a) Estimate the wavelength of the photon emitted as an electron falls from the $n=2$ shell to the $n=1$ shell in the gold atom $(Z=$
79). (b) About how much energy must bombarding electrons have to excite gold to radiate this emission line?
(a) As noted in Problem 44.1, to a first approximation the energies of the innermost electrons of a large- $Z$ atom are given by $\mathrm{E}_{n}=$ $-13.6 \mathrm{Z}^{2} / \mathrm{n}^{2} \mathrm{eV}$. Thus,
$$\Delta \mathrm{E}_{2,1}=13.6(79)^{2}\left(\frac{1}{1}-\frac{1}{4}\right)=63700 \mathrm{eV}$$
This corresponds to a photon with
$$\lambda=(1240 \mathrm{~nm})\left(\frac{1 \mathrm{eV}}{63700 \mathrm{eV}}\right)=0.0195 \mathrm{~nm}$$
It is clear from this result that inner-shell transitions in high-Z atoms give rise to the emission of X-rays.
(b) Before an $n=2$ electron can fall to the $n=1$ shell, an $n=1$ electron must be thrown to an empty state of large $n$, which we approximate as $n=\infty\left(\right.$ with $\mathrm{E}_{\infty}=0$ ). This requires an energy
$$\Delta \mathrm{E}_{1, \infty}=0-\frac{-13.6 \mathrm{Z}^{2}}{n^{2}}=\frac{13.6(79)^{2}}{1}=84.9 \mathrm{keV}$$
The bombarding electrons must thus have an energy of about $84.9 \mathrm{keV}$.

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Problem 5

Suppose electrons had no spin, so that the spin quantum number did not exist. If the Exclusion Principle still applied to the remaining quantum numbers, what would be the first three univalent atoms?
The electrons would take on the following quantum numbers:
$$\begin{array}{lllll}
\text { Electron 1: } & n=1, & \ell=0, & m_{\ell}=0 & \text { (univalent) } \\
\text { Electron 2: } & n=2, & \ell=0, & m_{\ell}=0 & \text { (univalent) } \\
\text { Electron 3: } & n=2, & \ell=1, & m_{\ell}=0 & \\
\text { Electron 4: } & n=2, & \ell=1, & m_{\ell}=+1 & \\
\text { Electron 5: } & n=2, & \ell=1, & m_{\ell}=-1 & \\
\text { Electron 6: } & n=3, & \ell=0, & m_{\ell}=0 & \text { (univalent) }
\end{array}$$
Each electron marked "univalent" is the first electron in a new shell. Since an electron is easily removed if it is the outermost electron in the atom, atoms with that number of electrons are univalent. They are the atoms with $Z=1$ (hydrogen), $Z=2$ (helium), and $Z=6$ (carbon). Can you show that $Z=15$ (phosphorus) would also be univalent?

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Problem 6

Electrons in an atom that have the same value for $\ell$ but different values for $m_{\ell}$ and $m_{s}$ are said to be in the same subshell. How many electrons exist in the $\ell=3$ subshell?
Because $m_{\ell}$ is restricted to the values $0, \pm 1, \pm 2, \pm 3$, and $m_{s}=\pm \frac{1}{2}$ only, the possibilities for $\ell=3$ are
$$\left(m_{t}, m_{s}\right)=\left(0, \pm \frac{1}{2}\right) \cdot\left(1, \pm \frac{1}{2}\right) \cdot\left(-1, \pm \frac{1}{2}\right) \cdot\left(2, \pm \frac{1}{2}\right) \cdot\left(-2, \pm \frac{1}{2}\right) \cdot\left(3, \pm \frac{1}{2}\right) \cdot\left(-3, \pm \frac{1}{2}\right)$$
which gives 14 possibilities. Therefore, 14 electrons can exist in this subshell.

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Problem 7

An electron beam in an X-ray tube is accelerated through $40 \mathrm{kV}$ and is incident on a tungsten target. What is the shortest wavelength emitted by the tube?
When an electron in the beam is stopped by the target, the photons emitted have an upper limit for their energy, namely, the energy of the incident electron. In this case, that energy is $40 \mathrm{keV}$. The corresponding photon has a wavelength given by
$$
\lambda=(1240 \mathrm{~nm})\left(\frac{1.0 \mathrm{eV}}{40000 \mathrm{eV}}\right)=0.031 \mathrm{~nm}
$$

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Problem 8

List all four quantum numbers for the single hydrogen $(Z=1)$ electron in the ground state. [Hint: It is in the 1 s orbital of the $K$ shell.]

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Problem 9

The single electron of a hydrogen atom can exist in a variety of excited states beyond the lowest-energy ground state. How many states would be available when the principal quantum number equals 4 and the orbital quantum number equals 2? [Hint: $n=4$ and $\ell=2 .]$

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Problem 10

State the quantum numbers $\left(n, \ell, m_{\ell}, m_{s}\right)$ for each electron in the ground state of helium $(Z=2)$. Explain your answer.

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Problem 11

Explain how it is that the maximum number of electrons in the $\ell$ th subshell is $2(2 \ell+1)$

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Problem 12

Verify that Fig. 44-1 shows that the number of electrons in a shell can be up to $2 n^{2}$.

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Problem 13

Argon has a ground state configuration of $1 s^{2} \cdot 2 s^{2} 2 p^{6} \cdot 3 s^{2} 3 p^{6}$ How many electrons does an argon atom possess? What can you say about argon and its last $p$ subshell?

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Problem 14

Aluminum has a ground state configuration of $1 s^{2} \cdot 2 s^{2} 2 p^{6} \cdot 3 s^{2} 3 p^{1}$. How many electrons does an aluminum atom possess? What is $\mathrm{Z}$ for aluminum? How many electrons have a principal quantum number of $2 ?$

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Problem 15

In the Periodic Table the far right column contains the noble gases, noble because they stay away from combining with other elements. Helium $(Z=2)$ has 2 electrons that fill a $1 s$ shell. Neon $(\mathrm{Z}=10)$ has 10 electrons $\left(1 s^{2} \cdot 2 s^{2} 2 p^{6}\right)$, and its outer $2 p$ subshell is filled. Argon $(\mathrm{Z}=18)$ is next $\left(1 s^{2} \cdot 2 s^{2} 2 p^{6} \cdot 3 s^{2} 3 p^{6}\right)$, and krypton $(Z$ $=36$ ) follows it in the column. In addition to the argon electron configuration, krypton adds 4s, $3 d$, and $4 p$ subshells. How many electrons are in each of these subshells? [Hint: The total number must be 36. Study Table 44-2.]

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Problem 16

Specify the ground state electron configuration for silicon (Si) for which $Z=14$. Explain your answer. Are all the shells filled?

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Problem 17

Silicon is a semiconductor, as is carbon (C) for which $Z=6$. Specify the ground state electron configuration. Why do they behave similarly?

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Problem 18

If there were no $m_{\ell}$ quantum number, what would be the first four
univalent atoms?

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Problem 19

Helium has a closed (completely filled) outer shell and is nonreactive because the atom does not easily lose an electron. Show why neon $(Z=10)$ is the next nonreactive element.

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Problem 20

It is desired to eject an electron from the $n=1$ shell of a uranium atom $(Z=92)$ by means of the atomic photoelectric effect. Approximately what is the longest-wavelength photon capable of doing this?

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