Content check — University Physics 3, Chapter 8: Atomic Structure

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Chapter 8: Atomic Structure

54 problems · 0 figures

Check Your Understanding 8.1 §8.1 source ↗

Can the magnitude of

L_{z}

ever be equal to L?

Answer

No. The quantum number

m =

− l,

− l +

1,…,

0,…,

l -

1,

l .

Thus, the magnitude of

L_{z}

is always less than L because

< \sqrt{l (l + 1)}

Check Your Understanding 8.2 §8.3 source ↗

If the Stern-Gerlach experiment yielded four distinct bands instead of two, what might be concluded about the spin quantum number of the charged particle?

Answer

s = 3 / 2 <

Check Your Understanding 8.3 §8.5 source ↗

X-rays are produced by bombarding a metal target with high-energy electrons. If the target is replaced by another with two times the atomic number, what happens to the frequency of X-rays?

Answer

frequency quadruples

8-1 §8.1 source ↗

Identify the physical significance of each of the quantum numbers of the hydrogen atom.

Answer

n (principal quantum number)

\to

total energy

l

(orbital angular quantum number)

\to

total absolute magnitude of the orbital angular momentum m (orbital angular projection quantum number)

\to

z-component of the orbital angular momentum

8-3 §8.1 source ↗

Distinguish between Bohr’s and Schrödinger’s model of the hydrogen atom. In particular, compare the energy and orbital angular momentum of the ground states.

Answer

The Bohr model describes the electron as a particle that moves around the proton in well-defined orbits. Schrödinger’s model describes the electron as a wave, and knowledge about the position of the electron is restricted to probability statements. The total energy of the electron in the ground state (and all excited states) is the same for both models. However, the orbital angular momentum of the ground state is different for these models. In Bohr’s model,

L (ground state) =

1

, and in Schrödinger’s model,

L (ground state) =

0

8-5 §8.2 source ↗

A hydrogen atom is placed in a magnetic field. Which of the following quantities are affected? (a) total energy; (b) angular momentum; (c) z-component of angular momentum; (d) polar angle.

Answer

a, c, d; The total energy is changed (Zeeman splitting). The work done on the hydrogen atom rotates the atom, so the z-component of angular momentum and polar angle are affected. However, the angular momentum is not affected.

8-7 §8.3 source ↗

Explain how a hydrogen atom in the ground state

(l = 0

) can interact magnetically with an external magnetic field.

Answer

Even in the ground state (

l = 0

), a hydrogen atom has magnetic properties due the intrinsic (internal) electron spin. The magnetic moment of an electron is proportional to its spin.

8-9 §8.3 source ↗

List all the possible values of s and

m_{s}

for an electron. Are there particles for which these values are different?

Answer

For all electrons,

s = ½

and

m_{s} = \pm ½ .

As we will see, not all particles have the same spin quantum number. For example, a photon as a spin 1 (

s = 1

), and a Higgs boson has spin 0 (

s = 0

8-11 §8.3 source ↗

What is spin-orbit coupling?

Answer

An electron has a magnetic moment associated with its intrinsic (internal) spin. Spin-orbit coupling occurs when this interacts with the magnetic field produced by the orbital angular momentum of the electron.

8-13 §8.4 source ↗

Compare the electron configurations of the elements in the same column of the periodic table.

Answer

Elements that belong in the same column in the periodic table of elements have the same fillings of their outer shells, and therefore the same number of valence electrons. For example:
Li:

1 s^{2} 2 s^{1}

(one valence electron in the

n = 2

shell)
Na:

1 s^{2} 2 s 2 p^{6} 3 s^{1}

(one valence electron in the

n = 2

shell)
Both, Li and Na belong to first column.

8-15 §8.5 source ↗

Atomic and molecular spectra are discrete. What does discrete mean, and how are discrete spectra related to the quantization of energy and electron orbits in atoms and molecules?

Answer

Atomic and molecular spectra are said to be “discrete,” because only certain spectral lines are observed. In contrast, spectra from a white light source (consisting of many photon frequencies) are continuous because a continuous “rainbow” of colors is observed.

8-17 §8.5 source ↗

NGC1763 is an emission nebula in the Large Magellanic Cloud just outside our Milky Way Galaxy. Ultraviolet light from hot stars ionize the hydrogen atoms in the nebula. As protons and electrons recombine, light in the visible range is emitted. Compare the energies of the photons involved in these two transitions.

Answer

UV light consists of relatively high frequency (short wavelength) photons. So the energy of the absorbed photon and the energy transition (

Δ E

) in the atom is relatively large. In comparison, visible light consists of relatively lower-frequency photons. Therefore, the energy transition in the atom and the energy of the emitted photon is relatively small.

8-19 §8.5 source ↗

How do the allowed orbits for electrons in atoms differ from the allowed orbits for planets around the sun?

Answer

For macroscopic systems, the quantum numbers are very large, so the energy difference (

Δ E

) between adjacent energy levels (orbits) is very small. The energy released in transitions between these closely space energy levels is much too small to be detected.

8-21 §8.6 source ↗

Why is a metastable state necessary for the production of laser light?

Answer

Laser light relies on the process of stimulated emission. In this process, electrons must be prepared in an excited (upper) metastable state such that the passage of light through the system produces de-excitations and, therefore, additional light.

8-23 §8.6 source ↗

How is a Blu-Ray player able to read more information that a CD player?

Answer

A Blu-Ray player uses blue laser light to probe the bumps and pits of the disc and a CD player uses red laser light. The relatively short-wavelength blue light is necessary to probe the smaller pits and bumps on a Blu-ray disc; smaller pits and bumps correspond to higher storage densities.

8-25 §8.1 source ↗

The wave function is evaluated at rectangular coordinates (

x, y, z

)

=

(2, 1, 1) in arbitrary units. What are the spherical coordinates of this position?

Answer

(r, θ, ϕ) =

(\sqrt{6,} 66 °, 27 °) .

8-27 §8.1 source ↗

What are the possible values of m for an electron in the

n = 4

state?

Answer

\pm 3, \pm 2, \pm 1, 0

are possible

8-29 §8.1 source ↗

How many possible states are there for the

l = 4

state?

Answer

8-31 §8.1 source ↗

The force on an electron is “negative the gradient of the potential energy function.” Use this knowledge and Equation 8.1 to show that the force on the electron in a hydrogen atom is given by Coulomb’s force law.

Answer

F = − k \frac{Q q}{r^{2}}

8-33 §8.1 source ↗

The wave function is evaluated at spherical coordinates

(r, θ, ϕ) =

(\sqrt{3}, 45 °, 45 °),

where the value of the radial coordinate is given in arbitrary units. What are the rectangular coordinates of this position?

Answer

(1, 1, 1)

8-35 §8.1 source ↗

Write an expression for the total number of states with orbital angular momentum l.

Answer

For the orbital angular momentum quantum number, l, the allowed values of:

m =

− l,

− l +

1,... 0,... l -

1,

l

.
With the exception of

m = 0

, the total number is just 2l because the number of states on either side of

m = 0

is just l. Including

m = 0

, the total number of orbital angular momentum states for the orbital angular momentum quantum number, l, is:

2 l + 1 .

Later, when we consider electron spin, the total number of angular momentum states will be found to twice this value because each orbital angular momentum states is associated with two states of electron spin: spin up and spin down).

8-37 §8.1 source ↗

What is the probability that the 1s electron of a hydrogen atom is found outside the Bohr radius?

Answer

The probability that the 1s electron of a hydrogen atom is found outside of the Bohr radius is

\int_{a_{0}}^{\infty} P (r) d r \approx 0.68

8-39 §8.1 source ↗

What is the maximum number of orbital angular momentum electron states in the

n = 2

shell of a hydrogen atom? (Ignore electron spin.)

Answer

For

n = 2

l = 0

(1 state), and

l = 1

(3 states). The total is 4.

8-41 §8.2 source ↗

Find the magnitude of the orbital magnetic dipole moment of the electron in in the 3p state. (Express your answer in terms of

μ_{B} .

)

Answer

The 3p state corresponds to

n = 3

l = 1

. Therefore,

μ =

μ_{B} \sqrt{1 (1 + 1)} μ =

μ_{B} \sqrt{2}

8-43 §8.2 source ↗

Estimate the ratio of the electron magnetic moment to the muon magnetic moment for the same state of orbital angular momentum. (Hint:

m_{μ} =

105.7 MeV / c^{2})

Answer

The ratio of their masses is 1/207, so the ratio of their magnetic moments is 207. The electron’s magnetic moment is more than 200 times larger than the muon.

8-45 §8.2 source ↗

For a 3d electron in an external magnetic field of

2.50 \times 10^{− 3} T

, find (a) the current associated with the orbital angular momentum, and (b) the maximum torque.

Answer

a. The 3d state corresponds to

n = 3

l = 2

. So,

I =

4.43 \times 10^{−7} A .

b. The maximum torque occurs when the magnetic moment and external magnetic field vectors are at right angles (

sin θ = 1)

. In this case:

| \vec{τ} | = μ B .

τ =

5.70 \times 10^{−26} N \cdot m .

8-47 §8.2 source ↗

Find the minimum torque magnitude

| \vec{τ}

| that acts on the orbital magnetic dipole of a 3p electron in an external magnetic field of

2.50 \times 10^{- 3} T

Answer

A 3p electron is in the state

n = 3

and

l = 1

. The minimum torque magnitude occurs when the magnetic moment and external magnetic field vectors are most parallel (antiparallel). This occurs when

m = \pm 1

.The torque magnitude is given by

| \vec{τ} | =

μ B sin θ,

Where

μ =

(1.31 \times 10^{−24} J/T) .

For

m = \pm 1,

we have:

| \vec{τ} | =

2.32 \times 10^{21} N \cdot m .

8-49 §8.2 source ↗

Show that $U = − \vec{μ} \cdot \vec{B}$ .

(Hint: An infinitesimal amount of work is done to align the magnetic moment with the external field. This work rotates the magnetic moment vector through an angle $- d θ$ (toward the positive z-direction), where $d θ$ is a positive angle change.)

Answer

An infinitesimal work dW done by a magnetic torque

τ

to rotate the magnetic moment through an angle

- d θ

d W = τ (− d θ)

,
where

τ = | \vec{μ} \times \vec{B} |

. Work done is interpreted as a drop in potential energy U, so

d W = − d U .

The total energy change is determined by summing over infinitesimal changes in the potential energy:

U = − μ B c o s θ

U = − \vec{μ} \cdot \vec{B} .

8-51 §8.3 source ↗

What are the possible polar orientations of the spin momentum vector for an electron?

Answer

Spin up (relative to positive z-axis):

θ = 55 ° .

Spin down (relative to positive z-axis):

θ =

{cos}^{−1} (\frac{S_{z}}{S}) =

{cos}^{−1} (\frac{- \frac{1}{2}}{\frac{\sqrt{3}}{2}}) =

{cos}^{−1} (\frac{−1}{\sqrt{3}}) =

125 ° .

8-53 §8.3 source ↗

A hydrogen atom is placed in an external uniform magnetic field (

B = 200 T

). Calculate the wavelength of light produced in a transition from a spin up to spin down state.

Answer

The spin projection quantum number is

m_{s} = \pm ½

, so the z-component of the magnetic moment is

μ_{z} = \pm μ_{B} .

The potential energy associated with the interaction between the electron and the external magnetic field is

U = \mp μ_{B} B .

The energy difference between these states is

Δ E = 2 μ_{B} B

, so the wavelength of light produced is

λ =

8.38 \times 10^{−5} m \approx

84 μ m

8-55 §8.3 source ↗

If the magnetic moment in the preceding problem is doubled, what happens to the frequency of light produced in a transition from a spin-up to spin-down state?

Answer

It is increased by a factor of 2.

8-57 §8.4 source ↗

(a) How many electrons can be in the $n = 4$ shell?

(b) What are its subshells, and how many electrons can be in each?

Answer

a. 32; b.

\begin{matrix} \underset{_____________________}{\begin{matrix} \underline{ℓ} & 2 (2 ℓ + 1) \\ 0 & s & 2 (0 + 1) & = & 2 \\ 1 & p & 2 (2 + 1) & = & 6 \\ 2 & d & 2 (4 + 1) & = & 10 \\ 3 & f & 2 (6 + 1) & = & 14 \end{matrix}} \\ 32 \end{matrix}

8-59 §8.4 source ↗

Unreasonable result. Which of the following spectroscopic notations are not allowed? (a)

5 s^{1}

(b)

1 d^{1}

(c)

4 s^{3}

(d)

3 p^{7}

(e)

5 g^{15}

. State which rule is violated for each notation that is not allowed.

Answer

a. and e. are allowed; the others are not allowed.
b.

l = 3

not allowed for

n =

1,

l \leq (n - 1) .

c. Cannot have three electrons in s subshell because

3 >

2 (2 l + 1) =

2 .

d. Cannot have seven electrons in p subshell (max of 6)

2 (2 l + 1) =

2 (2 + 1) =

6 .

8-61 §8.4 source ↗

Write the electron configuration for iron.

Answer

[Ar] 4 s^{2} 3 d^{6}

8-63 §8.4 source ↗

(a) If one subshell of an atom has nine electrons in it, what is the minimum value of l? (b) What is the spectroscopic notation for this atom, if this subshell is part of the

n = 3

shell?

Answer

a. The minimum value of

ℓ

l = 2

to have nine electrons in it.
b.

3 d^{9} .

8-65 §8.4 source ↗

Write the electron configuration for carbon.

Answer

[He] 2 s^{2} 2 p^{2}

8-67 §8.5 source ↗

What is the minimum frequency of a photon required to ionize: (a) a

{He}^{+}

ion in its ground state? (b) A

{Li}^{2+}

ion in its first excited state?

Answer

For

{He}^{+}

, one electron “orbits” a nucleus with two protons and two neutrons (

Z = 2

). Ionization energy refers to the energy required to remove the electron from the atom. The energy needed to remove the electron in the ground state of

{He}^{+}

ion to infinity is negative the value of the ground state energy, written:

E = −54.4 eV .

Thus, the energy to ionize the electron is

+54.4 eV.

Similarly, the energy needed to remove an electron in the first excited state of

{Li}^{2+}

ion to infinity is negative the value of the first excited state energy, written:

E = −30.6 eV .

The energy to ionize the electron is 30.6 eV.

8-69 §8.5 source ↗

The red light emitted by a ruby laser has a wavelength of 694.3 nm. What is the difference in energy between the initial state and final state corresponding to the emission of the light?

Answer

The wavelength of the laser is given by:

λ = \frac{h c}{− Δ E},

where

E_{γ}

is the energy of the photon and

Δ E

is the magnitude of the energy difference. Solving for the latter, we get:

Δ E =

−2.795 eV .

The negative sign indicates that the electron lost energy in the transition.

8-71 §8.5 source ↗

Estimate the wavelength of the

K_{α}

X-ray from calcium.

Answer

Δ E_{L \to K} \approx

{(Z - 1)}^{2} (10.2 eV) =

3.68 \times 10^{3} eV .

8-73 §8.5 source ↗

X-rays are produced by striking a target with a beam of electrons. Prior to striking the target, the electrons are accelerated by an electric field through a potential energy difference:

$Δ U = − e Δ V,$

where e is the charge of an electron and $Δ V$ is the voltage difference. If $Δ V = 15,000$ volts, what is the minimum wavelength of the emitted radiation?

Answer

According to the conservation of the energy, the potential energy of the electron is converted completely into kinetic energy. The initial kinetic energy of the electron is zero (the electron begins at rest). So, the kinetic energy of the electron just before it strikes the target is:

K = e Δ V .

If all of this energy is converted into braking radiation, the frequency of the emitted radiation is a maximum, therefore:

f_{max} = \frac{e Δ V}{h} .

When the emitted frequency is a maximum, then the emitted wavelength is a minimum, so:

λ_{min} =

0.1293 nm .

8-75 §8.5 source ↗

Suppose the experiment in the preceding problem is conducted with muons. What happens to the minimum wavelength?

Answer

A muon is 200 times heavier than an electron, but the minimum wavelength does not depend on mass, so the result is unchanged.

8-77 §8.5 source ↗

A color television tube generates some X-rays when its electron beam strikes the screen. What is the shortest wavelength of these X-rays, if a 30.0-kV potential is used to accelerate the electrons? (Note that TVs have shielding to prevent these X-rays from exposing viewers.)

Answer

4.13 \times 10^{−11} m

8-79 §8.5 source ↗

The maximum characteristic X-ray photon energy comes from the capture of a free electron into a K shell vacancy. What is this photon energy in keV for tungsten, assuming that the free electron has no initial kinetic energy?

Answer

72.5 keV

8-81 §8.5 source ↗

Compare the X-ray photon wavelengths for copper and gold.

Answer

The atomic numbers for Cu and Au are

Z = 29

and 79, respectively. The X-ray photon frequency for gold is greater than copper by a factor:

{(\frac{f_{Au}}{f_{Cu}})}^{2} =

{(\frac{79 - 1}{29 - 1})}^{2} \approx

8 .

Therefore, the X-ray wavelength of Au is about eight times shorter than for copper.

8-83 §8.6 source ↗

A carbon dioxide laser used in surgery emits infrared radiation with a wavelength of

10.6 μ m

. In 1.00 ms, this laser raised the temperature of

1.00 {cm}^{3}

of flesh to

100 °C

and evaporated it. (a) How many photons were required? You may assume that flesh has the same heat of vaporization as water. (b) What was the minimum power output during the flash?

Answer

a. If flesh has the same density as water, then we used ^{$1.34 \times 10^{23}$} photons. b. 2.52 MW

8-85 §8.AP source ↗

For a hydrogen atom in an excited state with principal quantum number n, show that the smallest angle that the orbital angular momentum vector can make with respect to the z-axis is

θ =

{cos}^{−1} (\sqrt{\frac{n - 1}{n}}) .

Answer

The smallest angle corresponds to

l = n - 1

and

m = l = n - 1

. Therefore

θ =

{cos}^{−1} (\sqrt{\frac{n - 1}{n}}) .

8-87 §8.AP source ↗

Sketch the potential energy function of an electron in a hydrogen atom. (a) What is the value of this function at

r = 0

? in the limit that

r = \infty

? (b) What is unreasonable or inconsistent with the former result?

Answer

a. According to Equation 8.1, when

r = 0

U (r) = − \infty

, and when

r =

+

\infty,

U (r) =

0 .

b. The former result suggests that the electron can have an infinite negative potential energy. The quantum model of the hydrogen atom avoids this possibility because the probability density at

r = 0

is zero.

8-89 §8.AP source ↗

Show that the maximum number of orbital angular momentum electron states in the nth shell of an atom is

n^{2}

. (Ignore electron spin.) (Hint: Make a table of the total number of orbital angular momentum states for each shell and find the pattern.)

Answer

A formal solution using sums is somewhat complicated. However, the answer easily found by studying the mathematical pattern between the principal quantum number and the total number of orbital angular momentum states.
For

n = 1

, the total number of orbital angular momentum states is 1; for

n = 2

, the total number is 4; and, when

n = 3

, the total number is 9, and so on. The pattern suggests the total number of orbital angular momentum states for the nth shell is

n^{2}

.
(Later, when we consider electron spin, the total number of angular momentum states will be found to be

2 n^{2}

, because each orbital angular momentum states is associated with two states of electron spin; spin up and spin down).

8-91 §8.AP source ↗

What is the maximum number of electron states in the

n = 5

shell?

Answer

8-93 §8.AP source ↗

Show that the maximum number of electron states in the nth shell of an atom is

2 n^{2}

Answer

The maximum number of orbital angular momentum electron states in the nth shell of an atom is

n^{2}

. Each of these states can be filled by a spin up and spin down electron. So, the maximum number of electron states in the nth shell is

2 n^{2}

8-95 §8.AP source ↗

Which of the following notations are allowed (that is, which violate none of the rules regarding values of quantum numbers)? (a)

1 s^{1};

(b)

1 d^{3};

(c)

4 s^{2};

(d)

3 p^{7};

(e)

6 h^{20}

Answer

a., c., and e. are allowed; the others are not allowed. b.

l > n

is not allowed.
d.

7 >

2 (2 l + 1)

8-97 §8.AP source ↗

The maximum characteristic X-ray photon energy comes from the capture of a free electron into a K shell vacancy. What is this photon frequency for tungsten, assuming that the free electron has no initial kinetic energy?

Answer

f = 1.8 \times 10^{9} Hz

8-99 §8.AP source ↗

Compare the X-ray photon wavelengths for copper and silver.

Answer

The atomic numbers for Cu and Ag are

Z = 29

and 47, respectively. The X-ray photon frequency for silver is greater than copper by the following factor:

{(\frac{f_{Ag}}{f_{Cu}})}^{2} =

2.7 .

Therefore, the X-ray wavelength of Ag is about three times shorter than for copper.

8-101 §8.AP source ↗

A student in a physics laboratory observes a hydrogen spectrum with a diffraction grating for the purpose of measuring the wavelengths of the emitted radiation. In the spectrum, they observe a yellow line and find its wavelength to be 589 nm. (a) Assuming that this is part of the Balmer series, determine

n_{i},

the principal quantum number of the initial state. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?

Answer

a. 3.24; b.

n_{i}

is not an integer. c. The wavelength must not be correct. Because

n_{i} > 2,

the assumption that the line was from the Balmer series is possible, but the wavelength of the light did not produce an integer value for

n_{i}

. If the wavelength is correct, then the assumption that the gas is hydrogen is not correct; it might be sodium instead.

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Problems and answers are adapted from University Physics Volume 3 by Samuel J. Ling, Jeff Sanny and William Moebs, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/university-physics-volume-3/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.