Content check — University Physics 3, Chapter 10: Nuclear Physics

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 10: Nuclear Physics

52 problems · 3 figures

Check Your Understanding 10.1 §10.1 source ↗

Nucleus X is two times larger than nucleus Y. What is the ratio of their atomic masses?

Answer

eight

Check Your Understanding 10.2 §10.2 source ↗

If the binding energy per nucleon is large, does this make it harder or easier to strip off a nucleon from a nucleus?

Answer

harder

Check Your Understanding 10.3 §10.3 source ↗

A radioactive nuclide has a high decay rate. What does this mean for its half-life and activity?

Answer

Half-life is inversely related to decay rate, so the half-life is short. Activity depends on both the number of decaying particles and the decay rate, so the activity can be great or small.

Check Your Understanding 10.4 §10.4 source ↗

In radioactive beta decay, does the atomic mass number, A, increase or decrease?

Answer

Neither; it stays the same.

Check Your Understanding 10.5 §10.5 source ↗

Which has a larger energy yield per fission reaction, a large or small sample of pure

^{235} U ?

Answer

the same

Check Your Understanding 10.6 §10.6 source ↗

Where does the energy from the Sun originate?

Answer

the conversion of mass to energy

Check Your Understanding 10.7 §10.7 source ↗

Radiation propagates in all directions from its source, much as electromagnetic radiation from a light bulb. Is activity concept more analogous to power, intensity, or brightness?

Answer

power

10-1 §10.1 source ↗

Define and make clear distinctions between the terms neutron, nucleon, nucleus, and nuclide.

Answer

The nucleus of an atom is made of one or more nucleons. A nucleon refers to either a proton or neutron. A nuclide is a stable nucleus.

10-3 §10.2 source ↗

Explain why a bound system should have less mass than its components. Why is this not observed traditionally, say, for a building made of bricks?

Answer

A bound system should have less mass than its components because of energy-mass equivalence

(E = m c^{2}) .

If the energy of a system is reduced, the total mass of the system is reduced. If two bricks are placed next to one another, the attraction between them is purely gravitational, assuming the bricks are electrically neutral. The gravitational force between the bricks is relatively small (compared to the strong nuclear force), so the mass defect is much too small to be observed. If the bricks are glued together with cement, the mass defect is likewise small because the electrical interactions between the electrons involved in the bonding are still relatively small.

10-5 §10.2 source ↗

To obtain the most precise value of the binding energy per nucleon, it is important to take into account forces between nucleons at the surface of the nucleus. Will surface effects increase or decrease estimates of BEN?

Answer

Nucleons at the surface of a nucleus interact with fewer nucleons. This reduces the binding energy per nucleon, which is based on an average over all the nucleons in the nucleus.

10-7 §10.3 source ↗

For the carbon dating described in this chapter, what important assumption is made about the time variation in the intensity of cosmic rays?

Answer

That it is constant.

10-9 §10.4 source ↗

What is the difference between

γ

rays and characteristic X-rays and visible light?

Answer

Gamma (γ) rays are produced by nuclear interactions and X-rays and light are produced by atomic interactions. Gamma rays are typically shorter wavelength than X-rays, and X-rays are shorter wavelength than light.

10-11 §10.4 source ↗

Consider Figure 10.12. If the magnetic field is replaced by an electric field pointed in toward the page, in which directions will the

α

β^{+}

-, and

γ

rays bend?

Figure shows a C-shaped material labeled lead. A small circle labeled radioactive source is shown in the hollow of the C-shape. Three rays radiate from this source towards the right. One curves upwards and is labeled alpha. One goes straight and is labeled gamma. The third curves downwards and is labeled beta minus. Magnetic field is shown as crosses. Two arrows originate from near the point where the rays emerge from the C-shape. The upwards pointing arrow is labeled F subscript alpha = q subscript alpha v B. The downwards pointing arrow is labeled F subscript beta = q subscript beta v B. — Figure 10.12 in the book ↗

Answer

Assume a rectangular coordinate system with an xy-plane that corresponds to the plane of the paper.

α

bends into the page (trajectory parabolic in the xz-plane);

β^{+}

bends into the page (trajectory parabolic in the xz-plane); and

γ

is unbent.

10-13 §10.5 source ↗

Should an atomic bomb really be called nuclear bomb?

Answer

Yes. An atomic bomb is a fission bomb, and a fission bomb occurs by splitting the nucleus of atom.

10-15 §10.5 source ↗

In what way is an atomic nucleus like a liquid drop?

Answer

Short-range forces between nucleons in a nucleus are analogous to the forces between water molecules in a water droplet. In particular, the forces between nucleons at the surface of the nucleus produce a surface tension similar to that of a water droplet.

10-17 §10.6 source ↗

Why does the fusion of light nuclei into heavier nuclei release energy?

Answer

The nuclei produced in the fusion process have a larger binding energy per nucleon than the nuclei that are fused. That is, nuclear fusion decreases average energy of the nucleons in the system. The energy difference is carried away as radiation.

10-19 §10.7 source ↗

Isotopes that emit

α

radiation are relatively safe outside the body and exceptionally hazardous inside. Explain why.

Answer

Alpha particles do not penetrate materials such as skin and clothes easily. (Recall that alpha radiation is barely able to pass through a thin sheet of paper.) However, when produce inside the body, neighboring cells are vulnerable.

10-21 §10.1 source ↗

Find the atomic numbers, mass numbers, and neutron numbers for (a)

_{29}^{58} C u,

(b)

_{11}^{24} N a,

(c)

_{84}^{210} P o,

(d)

_{20}^{45} C a,

and (e)

_{82}^{206} P b

Answer

Use the rule $A = Z + N .$

	Atomic Number (Z)	Neutron Number (N)	Mass Number (A)
(a)	29	29	58
(b)	11	13	24
(c)	84	126	210
(d)	20	25	45
(e)	82	124	206

10-23 §10.1 source ↗

The mass (M) and the radius (r) of a nucleus can be expressed in terms of the mass number, A. (a) Show that the density of a nucleus is independent of A. (b) Calculate the density of a gold (Au) nucleus. Compare your answer to that for iron (Fe).

Answer

r =

r_{0} A^{1 / 3},

ρ =

\frac{3 u}{4 π r_{0}^{3}}

;
b.

\begin{matrix} ρ & = & 2.3 \times 10^{17} {kg/m}^{3} \end{matrix}

10-25 §10.1 source ↗

Find the length of a side of a cube having a mass of 1.0 kg and the density of nuclear matter.

Answer

side length

= 1.6 μm

10-27 §10.2 source ↗

How much energy would be released if six hydrogen atoms and six neutrons were combined to form

_{6}^{12} C ?

Answer

92.4 MeV

10-29 §10.2 source ↗

^{56} Fe

is among the most tightly bound of all nuclides. It makes up more than

90 %

of natural iron. Note that

^{56} Fe

has even numbers of protons and neutrons. Calculate the binding energy per nucleon for

^{56} Fe

and compare it with the approximate value obtained from the graph in Figure 10.7.

Graph of binding energy per nucleon, MeV, versus mass number, A. The graph starts close to point 2,1 and peaks close to element 56 Fe, which has an MeV value of between 8 and 9. After this, the graph tapers off to an MeV value of roughly 7. 56 Fe is labeled most stable nucleus. A vertical bar at A = 60 is labeled region of very stable nuclides. Both sides of this bar have an arrow pointing to it. The left one is labeled fusion and the right one is labeled fission. — Figure 10.7 in the book ↗

Answer

8.790 MeV \approx

graph’s value

10-31 §10.2 source ↗

(a) Calculate BEN for

^{235} U

, the rarer of the two most common uranium isotopes; (b) Calculate BEN for

^{238} U

. (Most of uranium is

^{238} U

Answer

7.591 MeV \approx

graph’s value

; b. 7.570 MeV

10-33 §10.3 source ↗

A sample of radioactive material is obtained from a very old rock. A plot lnA verses t yields a slope value of

− 10^{−9} s^{−1}

(see Figure 10.10(b)). What is the half-life of this material?

Figure a shows a graph of A versus t. It starts at point A subscript 0 and reduces with time. The rate of reduction decreases slowly till A is very close to 0, making a curved plot on the graph. The plot is labeled A = A subscript 0 e to the power minus lambda t. Figure b shows a graph of ln A versus t. It starts at ln A subscript 0 and slopes downwards in a straight line. The slope of the line is labeled minus lambda t. — Figure 10.10 in the book ↗

Answer

The decay constant is equal to the negative value of the slope or

10^{−9} s^{−1} .

The half-life of the nuclei, and thus the material, is

T_{1 / 2} =

693 million years .

10-35 §10.3 source ↗

The half-life of strontium-91,

_{38}^{91} S r

is 9.70 h. Find (a) its decay constant and (b) for an initial 1.00-g sample, the activity after 15 hours.

Answer

a. The decay constant is

λ =

1.99 \times 10^{−5} s^{− 1}

. b. Since strontium-91 has an atomic mass of 90.90 g, the number of nuclei in a 1.00-g sample is initially

N_{0} =

6.63 \times 10^{21} nuclei .

The initial activity for strontium-91 is

\begin{matrix} A_{0} & = λ N_{0} \\ = 1.32 \times 10^{17} decays/s \end{matrix}

The activity at

t =

15.0 h =

5.40 \times 10^{4} s

A =

4.51 \times 10^{16} decays/s .

10-37 §10.3 source ↗

A radioactive sample initially contains

2.40 \times 10^{− 2}

mol of a radioactive material whose half-life is 6.00 h. How many moles of the radioactive material remain after 6.00 h? After 12.0 h? After 36.0 h?

Answer

1.20 \times 10^{− 2} mol

;

6.00 \times 10^{− 3} mol

;

3.75 \times 10^{− 4} mol

10-39 §10.3 source ↗

Calculate the activity

R

, in curies of 1.00 g of

^{226} R a .

(b) Explain why your answer is not exactly 1.00 Ci, given that the curie was originally supposed to be exactly the activity of a gram of radium.

Answer

a. 0.988 Ci; b. The half-life of

^{226} R a

is more precisely known than it was when the Ci unit was established.

10-41 §10.3 source ↗

World War II aircraft had instruments with glowing radium-painted dials. The activity of one such instrument was

1.0 \times 10^{5}

Bq when new. (a) What mass of

^{226} R a

was present? (b) After some years, the phosphors on the dials deteriorated chemically, but the radium did not escape. What is the activity of this instrument 57.0 years after it was made?

Answer

2.73 μ g

; b.

9.76 \times 10^{4} Bq

10-43 §10.3 source ↗

Armor-piercing shells with depleted uranium cores are fired by aircraft at tanks. (The high density of the uranium makes them effective.) The uranium is called depleted because it has had its

^{235} U

removed for reactor use and is nearly pure

^{238} U

. Depleted uranium has been erroneously called nonradioactive. To demonstrate that this is wrong: (a) Calculate the activity of 60.0 g of pure

^{238} U

. (b) Calculate the activity of 60.0 g of natural uranium, neglecting the

^{234} U

and all daughter nuclides.

Answer

7.46 \times 10^{5} Bq

; b.

7.75 \times 10^{5} Bq

10-45 §10.4 source ↗

(a) Calculate the energy released in the

α

decay of

^{238} U

. (b) What fraction of the mass of a single

^{238} U

is destroyed in the decay? The mass of

^{234} T h

is 234.043593 u. (c) Although the fractional mass loss is large for a single nucleus, it is difficult to observe for an entire macroscopic sample of uranium. Why is this?

Answer

a. 4.273 MeV; b.

1.927 \times 10^{−5}

; c. Since

^{238} U

is a slowly decaying substance, only a very small number of nuclei decay on human timescales; therefore, although those nuclei that decay lose a noticeable fraction of their mass, the change in the total mass of the sample is not detectable for a macroscopic sample.

10-47 §10.4 source ↗

(a) Write the complete

β^{-}

decay equation for

^{90} S r,

a major waste product of nuclear reactors. (b) Find the energy released in the decay.

Answer

_{38}^{90} S r_{52} \to

_{39}^{90} Y_{51} +

β^{−1} +

\bar{v_{e}}

; b. 0.546 MeV

10-49 §10.4 source ↗

Write the complete decay equation in the complete

_{Z}^{A} X_{N}

notation for the beta (

β^{-}

) decay of

^{3} H

(tritium), a manufactured isotope of hydrogen used in some digital watch displays, and manufactured primarily for use in hydrogen bombs.

Answer

_{1}^{3} H_{2} \to

_{2}^{3} H e_{1} +

β^{-} +

\bar{v_{e}}

10-51 §10.4 source ↗

An electron can interact with a nucleus through the beta-decay process:

$_{Z}^{A} X + e^{-} \to Y + v_{e}$ .

(a) Write the complete reaction equation for electron capture by $^{7} Be$ .

(b) Calculate the energy released.

Answer

_{4}^{7} B e_{3} +

e^{-} \to

_{3}^{7} L i_{4} +

v_{e}

; b. 0.862 MeV

10-53 §10.4 source ↗

A rare decay mode has been observed in which

^{222} R a

emits a

^{14} C

nucleus. (a) The decay equation is

^{222} Ra \to

^{A} X +

^{14} C

. Identify the nuclide

^{A} X

. (b) Find the energy emitted in the decay. The mass of

^{222} Ra

is 222.015353 u.

Answer

X =

_{82}^{208} P b_{126}

; b. 33.05 MeV

10-55 §10.5 source ↗

(a) Calculate the energy released in this rare neutron-induced fission $n +$ $^{238} U \to$ $^{96} S r +$ $^{140} X e +$ $3 n$ , given $m (^{96} S r) =$ $95.921750 u$ and $m (^{140} X e) =$ $139.92164$ .

(b) This result is about 6 MeV greater than the result for spontaneous fission. Why?

Answer

a. 177.1 MeV; b. This value is approximately equal to the average BEN for heavy nuclei. c.

\begin{matrix} n +_{92}^{238} U_{146} & \to_{38}^{96} S r_{58} +_{54}^{140} X e_{86} + 3 n, \\ A_{i} & = 239 = A_{f}, \\ Z_{i} & = 92 = 38 + 54 = Z_{f} \end{matrix}

10-57 §10.5 source ↗

The electrical power output of a large nuclear reactor facility is 900 MW. It has a $35.0 %$ efficiency in converting nuclear power to electrical power.

(a) What is the thermal nuclear power output in megawatts?

(b) How many $^{235} U$ nuclei fission each second, assuming the average fission produces 200 MeV?

Answer

2.57 \times 10^{3} MW

; b.

8.04 \times 10^{19} fissions/s

; c. 991 kg

10-59 §10.6 source ↗

Verify that the total number of nucleons, and total charge are conserved for each of the following fusion reactions in the proton-proton chain.

(i) $^{1} H +$ $^{1} H \to$ $^{2} H +$ $e^{+} +$ $v_{e}$ ,

(ii) $^{1} H +^{2} H \to^{3} H e + γ$ , and (iii) $^{3} H e +$ $^{3} H e \to$ $^{4} H e +$ $^{1} H +$ $^{1} H$ .

(List the value of each of the conserved quantities before and after each of the reactions.)

Answer

\begin{matrix} _{1}^{1} H +_{1}^{1} H & \to_{1}^{2} H + e^{+} + v_{e} \\ A_{i} & = 1 + 1 = 2; A_{f} = 2 Z_{i} = 1 + 1 = 2 \\ Z_{f} & = 1 + 1 = 2 \end{matrix}

;
ii.

\begin{matrix} _{1}^{1} H +_{1}^{2} H & \to_{2}^{3} H + γ \\ A_{i} & = 1 + 2 = 3; A_{f} = 3 + 0 = 3 Z_{i} = 1 + 1 = 2 \\ Z_{E} & = 1 + 1 = 2 \end{matrix}

;
iii.

\begin{matrix} _{2}^{3} H +_{2}^{3} H & \to_{2}^{4} H +_{1}^{1} H +_{1}^{1} H \\ A_{i} & = 3 + 3 = 6; A_{f} = 4 + 1 + 1 = 6 Z_{i} = 2 + 2 = 4 \\ Z_{f} & = 2 + 1 + 1 = 4 \end{matrix}

10-61 §10.6 source ↗

Show that the total energy released in the proton-proton chain is 26.7 MeV, considering the overall effect in

^{1} H +

^{1} H \to

^{2} H +

e^{+} +

v_{e}

^{1} H +^{2} H \to^{3} H e + γ

, and

^{3} H e +

^{3} H e \to

^{4} H e +

^{1} H +

^{1} H

. Be sure to include the annihilation energy.

Answer

26.73 MeV

10-63 §10.6 source ↗

The power output of the Sun is

4 \times 10^{26} W .

(a) If

90 %

of this energy is supplied by the proton-proton chain, how many protons are consumed per second? (b) How many neutrinos per second should there be per square meter at the surface of Earth from this process?

Answer

3 \times 10^{38} protons/s

; b.

6 \times 10^{14} {neutrinos/m}^{2} \cdot s

;
This huge number is indicative of how rarely a neutrino interacts, since large detectors observe very few per day.

10-65 §10.6 source ↗

(a) Calculate the energy released by the fusion of a 1.00-kg mixture of deuterium and tritium, which produces helium. There are equal numbers of deuterium and tritium nuclei in the mixture.

(b) If this process takes place continuously over a period of a year, what is the average power output?

Answer

a. The atomic mass of deuterium (

^{2} H

) is 2.014102 u, while that of tritium (

^{3} H

) is 3.016049 u, for a total of 5.032151 u per reaction. So a mole of reactants has a mass of 5.03 g, and in 1.00 kg, there are

(1000 g) / (5.03 g / mol) =

198.8 mol

of reactants. The number of reactions that take place is therefore

(198.8 mol) (6.02 \times 10^{23} {mol}^{−1}) =

1.20 \times 10^{26} reactions

.
The total energy output is the number of reactions times the energy per reaction:

E =

3.37 \times 10^{14} J;

b. Power is energy per unit time. One year has

3.16 \times 10^{7} s,

P = 10.7 MW .

We expect nuclear processes to yield large amounts of energy, and this is certainly the case here. The energy output of

3.37 \times 10^{14} J

from fusing 1.00 kg of deuterium and tritium is equivalent to 2.6 million gallons of gasoline and about eight times the energy output of the bomb that destroyed Hiroshima. Yet the average backyard swimming pool has about 6 kg of deuterium in it, so that fuel is plentiful if it can be utilized in a controlled manner.

10-67 §10.7 source ↗

Find the radiation dose in Gy for: (a) A 10-mSv fluoroscopic X-ray series. (b) 50 mSv of skin exposure by an

α

emitter. (c) 160 mSv of

β^{-}

and

γ

rays from the

^{40} K

in your body.

Answer

Gy = \frac{Sv}{RBE}

: a. 0.01 Gy; b. 0.0025 Gy; c. 0.16 Gy

10-69 §10.7 source ↗

In the 1980s, the term picowave was used to describe food irradiation in order to overcome public resistance by playing on the well-known safety of microwave radiation. Find the energy in MeV of a photon having a wavelength of a picometer.

Answer

1.24 MeV

10-71 §10.7 source ↗

One half the

γ

rays from

^{99 m} T c

are absorbed by a 0.170-mm-thick lead shielding. Half of the

γ

rays that pass through the first layer of lead are absorbed in a second layer of equal thickness. What thickness of lead will absorb all but one in 1000 of these

γ

rays?

Answer

1.69 mm

10-73 §10.7 source ↗

A plumber at a nuclear power plant receives a whole-body dose of 30 mSv in 15 minutes while repairing a crucial valve. Find the radiation-induced yearly risk of death from cancer and the chance of genetic defect from this maximum allowable exposure.

Answer

For cancer:

(3 rem) (\frac{10}{10^{6} rem \cdot y}) =

\frac{30}{10^{6} y},

The risk each year of dying from induced cancer is 30 in a million. For genetic defect:

(3 rem) (\frac{3.3}{10^{6} rem \cdot y}) =

\frac{9.9}{10^{6} y},

The chance each year of an induced genetic defect is 10 in a million.

10-75 §10.AP source ↗

The wiki-phony site states that the atomic mass of chlorine is 40 g/mol. Check this result. Hint: The two, most common stable isotopes of chlorine are:

_{17}^{35} Cl

and

_{17}^{37} Cl

. (The abundance of Cl-35 is

75.8 %

, and the abundance of Cl-37 is

24.2 %

Answer

atomic mass (Cl) =

35.5 g/mol

10-77 §10.AP source ↗

A nuclear physicist finds

1.0 μ g

^{236} U

in a piece of uranium ore (T_1/2 =

2.348 \times 10^{7} y

). (a) Use the decay law to determine how much

^{236} U

would had to have been on Earth when it formed

4.543 \times 10^{9} y

ago for

1.0 μ g

to be left today. (b) What is unreasonable about this result? (c) How is this unreasonable result resolved?

Answer

1.71 \times 10^{58} kg

; b. This mass is impossibly large; it is greater than the mass of the entire Milky Way galaxy. c.

^{236} U

is not produced through natural processes operating over long times on Earth, but through artificial processes in a nuclear reactor.

10-79 §10.AP source ↗

According to your lab partner, a 2.00-cm-thick sodium-iodide crystal absorbs all but

10 %

of rays from a radioactive source and a 4.00-cm piece of the same material absorbs all but

5 % ?

Is this result reasonable?

Answer

10 %

of rays are left after 2.00 cm, then only

{(0.100)}^{2} =

0.01 =

1 %

are left after 4.00 cm. This is much smaller than your lab partner’s result (

5 %

10-81 §10.AP source ↗

The ceramic glaze on a red-orange “Fiestaware” plate is

U_{2} O_{3}

and contains 50.0 grams of

^{238} U

, but very little

^{235} U

. (a) What is the activity of the plate? (b) Calculate the total energy that will be released by the

^{238} U

decay. (c) If energy is worth 12.0 cents per

kW \cdot h

, what is the monetary value of the energy emitted? (These brightly-colored ceramic plates went out of production some 30 years ago, but are still available as collectibles.)

Answer

1.68 \times 10^{−5} Ci

; (b) From Appendix B, the energy released per decay is 4.27 MeV, so

8.65 \times 10^{10} J

; (c) The monetary value of the energy is

$ 2.9 \times 10^{3}

10-83 §10.AP source ↗

A piece of wood from an ancient Egyptian tomb is tested for its carbon-14 activity. It is found to have an activity per gram of carbon of

A =

10 decay/min \cdot g

. What is the age of the wood?

Answer

We know that

λ =

3.84 \times 10^{−12} s^{− 1}

and

A_{0} =

0.25 decays / s \cdot g =

15 decays / min \cdot g .

Thus, the age of the tomb is

t =

-

\frac{1}{3.84 \times 10^{−12} s^{− 1}} ln \frac{10 decays / min \cdot g}{15 decays/ min \cdot g} =

1.06 \times 10^{11} s \approx

3350 y .

10-85 §10.CP source ↗

The Galileo space probe was launched on its long journey past Venus and Earth in 1989, with an ultimate goal of Jupiter. Its power source is 11.0 kg of $^{238} Pu,$ a by-product of nuclear weapons plutonium production. Electrical energy is generated thermoelectrically from the heat produced when the 5.59-MeV $α$ particles emitted in each decay crash to a halt inside the plutonium and its shielding. The half-life of $^{238} Pu$ is 87.7 years.

(a) What was the original activity of the $^{238} Pu$ in becquerels?

(b) What power was emitted in kilowatts?

(c) What power was emitted 12.0 y after launch? You may neglect any extra energy from daughter nuclides and any losses from escaping $γ$ rays.

Answer

6.97 \times 10^{15} Bq

; b. 6.24 kW; c. 5.67 kW

10-87 §10.CP source ↗

Engineers are frequently called on to inspect and, if necessary, repair equipment in nuclear power plants. Suppose that the city lights go out. After inspecting the nuclear reactor, you find a leaky pipe that leads from the steam generator to turbine chamber. (a) How do the pressure readings for the turbine chamber and steam condenser compare? (b) Why is the nuclear reactor not generating electricity?

Answer

a. Due to the leak, the pressure in the turbine chamber has dropped significantly. The pressure difference between the turbine chamber and steam condenser is now very low. b. A large pressure difference is required for steam to pass through the turbine chamber and turn the turbine.

10-89 §10.CP source ↗

For the reaction,

n +^{3} He \to^{4} He + γ

, find the amount of energy transfers to

^{4} H e

and

γ

(on the right side of the equation). Assume the reactants are initially at rest. (Hint: Use conservation of momentum principle.)

Answer

The energies are

\begin{matrix} E_{γ} & = 20.6 MeV \\ E_{^{4} H e} & = 5.68 \times 10^{−2} MeV \end{matrix}

. Notice that most of the energy goes to the

γ

ray.

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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.