30 Instructor Solutions Manual

30  14  Download (0)

Full text

(1)

Responses to Questions

1. Different isotopes of a given element have the same number of protons and electrons. Because they have the same number of electrons, they have almost identical chemical properties. Each isotope has a different number of neutrons from other isotopes of the same element. Accordingly, they have different atomic masses and mass numbers. Since the number of neutrons is different, they may have different nuclear properties, such as whether they are radioactive or not.

2. Identify the element based on the atomic number. (a) Uranium (Z = 92)

(b) Nitrogen (Z = 7) (c) Hydrogen (Z = 1) (d) Strontium (Z = 38) (e) Fermium (Z = 100)

3. The number of protons is the same as the atomic number, and the number of neutrons is the mass number minus the number of protons.

(a) Uranium: 92 protons, 140 neutrons (b) Nitrogen: 7 protons, 11 neutrons (c) Hydrogen: 1 proton, 0 neutrons (d) Strontium: 38 protons, 48 neutrons (e) Fermium: 100 protons, 152 neutrons

4. With 87 nucleons and 50 neutrons, there must be 37 protons. This is the atomic number, so the element is rubidium. The nuclear symbol is 8737Rb.

5. The atomic mass of an element as shown in the Periodic Table is the average atomic mass of all naturally occurring isotopes. For example, chlorine occurs as roughly 75% 1735Cl and 25% 1737Cl, so its atomic mass is about 35.5 ( 0.75 35 0.25 37).= × + × Other smaller effects would include the fact that the masses of the nucleons are not exactly 1 atomic mass unit and that some small fraction of the mass energy of the total set of nucleons is in the form of binding energy.

(2)

6. The alpha particle is a very stable nucleus. It has less energy when bound together than when split apart into separate nucleons.

7. The strong force and the electromagnetic (EM) force are two of the four fundamental forces in nature. They are both involved in holding atoms together: the strong force binds quarks into nucleons and binds nucleons together in the nucleus; the EM force is responsible for binding negatively charged electrons to positively charged nuclei and for binding atoms into molecules. The strong force is the strongest fundamental force; the EM force is about 100 times weaker at distances on the order of

17

10− m. The strong force operates at short range and is negligible for distances greater than about the size of the nucleus. The EM force is a long-range force that decreases as the inverse square of the distance between the two interacting charged particles. The EM force operates only between charged particles. The strong force is always attractive; the EM force can be attractive or repulsive. Both of these forces have mediating field particles associated with them—the gluon for the strong force and the photon for the EM force.

8. Quoting from Section 30–3, “… radioactivity was found in every case to be unaffected by the strongest physical and chemical treatments, including strong heating or cooling and the action of strong

chemical reagents.” Chemical reactions are a result of electron interactions, not nuclear processes. The absence of effects caused by chemical reactions is evidence that the radioactivity is not due to electron interactions. Another piece of evidence is the fact that the α particle emitted in many radioactive decays is much heavier than an electron and has a different charge than the electron, so it can’t be an electron. Therefore, it must be from the nucleus. Finally, the energies of the electrons or photons emitted from radioactivity are much higher than those corresponding to electron orbital transitions. All of these observations support radioactivity being a nuclear process.

9. The resulting nuclide for gamma decay is the same isotope in a lower energy state. 6429Cu* → 6429Cu+γ

The resulting nuclide for β− decay is an isotope of zinc, 6430Zn. 6429Cu → 6430Zn e+ −+v

The resulting nuclide for β+ decay is an isotope of nickel, 6428Ni. 6429Cu → 6428Ni e+ ++v

10. 23892U decays by alpha emission into 23490Th, which has 144 neutrons.

(3)

12. (a) 4520Ca → 4521Sc e+ −+v Scandium-45 is the missing nucleus. (b) 5829Cu → 5829Cu+γ Copper-58 is the missing nucleus.

(c) 4624Cr → 4623V e+ ++v The positron and the neutrino are the missing particles. (d) 23494Pu → 23092U+α Uranium-230 is the missing nucleus.

(e) 23993Np → 23994Pu e+ −+v The electron and the antineutrino are the missing particles. 13. The two extra electrons held by the newly formed thorium will be very loosely held, as the number of

protons in the nucleus will have been reduced from 92 to 90, reducing the nuclear charge. It will be easy for these extra two electrons to escape from the thorium atom through a variety of mechanisms. They are in essence “free” electrons. They do not gain kinetic energy from the decay. They might get captured by the alpha nucleus, for example.

14. When a nucleus undergoes either β− or β+ decay, it becomes a different element, since it has converted either a neutron to a proton or a proton to a neutron. Thus its atomic number (Z) has changed. The energy levels of the electrons are dependent on Z, so all of those energy levels change to become the energy levels of the new element. Photons (with energies on the order of tens of eV) are likely to be emitted from the atom as electrons change energies to occupy the new levels.

15. In alpha decay, assuming the energy of the parent nucleus is known, the unknowns after the decay are the energies of the daughter nucleus and the alpha. These two values can be determined by energy and momentum conservation. Since there are two unknowns and two conditions, the values are uniquely determined. In beta decay, there are three unknown postdecay energies since there are three particles present after the decay. The conditions of energy and momentum conservation are not sufficient to exactly determine the energy of each particle, so a range of values is possible.

16. In electron capture, the nucleus will effectively have a proton change to a neutron. This isotope will then lie to the left and above the original isotope. Since the process would only occur if it made the nucleus more stable, it must lie BELOW the line of stability in Fig. 30–2.

17. Neither hydrogen nor deuterium can emit an α particle. Hydrogen has only one nucleon (a proton) in its nucleus, and deuterium has only two nucleons (one proton and one neutron) in its nucleus. Neither one has the necessary four nucleons (two protons and two neutrons) to emit an α particle.

18. Many artificially produced radioactive isotopes are rare in nature because they have decayed away over time. If the half-lives of these isotopes are relatively short in comparison with the age of Earth (which is typical for these isotopes), then there won’t be any significant amount of these isotopes left to be found in nature. Also, many of these isotopes have a very high energy of formation, which is generally not available under natural circumstances.

19. After two months the sample will not have completely decayed. After one month, half of the sample will remain, and after two months, one-fourth of the sample will remain. Each month, half of the remaining atoms decay.

(4)

21. There are a total of 4 protons and 3 neutrons in the reactant. The α particle has 2 protons and 2 neutrons, so 2 protons and 1 neutron are in the other product particle. It must be 32He.

63Li+11p → 42α+32He

22. The technique of 146C would not be used to measure the age of stone walls and tablets. Carbon-14 dating is only useful for measuring the age of objects that were living at some earlier time.

23. The decay series of Fig. 30–11 begins with a nucleus that has many more neutrons than protons and lies far above the line of stability in Fig. 30–2. The alpha decays remove both 2 protons and 2 neutrons, but smaller stable nuclei have a smaller percentage of neutrons. In β+ decay, a proton is converted to a neutron, which would take the nuclei in this decay series even farther from the line of stability. Thus, β+ decay is not energetically preferred.

24. There are four alpha particles and four β− particles (electrons) emitted, no matter which decay path is chosen. The nucleon number drops by 16 as 22286Rn decays into 20682Pb, indicating that four alpha decays occurred. The proton number only drops by four, from Z = 86 to Z = 82, but four alpha decays would result in a decrease of eight protons. Four β− decays will convert four neutrons into protons, making the decrease in the number of protons only four, as required. (See Fig. 30–11.)

25. (i) Since the momentum before the decay was 0, the total momentum after the decay will also be 0. Since there are only two decay products, they must move in opposite directions with equal magnitude of momentum. Thus, (c) is the correct choice: both the same.

(ii) Since the products have the same momentum, the one with the smallest mass will have the greater velocity. Thus (b) is the correct choice: the alpha particle.

(iii) We assume that the products are moving slowly enough that classical mechanics can be used. In that case, KE= p2/2 .m Since the particles have the same momentum, the one with the smallest mass will have the greater kinetic energy. Thus, (b) is the correct choice: the alpha particle.

Responses to MisConceptual Questions

1. (a) A common misconception is that the elements of the Periodic Table are distinguished by the number of electrons in the atom. This misconception arises because the number of electrons in a neutral atom is the same as the number of protons in its nucleus. Elements can be ionized by adding or removing electrons, but this does not change what type of element it is. When an element undergoes a nuclear reaction that changes the number of protons in the nucleus, the element does transform into a different element.

(5)

3. (c) Large nuclei are typically unstable, so increasing the number of nuclei does not necessarily make the nucleus more stable. Nuclei such as 148O have more protons than neutrons, yet 148O is not as stable as 168O. Therefore, having more protons than neutrons does not necessarily make a nucleus more stable. Large unstable nuclei, such as 23894Pu, have a much larger total binding energy than small stable nuclei such as 42He, so the total binding energy is not a measure of stability. However, stable nuclei typically have large binding energies per nucleon, which means that each nucleon is more tightly bound to the others.

4. (e) The Coulomb repulsive force does act inside the nucleus, pushing the protons apart. Another larger attractive force is thus necessary to keep the nucleus together. The force of gravity is far too small to hold the nucleus together. Neutrons are not negatively charged. It is the attractive strong nuclear force that overcomes the Coulomb force to hold the nucleus together.

5. (b) The exponential nature of radioactive decay is a concept that can be misunderstood. It is sometimes thought that the decay is linear, such that the time for a substance to decay

completely is twice the time for half of the substance to decay. However, radioactive decay is not linear but exponential. That is, during each half-life, half of the remaining substance decays. If half the original decays in the first half-life, then half remains. During the second half-life, half of what is left decays, which would be one-quarter of the initial substance. In each subsequent half-life, half of the remaining substance decays, so it takes many half-lives for a substance to effectively decay away. The decay constant is inversely related to the half-life.

6. (e) The half-life is the time it takes for half of the substance to decay away. The half-life is a constant determined by the composition of the substance and not by the quantity of the initial substance. As the substance decays away, the number of nuclei is decreasing and the activity (number of decays per second) is decreasing, but the half-life remains constant.

7. (d) A common misconception is that it would take twice the half-life, or 20 years, for the substance to completely decay. This is incorrect because radioactive decay is an exponential process. That is, during each half-life, 1/2 of the remaining substance decays. After the first 10 years, 1/2 remains. After the second 10-year period, 1/4 remains. After each succeeding half-life another half of the remaining substance decays, leaving 1/8, then 1/16, then 1/32, and so forth. The decays stop when none of the substance remains, and this time cannot be exactly determined. 8. (e) During each half-life, 1/2 of the substance decays, leaving 1/2 remaining. After the first day, 1/2

remains. After the second day, 1/2 of 1/2, or 1/4, remains. After the third day, 1/2 of 1/4, or 1/8, remains. Since 1/8 remains, 7/8 of the substance has decayed.

9. (c) A common misconception is that after the second half-life none of the substance remains. However, during each half-life, 1/2 of the remaining substance decays. After one half-life, 1/2 remains. After two half-lives, 1/4 remains. After three half-lives, 1/8 remains.

(6)

11. (d) The element with the largest decay constant will have the shortest half-life. Converting each of the choices to decays per second yields the following: (a) 100/s, (b) 1.6 10 /s,× 8 (c) 2.5 10 /s,× 10 (d) 8.6 10 /s.× 13 Answer (d) has the largest decay rate, so it will have the smallest half-life. 12. (d) If U-238 only decayed by beta decay, the number of nucleons would not change. Pb-206 has 32

nucleons less than U-238, so all decays cannot be beta decays. In each alpha decay, the number of nucleons decreases by four and the number of protons decreases by two. U-238 has 92 protons and Pb-206 has 82 protons, so if only alpha decays occurred, then the number of nucleons would need to decrease by 20. But Pb-206 has 32 fewer nucleons than U-238. Therefore, the sequence is a combination of alpha and beta decays. Gamma decays are common, since some of the alpha and beta decays leave the nuclei in excited states where energy needs to be released.

13. (d) A common misconception is that carbon dating is useful for very long time periods. C-14 has a half-life about 6000 years. After about 10 half-lives only about 1/1000 of the substance remains. It is difficult to obtain accurate measurements for longer time periods. For example, after 600,000 years, or about 100 half-lives, only one part in 1030 remains.

14. (b) The half-life of radon remains constant and is not affected by temperature. Radon that existed several billion years ago will have completely decayed away. Small isotopes, such as carbon-14, can be created from cosmic rays, but radon is a heavy element. Lightning is not energetic enough to affect the nucleus of the atoms. Heavy elements such as plutonium and uranium can decay into radon and thus are the source of present-day radon gas.

15. (d) The gravitational force between nucleons is much weaker (about 50 orders of magnitude weaker) than the repulsive Coulomb force; therefore, gravity cannot hold the nucleus together. Neutrons are electrically neutral and therefore cannot overcome the Coulomb force. Covalent bonds exist between the electrons in molecules, not among the nucleons. The actual force that holds the nucleus together is the strong nuclear force, but this was not one of the options.

16. (a) The nature of mass and energy in nuclear physics is often misunderstood. When the neutron and proton are close together, they bind together by releasing mass energy that is equivalent to the binding energy. This energy comes from a reduction in their mass. Therefore, when the neutron and proton are far from each other, their net mass is greater than their net mass when they are bound together.

Solutions to Problems

1. Convert the units from MeV/c2 to atomic mass units.

(139 MeV/ )2 1 u 2 0.149 u 931.5 Me V/

m c

c

⎛ ⎞

= ⎜=

⎝ ⎠

2. The α particle is a helium nucleus and has A = 4. Use Eq. 30–1.

(7)

3. The radii of the two nuclei can be calculated with Eq. 30–1. Take the ratio of the two radii. 1/3 15 1/3 238 15 1/3 232

(1.2 10 m)(238) 238

1.00855 232

(1.2 10 m)(232) r r − − × ⎛ ⎞ = =⎜ = ×

So the radius of 23892U is 0.855% larger than the radius of 23292U. 4. Use Eq. 30–1 for both parts of this problem.

(a) r=(1.2 10× −15 m)A1/3=(1.2 10× −15 m)(112)1/3= 5.8 10× −15 m =5.8 fm

(b)

3

3 15

15 1/3

15 15

3.7 10 m

(1.2 10 m) 29.3 29

1 2 10 m 1.2 10 m

r

r= × − AA= =⎜⎛ × − ⎞⎟ = ≈

. × ×

⎝ ⎠

5. To find the mass of an α particle, subtract the mass of the two electrons from the mass of a helium atom:

He e

2

2 2

2

931 5 MeV/

(4.002603 u) 2(0 511 MeV/ ) 3727 MeV/ 1 u

m m m

c

c c

α = −

.

= ⎜− . =

⎝ ⎠

6. Each particle would exert a force on the other through the Coulomb electrostatic force (given by Eq. 16–1). The distance between the particles is twice the radius of one of the particles.

( )

2

9 2 2 19

2 2

1/3 15

(8.988 10 N m /C ) (2)(1 60 10 C)

63.41 N 63 N (2 ) (2) 4 (1 2 10 m)

Q Q F k r α α α − − ⎡ ⎤ × ⋅ . × = = = ≈ ⎡ . × ⎤ ⎣ ⎦

The acceleration is found from Newton’s second law. We use the mass of a “bare” alpha particle as calculated in Problem 5.

27 27 2

2

2 63.41 N

9 5 10 m/s 1 6605 10 kg

3727 MeV/

931 5 MeV/ F

F ma a

m c c − = → = = = . × ⎛ . × ⎞ ⎜ ⎟ ⎜ . ⎟ ⎝ ⎠

7. First, we calculate the density of nuclear matter. The mass of a nucleus with mass number A is approximately (A u) and its radius is r=(1.2 10× −15m)A1/3. Calculate the density.

27 27

17 3

3 15 3

4 4

3 3

(1.6605 10 kg/u) (1.6605 10 kg/u)

2.294 10 kg/m (1.2 10 m)

m A A

V r A

ρ π π − − − × × = = = = × ×

We see that this is independent of A. The value has 2 significant figures. (a) We set the density of the Earth equal to the density of nuclear matter.

Earth Earth nuclear 4 3

matter 3 Earth 1/3

1/3 24

Earth

Earth 4 4 17 3

nuclear

3 3

matter

5 98 10 kg

183.9 m 180 m (2 294 10 kg/m )

(8)

(b) Set the density of Earth equal to the density of uranium, and solve for the radius of the uranium. Then compare that with the actual radius of uranium, using Eq. 30–1, with A = 238.

Earth U Earth U

3 3 3 3

4 4

Earth U Earth U

3 3

1/3

1/3 27

6 10

U

U Earth 24

Earth 10

4

15 1/3

(238 u)(1.6605 10 kg/u)

(6.38 10 m) 2.58 10 m

(5.98 10 kg) 2.58 10 m 3.5 10

(1.2 10 )(238)

M m M m

R r R r

m r R M ρ π π − − − − = = → = → ⎡ ⎤ ⎛ ⎞ × = ⎜ ⎟ = × ⎢ ⎥ = × × ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ × = × ×

8. Use Eq. 30–1 to find the value for A. We use uranium-238 since it is the most common isotope.

15 1/3 3 unknown 15 1/3 U

(1.2 10 m)

0.5 238(0.5) 29.75 30 (1.2 10 m)(238)

r A A r − − × = = → = = ≈ ×

From Appendix B, a stable nucleus with A≈30 is 1531P .

9. Use conservation of energy. Assume that the centers of the two particles are located a distance from each other equal to the sum of their radii, and use that distance to calculate the initial electrical PE. Then we also assume, since the nucleus is much heavier than the alpha, that the alpha has all of the final KE when the particles are far apart from each other (so they have no PE).

Fm

KEi PEi KEf PEf KE

0 Fm

19 2

9 2 2 7

KE

1/3 1/3 15 19

7

1

0 0

4 ( )

(2)(100)(1.60 10 C)

(8.99 10 N m /C ) 3.017 10 eV

(4 257 )(1.2 10 m)(1.60 10 J/eV) 3.0 10 eV 30 MeV

q q r r α α α α πε − − − + = + → + = + → + × = × ⋅ = × + × × ≈ × =

10. (a) The hydrogen atom is made of a proton and an electron. Use values from Appendix B.

p

H

1 007276 u 0.9994553 99.95% 1 007825 u

m m

.

= = ≈

.

(b) Compare the volume of the nucleus to the volume of the atom. The nuclear radius is given by Eq. 30–1. For the atomic radius we use the Bohr radius, given in Eq. 27–13.

3 3 3 4 15 nucleus 14

nucleus 3 nucleus

3 10

4

atom 3 atom atom

(1.2 10 m) 1.2 10 (0.53 10 m)

r

V r

V r r

π π − − − ⎡ ⎤ ⎛ ⎞ × = =⎜ ⎟ =⎢ ⎥ = × × ⎢ ⎥ ⎝ ⎠ ⎣ ⎦

11. Electron mass is negligible compared to nucleon mass, and one nucleon weighs about 1.0 atomic mass unit. Therefore, in a 1.0-kg object, we find the following:

26

26 (1.0 kg)(6.02 10 u/kg)

6.0 10 nucleons 1.0 u/nucleon

N= × ≈ ×

(9)

12. The initial kinetic energy of the alpha must be equal to the electrical potential energy when the alpha just touches the uranium. The distance between the two particles is the sum of their radii.

U KEi PEi KEf PEf KE

U

19 2

9 2 2 7

KE

1/3 1/3 15 19

0 0

( )

(2)(92)(1 60 10 C)

(8.99 10 N m /C ) 2 852 10 eV

(4 232 )(1.2 10 m)(1.60 10 J/eV) 29 MeV

Q Q k

r r α α

α

α

− −

+ = + → + = + →

+ . ×

= × ⋅ = . ×

+ × ×

13. The text states that the average binding energy per nucleon between A≈40 and A≈80 is about 8.7 MeV. Multiply this by the number of nucleons in the nucleus.

(63)(8.7 MeV) 548.1 MeV= ≈ 550 MeV

14. (a) From Fig. 30–1, we see that the average binding energy per nucleon at A=238 is 7.5 MeV. Multiply this by the 238 nucleons.

(238)(7.5 MeV)=1.8 10 MeV× 3

(b) From Fig. 30–1, we see that the average binding energy per nucleon at A=84 is 8 7 MeV. .

Multiply this by the 84 nucleons. (84)(8.7 MeV)= 730 MeV

15. 157N consists of seven protons and eight neutrons. We find the binding energy from the masses of the components and the mass of the nucleus, from Appendix B.

( ) ( ) ( )

1 1 15 2

1 0 7

2 2

Binding energy 7 H 8 n N

931.5 MeV/ [7(1.007825 u) 8(1.008665 u) (15.000109 u)]

u 115.49 MeV

Binding energy per nucleon (115.49 MeV)/15 7.699 MeV

m m m c

c c

⎡ ⎤

= + −

⎛ ⎞

= + − ⎜

⎝ ⎠

=

= =

16. Deuterium consists of one proton, one neutron, and one electron. Ordinary hydrogen consists of one proton and one electron. We use the atomic masses from Appendix B, and the electron masses cancel.

( ) ( ) ( )

1 1 2 2

1 0 1

2 2

Binding energy H n H

931.5 MeV/ [(1 007825 u) (1 008665 u) (2 014102 u)]

u 2 224 MeV

m m m c

c c

⎡ ⎤

= + −

⎛ ⎞

= . + . − . ⎜

⎝ ⎠

(10)

17. We find the binding energy of the last neutron from the masses of the isotopes.

( ) ( ) ( )

22 1 23 2

11 0 11

2 2

Binding energy Na n Na

[(21 994437 u) (1 008665 u) (22 989769 u)] (931.5 MeV/ ) 12.42 MeV

m m m c

c c

⎡ ⎤

= + −

= . + . − .

=

18. (a) 73Li consists of three protons and three neutrons. We find the binding energy from the masses, using hydrogen atoms in place of protons so that we account for the mass of the electrons.

( ) ( ) ( )

1 1 7 2

1 0 3

2 2

Binding energy 3 H 4 n Li

931 5 MeV/ [3(1 007825 u) 4(1 008665 u) (7 016003 u)]

u 39 25 MeV

Binding energy 39 25 MeV

5 607 MeV/nucleon nucleon 7 nucleons

m m m c

c c

⎡ ⎤

= + −

.

= . + . − . ⎜

⎝ ⎠

= . .

= = .

(b) 19578Pt consists of 78 protons and 117 neutrons. We find the binding energy as in part (a).

( )

1

( ) ( )

1 195 2

1 0 78

2 2

Binding energy 78 H 117 n Pt

931.5 MeV/ [78(1.007825 u) 117(1.008665 u) (194.964792 u)]

u 1545.7 MeV 1546 MeV

Binding energy 1545.7 MeV

7.927 MeV/nucleon nucleon 195 nucleons

m m m c

c c

⎡ ⎤

= + −

⎛ ⎞

= + − ⎜

⎝ ⎠

= ≈

= =

19. 2311Na consists of 11 protons and 12 neutrons. We find the binding energy from the masses.

( )

1

( ) ( )

1 23 2

1 0 11

2 2

Binding energy 11 H 12 n Na

931.5 MeV/ [11(1.007825 u) 12(1.008665 u) (22.989769 u)]

u 186.6 MeV

Binding energy 186.6 MeV

8.113 MeV/nucleon nucleon 23

m m m c

c c

⎡ ⎤

= + −

⎛ ⎞

= + − ⎜

⎝ ⎠

=

= =

We do a similar calculation for 2411Na, consisting of 11 protons and 13 neutrons.

( )

1

( ) ( )

1 24 2

1 0 11

2 2

Binding energy 11 H 13 n Na

931.5 MeV/ [11(1.007825 u) 13(1.008665 u) (23.990963 u)]

u 193.5 MeV

Binding energy 193.5 MeV 8.063 MeV/nucleon nucleon 24

m m m c

c c

⎡ ⎤

= + −

⎛ ⎞

= + − ⎜

⎝ ⎠

=

(11)

By this measure, the nucleons in 2311Na are more tightly bound than those in 2411Na. Thus we expect 23

11Na to be more stable than 2411Na.

20. We find the required energy by calculating the difference in the masses.

(a) Removal of a proton creates an isotope of carbon. To balance electrons, the proton is included as a hydrogen atom: 157N → 11H+146C.

( ) ( ) ( )

14 1 15 2

6 1 7

2

Energy needed C H N

931 5 MeV/ [(14 003242 u) (1 007825 u) (15 000109 u)]

u 10.21 MeV

m m m c

c

⎡ ⎤

= + −

.

= . + . − . ⎜

⎝ ⎠

=

(b) Removal of a neutron creates another isotope of nitrogen: 157N → 01n+147N.

( ) ( ) ( )

14 1 15 2

7 0 7

2

Energy needed N n N

931.5 MeV/ [(14 003074 u) (1.008665 u) (15.000109 u)]

u 10.83 MeV

m m m c

c

⎡ ⎤

= + −

⎛ ⎞

= . + − ⎜

⎝ ⎠

=

The nucleons are held by the attractive strong nuclear force. It takes less energy to remove the proton because there is also the repulsive electric force from the other protons.

21. (a) We find the binding energy from the masses.

( ) ( )

4 8 2

2 4

2 2

Binding energy 2 He Be

931.5 MeV/ [2(4.002603 u) (8.005305 u)]

u 0.092 MeV

m m c

c c

⎡ ⎤

=

⎛ ⎞

= − ⎜

⎝ ⎠

= −

Because the binding energy is negative, the nucleus is unstable. It will be in a lower energy state as two alphas instead of a beryllium.

(b) We find the binding energy from the masses.

( ) ( )

4 12 2

2 6

2 2

Binding energy 3 He C

931.5 MeV/

[3(4 002603 u) (12 000000 u)] 7.3 MeV u

m m c

c c

⎡ ⎤

=

⎛ ⎞

= . − . ⎜= +

⎝ ⎠

Because the binding energy is positive, the nucleus is stable. 22. The wavelength is determined from the energy change between the states.

34 8

12 13

(6.63 10 J s)(3.00 10 m/s)

2.6 10 m (0 48 MeV)(1 60 10 J/MeV)

c hc

E hf h

E

λ λ

− −

× ⋅ ×

Δ = = → = = = ×

(12)

23. For the decay 116C → 105B+11p, we find the difference of the initial and the final masses. We use hydrogen so that the electrons are balanced.

( ) ( ) ( )

11 10 1

6C 5B 1H

(11 011434 u) (10 012937 u) (1 007825 u) 0.009328 u

m m m m

Δ = − −

= . − . − . = −

Since the final masses are more than the original mass, energy would not be conserved.

24. The decay is 13H → 32He+01e+v. When we add one electron to both sides to use atomic masses, we see that the mass of the emitted β particle is included in the atomic mass of 23He. The energy released is the difference in the masses.

( ) ( )

3 3 2

1 2

2 2

Energy released H He

931.5 MeV/

[(3 016049 u) (3 016029 u)] 0 019 MeV u

m m c

c c

⎡ ⎤

=

⎛ ⎞

= . − . ⎜= .

⎝ ⎠

25. The decay is 01n → 11p+−01e+v. The electron mass is accounted for if we use the atomic mass of 11H as a product. If we ignore the recoil of the proton and the neutrino, and any possible mass of the neutrino, then we get the maximum kinetic energy.

( ) ( )

2

1 1 2 2

KEmax 0n 1H [(1 008665 u) (1 007825 u)] 931.5 MeV/ u 0.782 MeV

c

m m c c ⎛ ⎞

⎡ ⎤

= = . − . ⎜

⎝ ⎠

=

26. For each decay, we find the difference of the final masses and the initial mass. If the final mass is more than the initial mass, then the decay is not possible.

(a) Δ =m m

( ) ( ) ( )

23292U +m 10n −m 23292U =232 037156 u 1 008665 u 233 039636 u 0 006185 u. + . − . = . Because an increase in mass is required, the decay is not possible.

(b) Δ =m m

( ) ( ) ( )

137N +m 10n −m 137N = .13 005739 u 1.008665 u 14.003074 u 0 011330 u+ − = .

Because an increase in mass is required, the decay is not possible.

(c) Δ =m m

( ) ( ) ( )

1939K +m 10n −m 1940K =38.963706 u 1.008665 u 39.963998 u 0.008373 u+ − = Because an increase in mass is required, the decay is not possible.

27. (a) From Appendix B, 2411Na is a β−emitter .

(13)

( ) (

24 24

)

2

KE 11 12

2 2

Na Mg

931.5 MeV/

[(23 990963 u) (23 985042 u)] 5.515 MeV u

m m c

c c

β− =⎡ − ⎤

⎛ ⎞

= . − . ⎜=

⎝ ⎠

28. (a) We find the final nucleus by balancing the mass and charge numbers.

( ) (U) (He) 92 2 90 ( ) (U) (He) 238 4 234

Z X Z Z

A X A A

= − = − =

= − = − =

Thus the final nucleus is 23490Th.

(b) If we ignore the recoil of the thorium, then the kinetic energy of the α particle is equal to the difference in the mass energies of the components of the reaction. The electrons are balanced.

( ) (

) ( )

(

) ( ) ( )

238 234 4 2

KE 92 90 2

KE

234 238 4

90 92 2 2

2 2

U Th He

Th U He

4 20 MeV 1 u 238 050788 u 4 002603 u

931 5 MeV/ 234.043676 u

m m m c

m m m

c

c c

⎡ ⎤

= − −

= − −

. ⎛ ⎞⎤

=⎢ . − . − ⎜

.

⎢ ⎝ ⎠⎥

⎣ ⎦

=

This answer assumes that the 4.20-MeV value does not limit the significant figures of the answer.

29. The reaction is 6027Co→6028Ni+β−+v. The kinetic energy of the β− will be maximum if the (essentially) massless neutrino has no kinetic energy. We also ignore the recoil of the nickel.

( ) ( )

60 60 2

KE 27 28

2 2

Co Ni

931.5 MeV/

[(59 933816 u) (59 930786 u)] 2.822 MeV u

m m c

c c

β =⎡ − ⎤

⎛ ⎞

= . − . ⎜=

⎝ ⎠

30. We add three electron masses to each side of the reaction 74Be+01e→37Li+v. Then for the mass of the product side, we may use the atomic mass of 37Li. For the reactant side, including the three electron masses and the mass of the emitted electron, we may use the atomic mass of 74Be. The energy released is the Q-value.

( ) ( )

7 7 2

4 3

2 2

Be Li

931.5 MeV/

[(7 016929 u) (7 016003 u)] 0 863 MeV u

Q m m c

c c

⎡ ⎤

=

⎛ ⎞

= . − . ⎜= .

(14)

31. For alpha decay we have 21884Po→21482Pb+42He. We find the Q-value, which is the energy released.

(

218

) (

214

) ( )

4 2

84 82 2

2 2

Po Pb He

931 5 MeV/ [218 008973 u 213 999806 u 4 002603 u]

u 6.114 MeV

Q m m m c

c c

⎡ ⎤

= − −

.

= . − . − . ⎜

⎝ ⎠

=

For beta decay we have 21884Po→21882At+−01e+v. We assume that the neutrino is massless, and find the Q-value. The original 84 electrons plus the extra electron created in the beta decay means that there are 85 total electrons on the right side of the reaction, so we can use the mass of the astatine atom and “automatically” include the mass of the beta decay electron.

(

218

) (

218

)

2

84 85

2 2

Po At

931 5 MeV/

[218 008973 u 218 008695 u] 0 259 MeV u

Q m m c

c c

⎡ ⎤

=

.

= . − . ⎜= .

⎝ ⎠

32. (a) We find the final nucleus by balancing the mass and charge numbers.

( ) (P) (e) 15 ( 1) 16 ( ) (P) (e) 32 0 32

Z X Z Z

A X A A

= − = − − =

= − = − =

Thus the final nucleus is 1632S .

(b) If we ignore the recoil of the sulfur and the energy of the neutrino, then the maximum kinetic energy of the electron is the Q-value of the reaction. The reaction is 3215P→1632S+β−+v. We add 15 electrons to each side of the reaction, and then we may use atomic masses. The mass of the emitted beta is accounted for in the mass of the sulfur.

( ) ( )

( ) ( )

32 32 2

KE 15 16

KE

32 32

16 15 2 2 2

P S

1.71 MeV 1 u

S P (31.973908 u) 31.97207 u

931.5 MeV/

m m c

m m

c c c

⎡ ⎤

=

⎡ ⎤

= − =⎢ − ⎥=

⎢ ⎥

⎣ ⎦

33. We find the energy from the wavelength.

34 8

13 13

(6 63 10 J s)(3.00 10 m/s)

10.8 MeV (1.15 10 m)(1.602 10 J/MeV)

hc E

λ

− −

. × ⋅ ×

= = =

× ×

This has to be a γ ray from the nucleus rather than a photon from the atom. Electron transitions do not involve this much energy. Electron transitions involve energies on the order of a few eV.

(15)

KE

K K K

2 2

5

KEK 2 2

2 K

2

(1 46 MeV) 2.86 10 MeV 28.6 eV

2 931.5 MeV/

2(39.963998 u)

u

E

p p m

c E

m c c

c

γ γ

γ −

= = = →

.

= = = × =

⎛ ⎞

⎜ ⎟

⎜ ⎟

⎝ ⎠

35. The kinetic energy of the β+ particle will be its maximum if the (almost massless) neutrino has no kinetic energy. We ignore the recoil of the boron. Note that if the mass of one electron is added to the mass of the boron, then we may use atomic masses. We also must include the mass of the β+. (See Problem 36 for details.)

(

)

( ) ( ) ( ) ( )

( ) ( ) ( )

11 11 0 0

6 5 1 1

11 11 0 0 2 11 11 0 2

KE 6 5 1 1 6 5 1

2 2

C B e

C B e C B 2 e

931 5 MeV/

[(11 011434 u) (11 009305 u) 2(0 00054858 u)] 0 9612 MeV u

v

m m m m c m m m c

c c

β

β

+ −

+

− −

→ + + +

⎡ ⎤ ⎡ ⎤

= − − − = − −

.

= . − . − . ⎜= .

⎝ ⎠

If the β+ has no kinetic energy, then the maximum kinetic energy of the neutrino is also 0.9612 MeV . The minimum energy of each is 0, when the other has the maximum.

36. For the positron emission process, ZAN→ZA1N′ +e++v. We must add Z electrons to the nuclear mass of N to be able to use the atomic mass, so we must also add Z electrons to the reactant side. On the reactant side, we use Z−1 electrons to be able to use the atomic mass of N .′ Thus we have 1 “extra” electron mass and the β particle mass, which means that we must include 2 electron masses on the right-hand side. We find the Q-value given this constraint.

Q=[MP−(MD+2m ce)] 2=(MP−MD−2m ce) 2

37. We assume that the energies are low enough that we may use classical kinematics. In particular, we will use p= 2mKE. The decay is 23892U→23490Th+42He. If the uranium nucleus is at rest when it decays, then the magnitude of the momentum of the two daughter particles must be the same.

2 2 Th

KE KE

Th Th

Th Th Th Th

2 4 u

; (4.20 MeV) 0.0718 MeV

2 2 2 234 u

p m K m

p

p p

m m m m

α α α α

α = = = = = α =⎛⎜ ⎞⎟ =

⎝ ⎠

The Q-value is the total kinetic energy produced.

(16)

38. (a) The decay constant can be found from the half-life, using Eq. 30–6.

9 10 1 18 1

1/2

ln 2 ln 2 1.5 10 yr 4.9 10 s 4.5 10 yr

T

λ= = = × − − = × − −

×

(b) The half-life can be found from the decay constant, using Eq. 30–6.

1/2 ln 2 ln 25 1 21,660 s 6.0 h 3.2 10 s

T

λ − −

= = = =

×

39. We find the half-life from Eq. 30–5 and Eq. 30–6.

1/ 2

ln 2

0 0 1/2

0

ln 2 ln 2

(3.6 h) 1.2 h 140

ln ln

1120

t T t

R R e R e T t

R R λ

= = 2 → = − = − =

40. We use Eq. 30–4 and Eq. 30–6 to find the fraction remaining.

1/ 2

(ln 2)(2 5 yr)(12 mo/yr) ln 2

9 mo 0

0

0.0992 0.1 10%

t T

t N t

N N e e e e

N

λ λ

⎡ . ⎤

− ⎢ ⎥

− − ⎣ ⎦

= → = = = = ≈ ≈

Only 1 significant figure was kept since the Problem said “about” 9 months.

41. The activity at a given time is given by Eq. 30–3b. We also use Eq. 30–6. The half-life is found in Appendix B.

7 20 9

1/2

ln 2 ln 2 (6.5 10 nuclei) 2.5 10 decays/s (5730 yr)(3.16 10 s/yr)

dN N N

dt =λ =T = × × = ×

42. For every half-life, the sample is multiplied by one-half.

( ) ( )

1 1 5

2 2

0

0.03125 1/32 3.125%

n

N

N = = = = =

43. We need the decay constant and the initial number of nuclei. The half-life is found in Appendix B. Use Eq. 30–6 to find the decay constant.

7 1

1/2

ln 2 ln 2 9.99668 10 s

(8.0252 days)(24 h/day)(3600 s/h) T

λ= = = × − −

6

23 18

0 (782 10 g) (6.02 10 atoms/mol) 3.5962 10 nuclei (130.906 g/mol)

N =⎡⎢ × − ⎤⎥ × = ×

⎢ ⎥

⎣ ⎦

(a) We use Eq. 30–3b and Eq. 30–4 to evaluate the initial activity.

7 1 18 0 12

0 12

Activity (9.99668 10 s )(3.5962 10 ) 3.5950 10 decays/s 3.60 10 decays/s

t

N N e λ e

λ λ − − − −

= = = × × = ×

(17)

(b) We evaluate Eq. 30–5 at t=1.5 h.

7 1 3600 s (9 99668 10 s )(1 5 h)

12 h

0 12

(3.5950 10 decays/s) 3.58 10 decays/s

t

R R e λ e

− − ⎛ ⎞

− . × . ⎜ ⎟

− ⎝ ⎠

= = ×

≈ ×

(c) We evaluate Eq. 30–5 at t = 3.0 months. We use a time of 1/4 year for the 3.0 months.

7 1 7

12 (9 99668 10 s )(0 25yr)(3 156 10 s/yr) 0

9 (3.5950 10 decays/s) 1.3497 decays/s 1.34 10 decays/s

t

R R e= −λ = × e− . × − − . . ×

= ≈ ×

44. We find the number of nuclei from the activity of the sample, using Eq. 30–3b and Eq. 30–6. The half-life is found in Appendix B.

1/2 9 7

19 1/2

ln 2

(4.468 10 yr)(3.156 10 s/yr)

(420 decays/s) 8.5 10 nuclei

ln 2 ln 2

N

N N

t T

T N

N

t

λ

Δ = = →

Δ

Δ × ×

= = = ×

Δ

45. Each α emission decreases the mass number by 4 and the atomic number by 2. The mass number changes from 235 to 207, a change of 28. Thus there must be 7αparticles emitted. With the 7 α emissions, the atomic number would have changed from 92 to 78. Each β− emission increases the atomic number by 1, so to have a final atomic number of 82, there must be 4β− particles emitted.

46. We use the decay constant often, so we calculate it here: 1 1/2

ln 2 ln 2

0.022505 s . 30.8 s

T

λ= = =

(a) We find the initial number of nuclei from an estimate of the atomic mass.

6

23 16 16

0 (8.7 10(124 g/mol)g)(6.02 10 atoms/mol) 4.223 10 4.2 10 nuclei

N = × − × = × ≈ ×

(b) Evaluate Eq. 30–4 at t=2.6 min .

N N e0 λt (4.223 10 )16 e (0 022505 s )(2 6 min)(60 s/ min)1 1.262 1015 1.3 10 nuclei15

− − . .

= = × = × ≈ ×

(c) The activity is found from using the absolute value of Eq. 30–3b.

λN=(0.022505 s )(1.262 10 ) 2.840 10 decays/s−1 × 15 = × 13 ≈ 2.8 10 decays/s× 13 (d) We find the time from Eq. 30–4.

0

1 16

0

1 1 decay/s ln

ln

(0 022505 s )(4 223 10 ) decays/s

1532 s 26 min 0.022505 s

t

N N e

N N t

λ

λ λ

λ λ

λ

= →

⎡ ⎤

⎛ ⎞

⎢ ⎥

⎜ ⎟

. . ×

⎢ ⎥

⎝ ⎠ ⎣ ⎦

(18)

47. Find the initial number of nuclei from the initial decay rate (activity) and then the mass from the number of nuclei.

5

0 0

1/2

9 7

5 1 5 1 22

1/2 0

22

0 23

ln 2 initial decay rate 2.0 10 decays/s

(1.248 10 yr)(3.156 10 s/yr)

(2.4 10 s ) (2.4 10 s ) 1.364 10 nuclei

ln 2 ln 2

(atomic weight) g/mol (39. (1.364 10 nuclei) 6.02 10 nuclei/mol

N N T T N m N λ − − = × = = → × × = × = × = × = = × × 23 963998 g) 0.91 g (6.02 10 )× =

48. The number of nuclei is found from the mass and the atomic weight. The activity is then found from the number of nuclei and the half-life.

6 23 17 17 10 6 1/2

(6.7 10 g)

(6.02 10 atoms/mol) 1.261 10 nuclei (31.9739 g/mol)

ln 2 ln 2 (1.261 10 ) 7.1 10 decays/s (1.23 10 s)

N N N T λ − ⎡ × ⎤ =⎢ ⎥ × = × ⎢ ⎥ ⎣ ⎦ ⎡ ⎤ = =⎢ ⎥ × = × × ⎢ ⎥ ⎣ ⎦

49. (a) The decay constant is found from Eq. 30–6.

5 7 13 1 13 1

1/2

ln 2 ln 2

1.381 10 s 1.38 10 s (1.59 10 yr)(3.156 10 s/yr)

T

λ= = = × − − × − −

× ×

(b) The activity is the decay constant times the number of nuclei.

13 1 18 5

7

60 s (1.381 10 s )(4.50 10 ) 6.215 10 decays/s

1 min 3.73 10 decays/ min

N

λ = × − − × = ×

⎝ ⎠

= ×

50. We use Eq. 30–5.

( )

( )

( )

1 1 1

0 0

6 6 6

1/2

1/2 1 1

6 6

ln 2 ln

ln 2 ln 2

(9.4 min) 3.6 min

ln ln

t t

R R R e e t t

T

T t

λ λ λ

− −

= = → = → = − = →

= = − =

2

2

51. Because the fraction of atoms that are 146C is so small, we use the atomic weight of 126C to find the number of carbon atoms in the sample. The activity is found from Eq. 30–3b.

14 12 6 6 23 13 12 12 13 7 1/2

1.3 1.3 (345 g)

(6.02 10 nuclei/mol) 2.250 10 nuclei C 10 C 10 (12 g/mol)

ln 2 ln 2

(2.250 10 ) 86 decays/s (5730 yr)(3.156 10 s/yr)

(19)

52. We first find the number of nuclei from the activity and then find the mass from the number of nuclei and the atomic weight. The half-life is found in Appendix B.

2 1/2

9 7 2

19 ln 2

Activity 4.20 10 decays/s

(4.468 10 yr)(3.156 10 s/yr)(4.20 10 decays/s) 8.544 10 nuclei ln 2

N N

T

N

λ

= = = × →

× × ×

= = ×

19

2 23

(8.544 10 nuclei)

(238.05 g/mol) 3.38 10 g 33.8 mg (6.02 10 atoms/mol)

m=⎡⎢ × ⎤⎥ = × − =

×

⎢ ⎥

⎣ ⎦

53. We assume that the elapsed time is much smaller than the half-life, so we can approximate the decay rate as being constant. We also assume that the 8738Sr is stable, and there was none present when the rocks were formed. Thus every atom of 8737Rb that decayed is now an atom of 8738Sr.

10

9 Sr 1/2

Sr Rb Rb

Rb

4.75 10 yr

(0.0260) 1.78 10 yr

ln 2 ln 2

N T

N N N t t

N

λ ×

= −Δ = Δ → Δ = = = ×

This is ≈4% of the half-life, so our original assumption is valid.

54. We are not including the neutrinos that will be emitted during beta decay. First sequence: 23290Th → 22888Ra +42α; 228 228

88Ra → 89Ac +β−; 22889Ac → 22890Th +β−; 22890Th → 22488Ra +42α; 224 220 4

88Ra → 86Rn +2α

Second sequence: 23592U → 23190Th +42α; 23190Th 23191Pa +β−; 231 227 4 91Pa → 89Ac +2α; 22789Ac 22790Th +β−; 227 223 4

90Th → 88Ra +2α

55. Because the fraction of atoms that are 146C is so small, we use the atomic weight of 126C to find the number of carbon atoms in 73 g. We then use the given ratio to find the number of 146C atoms present when the club was made. Finally, we use the activity as given in Eq. 30–5 to find the age of the club.

(

)

(

)

12 6

14 6

14 14

6 6

23 24

12 24 12

today 0

(73 g) (6.02 10 atoms/mol) 3.662 10 atoms C (12 g/mol)

(1.3 10 )(3.662 10 ) 4.761 10 nuclei C

C C t

N

N

N N e λ

λ λ

⎡ ⎤

=⎢ ⎥ × = ×

⎣ ⎦

= × × = ×

= →

(

)

(

)

(

)

14 14

6 6

14

14 6

6

today 1/2 today

0

1/2 0

12 7

0

C C

1

ln ln

ln 2 ln 2

C C

5730 yr (7.0 decays/s)

ln 7921 yr 7900 yr

ln 2 ln 2

(4.761 10 nuclei) (5730 yr)(3.156 10 s/yr)

N N

T t

N N

T

λ λ

λ λ

= − = −

⎛ ⎞

⎜ ⎟

⎝ ⎠

= − = ≈

⎡ ⎤

×

⎢ ⎥

×

⎢ ⎥

(20)

56. The decay rate is given by Eq. 30–3b, N N.

t λ

Δ = −

Δ We assume equal numbers of nuclei decaying by α emission.

1/2 4

214 7

218 218 218 218 214 214 214 1/2

218 214

(1.6 10 s)

8.6 10 (3.1 min)(60 s/ min)

N T

t N

N N T

t

λ λ

λ λ

Δ

⎛ ⎞

Δ

×

⎝ ⎠ = = = = = ×

Δ −

⎛ ⎞

Δ

⎝ ⎠

57. The activity is given by Eq. 30–5. The original activity is λN0, so the activity 31.0 hours later is 0

0.945λN .

0 0

1/2 1/2

ln 2

0.945 ln 0.945

ln 2 1 d

(31.0 h) 379.84 h 15.8 d

ln 0.945 24 h

t

N N e t t

T

T

λ

λ =λ − = − = −λ

⎛ ⎞

= − = ⎜ ⎟=

⎝ ⎠

58. The activity is given by Eq. 30–5.

(a) 0 1/2 2

0 0

1ln ln 53 d ln 25 decays/s 201.79 d 2.0 10 d ln 2 ln 2 350 decays/s

t R T R

R R e t

R R

λ

λ

− ⎛ ⎞

= → = − = − = − ⎜ ⎟= ≈ ×

⎝ ⎠

(b) We find the mass from the activity. Note that NA is used to represent Avogadro’s number, and A is the atomic weight.

0 A

0 0

1/2

14 0 1/2

23 A

ln 2

(350 decays/s)(53 d)(86, 400 s/d)(7.017 g/mole) 2.7 10 g ln 2 (6.02 10 nuclei/mole) ln 2

m N

R N

T A

R T A m

N

λ

= = →

= = = ×

×

59. The number of radioactive nuclei decreases exponentially, and every radioactive nucleus that decays becomes a daughter nucleus.

N=N e0 −λt; ND =N0− =N N0(1−e−λt) 60. The activity is given by Eq. 30–5, with R= .0 01050 .R0

0

0 1/2

1/2 0

ln / ln 2 ln 2 (4.00 h) ln 2

0 6085 h 36 5 min ln / ln 0.01050

t R R t

R R e T

t T R R

λ λ

= → = − = → = − = − = . = .

From Appendix B we see that the isotope is 21182Pb .

61. Because the carbon is being replenished in living trees, we assume that the amount of 146C is constant until the wood is cut, and then it decays. Use Eq. 30–4 and Eq. 30–7 to find the age of the tool.

1/ 2 0 693

4 1/2

0 0 0.045 0 0.693ln(0.045) (5730 yr) ln (0.045)0.693 2.6 10 yr

t T

t T

N N e λ N e N t

. − −

(21)

62. (a) The mass number is found from the radius, using Eq. 30–1.

3 3

15 1/3 55 55

15 15

5000 m

(1 2 10 m) 7.23 10 7 10

1.2 10 m 1.2 10 m

r

r= . × − AA=⎛⎜ ⎟ =⎜ ⎟⎞ = × ≈ ×

× ×

⎝ ⎠ ⎝ ⎠

(b) The mass of the neutron star is the mass number times the atomic mass unit conversion in kg.

27 55 27 29 29

(1 66 10 kg/u) (7.23 10 u)(1.66 10 kg/u) 1.20 10 kg 1 10 kg

m=A . × − = × × − = × ≈ ×

Note that this is about 6% of the mass of the Sun.

(c) The acceleration of gravity on the surface of the neutron star is found from Eq. 5–5 applied to the neutron star.

11 2 2 29

11 2 11 2

2 2

(6.67 10 N m /kg )(1.20 10 kg) 3.20 10 m/s 3 10 m/s (5000 m)

Gm g

r

× ⋅ ×

= = = × ≈ ×

63. Because the tritium in water is being replenished, we assume that the amount is constant until the wine is made, and then it decays. We use Eq. 30–4.

0 0 1/2

1/2 0

ln

ln 2 (12.3 yr) ln 0.10

ln 41 yr

ln 2 ln 2

t

N

N T N

N N e t

t T N

λ λ

= → = − = → = − = − =

64. (a) We assume a mass of 70 kg of water and find the number of protons, given that there are 10 protons in a water molecule.

3

23 protons

28

(70 10 g water) molecules water 10 protons 6 02 10

(18 g water/mol water) mol water water molecule 2.34 10 protons

N =⎡⎢ × ⎤⎥ ⎜⎛ . × ⎞⎛⎟⎜⎠⎝ ⎞⎟

⎢ ⎥

⎣ ⎦

= ×

We assume that the time is much less than the half-life so that the rate of decay is constant.

1/2 33

4 1/2

28 ln 2

1 proton 10 yr 6.165 10 yr 60, 000 yr ln 2 2.34 10 protons ln 2

N

N N

t T

T N t

N

λ ⎛ ⎞

Δ = =⎜ ⎟ →

Δ ⎝ ⎠

⎛ ⎞

⎛ ⎞

Δ

Δ = ⎜ ⎟= ⎜= × ≈

×

⎝ ⎠ ⎝ ⎠

This is about 880 times a normal life expectancy.

(b) Instead of waiting 61,650 consecutive years for one person to experience a proton decay, we could interpret that number as there being 880 people, each living 70 years, to make that 61,650 years (since 880 70 61,650).× ≈ We would then expect one person out of every 880 to experience a proton decay during their lifetime. Divide 7 billion by 880 to find out how many people on Earth would experience proton decay during their lifetime.

7 109 7.95 106 8 million people 880

(22)

65. We assume that all of the kinetic energy of the alpha particle becomes electrostatic potential energy at the distance of closest approach. Note that the distance found is the distance from the center of the alpha to the center of the gold nucleus.

Au KEi PEi KEf PEf KE

19 2

9 2 2 14

Au

13 KE

14

0 0

(2)(79)(1.60 10 C)

(8.988 10 N m /C ) 2.951 10 m

(7.7 MeV)(1.60 10 J/MeV) 3.0 10 m

Q Q k

r Q Q

r k

α α

α α

− −

+ = + → + = + →

×

= = × ⋅ = ×

× ≈ ×

We use Eq. 30–1 to compare to the size of the gold nucleus.

approach 1/3 1415

Au

2.951 10 m 4.2 197 (1.2 10 m)

r r

− − ×

= =

×

So the distance of approach is about 4.2× the radius of the gold nucleus.

66. Use Eq. 30–5 and Eq. 30–6.

1/ 2

ln 2

0 0 0

1/2

ln (0 0200)

0.0200 5.64

ln 2

t T

t t

R R e R e R

T

λ −

− .

= = = → = − =

It takes 5.64 half-lives for a sample to drop to 2.00% of its original activity.

67. The number of 4019K nuclei can be calculated from the activity, using Eq. 30–3b and Eq. 30–5. The half-life is found in Appendix B. A subscript is used on the variable N to indicate the isotope.

7 9

18 1/2

40 40

decays 3.156 10 s

42 (1.248 10 yr)

s 1 yr

2.387 10 nuclei

ln 2 ln 2

RT R

R λN N

λ

×

⎛ ⎞ ×

⎜ ⎟

⎝ ⎠

= → = = = = ×

The mass of 4019K in the milk is found from the atomic mass and the number of nuclei. m40=(2.387 10 )(39.964 u)(1.66 10× 18 × −27 kg/u) 1.583 10= × −7kg≈ 0.16 mg

From Appendix B, in a natural sample of potassium, 0.0117% is 1940K and 93.2581% is 1939K. Find the number of 1939K nuclei and then use the atomic mass to find the mass of 1939K in the milk.

40 total 39 total

18 22

39 40

(0.0117%) ; (93 2581%) (93.2581%) (93.2581%)

(2.387 10 nuclei) 1.903 10 nuclei (0.0117%) (0.0117%)

N N N N

N N

= = . →

⎡ ⎤ ⎡ ⎤

=⎢ ⎥ =⎢ ⎥ × = ×

⎣ ⎦ ⎣ ⎦

The mass of 1939K is the number of nuclei times the atomic mass.

22 27 3

39 (1.903 10 )(38.964 u)(1.66 10 kg/u) 1.231 10 kg 1.2 g

(23)

68. In the Periodic Table, Sr is in the same column as Ca. If Sr is ingested, then the body may treat it chemically as if it were Ca, which means it might be stored by the body in bones. Use Eq. 30–4 to find the time to reach a 1% level.

0 1/2

0

1/2 0

ln

ln 2 (29 yr) ln 0.01

ln 192.67 yr 200 yr

ln 2 ln 2

t

N

N T N

N N e t

t T N

λ λ

= → = − = → =2 = − = ≈

Assume that both the Sr and its daughter undergo beta decay, since the Sr has too many neutrons. 3890Sr → 9039Y+01e+ν; 3990Y → 4090Zr+01e+ν

69. The number of nuclei is found from Eq. 30–5 and Eq. 30–3b. The mass is then found from the number of nuclei and the atomic weight. The half-life is given.

1/2 1/2

1/2

4 27

14 11

ln 2

; (atomic weight) (atomic weight)

ln 2 ln 2

24 h 3600 s (87.37 d)

1 d 1 h

(4.28 10 decays/s)(34.969 u)(1.66 10 kg/u) ln 2

2.71 10 kg 2.71 10 g

T T

R N N N R m N R

T

m

λ

− −

= = → = = =

⎛ ⎞⎛ ⎞

⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

= × ×

= × = ×

70. (a) We find the daughter nucleus by balancing the mass and charge numbers. ( ) (Os) (e ) 76 ( 1) 77

Z X =ZZ − = − − = ( ) (Os) (e ) 191 0 191

A X = AA − = − =

The daughter nucleus is 19177Ir . (b) See the diagram.

(c) Because there is only one β energy, the

β decay must be to the higher excited state.

71. (a) The number of nuclei is found from the mass of the sample and the atomic mass. The activity is found from the half-life and the number of nuclei, using Eq. 30–3b and Eq. 30–5.

23 21

21 15

4 1.0 g

(6.02 10 nuclei/mol) 4.599 10 nuclei 130.91 g/mol

0.693

(4.599 10 ) 4.6 10 decays/s (8.02 d)(8.64 10 s/d)

N

R λN

⎛ ⎞

=⎜ ⎟ × = ×

⎝ ⎠

⎡ ⎤

= =⎢ ⎥ × = ×

×

⎢ ⎥

⎣ ⎦

(b) Follow the same procedure as in part (a).

23 21

21 4

9 7

1.0 g (6.02 10 nuclei/mol) 2.529 10 nuclei 238.05 g/mol

0.693 (2.529 10 ) 1.2 10 decays/s (4.47 10 yr)(3.16 10 s/yr)

N

R λN

⎛ ⎞

=⎜ ⎟ × = ×

⎝ ⎠

⎡ ⎤

= =⎢ ⎥ × = ×

× ×

⎢ ⎥

⎣ ⎦

191 76Os

191 77Ir*

γ(0.042 MeV)

γ(0.129 MeV)

β–(0.14 MeV) 191

77Ir 191

(24)

72. From Fig. 30–1, the average binding energy per nucleon at A=63 is ~8.6 MeV. Use the average atomic weight as the average number of nucleons for the two stable isotopes of copper to find the total binding energy for one copper atom.

(63 5 nucleons)(8.6 MeV/nucleon) 546.1 MeV. = ≈ 550 MeV

Convert the mass of the penny to a number of atoms and then use the above value to calculate the energy needed.

23 6 19

12

Energy ( atoms)(energy/atom)

(3.0 g) (6.02 10 atoms/mol)(546.1 10 eV/atom)(1.60 10 J/eV) (63.5 g/mol)

2.5 10 J

= #

⎡ ⎤

=⎢ ⎥ × × ×

⎣ ⎦

= ×

73. (a)

( ) ( ) ( )

4 4 4

2 2 2

2 2

He He He 4.002603 u 4 0.002603 u (0.002603 u)(931.5 MeV/u ) 2.425 MeV/

m A

c c

Δ = − = − =

= =

(b) Δ

( ) ( ) ( )

126 C =m 126 C −A 126 C =12.000000 u 12− = 0

(c)

( ) ( ) ( )

8638 8638 8638

2 2

Sr Sr Sr 85.909261 u 86 0.090739 u ( 0.090739 u)(931.5 MeV/u ) 84.52 MeV/

m A

c c

Δ = − = − = −

= − = −

(d)

( ) ( ) ( )

23592 23592 23592

2 2

U U U 235.043930 u 235 0.043930 u (0.043930 u)(931 5 MeV/u ) 40.92 MeV/

m A

c c

Δ = − = − =

= . =

(e) From Appendix B, we see the following:

0 for 0 8 and 85; 0 for 9 84

Z Z

Z

Δ ≥ ≤ ≤ ≥

Δ < ≤ ≤

0 for 0 15 and 218; 0 for 16 218

A A

A

Δ ≥ ≤ ≤ ≥

Δ < ≤ <

74. The reaction is 11H+01n→21H. If we assume that the initial kinetic energies are small, then the energy of the gamma is the Q-value of the reaction.

( ) ( ) ( )

1 1 2 2

1 0 1

2 2

H n H

[(1.007825 u) (1 008665 u) (2 014102 u)] (931.5 MeV/u ) 2.224 MeV

Q m m m c

c c

⎡ ⎤

= + −

(25)

75. The mass of carbon 60,000 years ago was 1.0 kg. Find the number of carbon atoms at that time and then find the number of 146C atoms at that time. Use that with the half-life to find the present activity, using Eq. 30–5 and Eq. 30–6.

1/ 2

3 23

25

25 12 13 14

14 6 0

ln 2 ln 2(60,000 yr)

13

0 7

1/2 1/2

(1.0 10 g)(6.02 10 atoms/mol)

5.017 10 C atoms 12 g/mol

(5.017 10 )(1.3 10 ) 6.522 10 nuclei of C

ln 2 ln 2 ln 2

(6.522 10 ) (5730 yr)(3.16 10 s/yr)

t T

N

N N

R N N N e e

T T

λ

− −

× ×

= = ×

= × × = × =

= = = = ×

×

(5730 yr) 0.1759 decays/s= ≈ 0.2 decays/s

76. The energy to remove the neutron would be the difference in the masses of the 42He and the combination of 32He n.+ It is also the opposite of the Q-value for the reaction. The number of electrons doesn’t change, so atomic masses can be used for the helium isotopes.

2

2 2

He ( He-3 n He-4) (3.016029 u 1.008665 u 4.002603 u) 931.5 MeV/u 20.58 MeV

c

Q m m m c c ⎛ ⎞

− = + − = + − ⎜

⎝ ⎠

=

Repeat the calculation for the carbon isotopes.

2

2 2

C ( C-12 n C-13) (12 000000 u 1.008665 u 13.003355 u) 931.5 MeV/ u 4.946 MeV

c

Q m m m c c ⎛ ⎞

− = + − = . + − ⎜

⎝ ⎠

=

The helium value is 4 16. × greater than the carbon value.

77. (a) Take the mass of the Earth and divide it by the mass of a nucleon to find the number of nucleons. Then use Eq. 30–1 to find the radius.

1/3 24

15 1/3 15

27 5.98 10 kg

(1.2 10 m) (1.2 10 m) 183.6 m 180 m 1.67 10 kg

r= × − A = × − ⎡⎢ × ⎤⎥ = ≈

×

⎢ ⎥

⎣ ⎦

(b) Follow the same process as above, but this time use the Sun’s mass. 1/3 30

15 1/3 15 4

27 1.99 10 kg

(1.2 10 m) (1.2 10 m) 12,700 m 1.3 10 m 1.67 10 kg

r= × − A = × − ⎡⎢ × ⎤⎥ = ≈ ×

×

⎢ ⎥

Figure

Updating...

References