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

Chapter 31: Radioactivity and Nuclear Physics

49 problems · 2 figures

31-1 §31.2 source ↗

The energy of 30.0 eV is required to ionize a molecule of the gas inside a Geiger tube, thereby producing an ion pair. Suppose a particle of ionizing radiation deposits 0.500 MeV of energy in this Geiger tube. What maximum number of ion pairs can it create?

Answer

1.67 × 10 4

31-5 §31.3 source ↗

Verify that a 2.3×1017kg mass of water at normal density would make a cube 60 km on a side, as claimed in Example 31.1. (This mass at nuclear density would make a cube 1.0 m on a side.)
Example 31.1 How Small and Dense Is a Nucleus? in the book ↗

Answer

m = ρV = ρd 3 ⇒ a = m ρ 1/3 = 2.3 × 10 17 kg 1000 kg/m 3 1 3 = 61 × 10 3 m = 61 km

31-7 §31.3 source ↗

What is the radius of an α particle?

Answer

1.9 fm

31-9 §31.3 source ↗

(a) Calculate the radius of 58Ni, one of the most tightly bound stable nuclei.

(b) What is the ratio of the radius of 58Ni to that of 258Ha, one of the largest nuclei ever made? Note that the radius of the largest nucleus is still much smaller than the size of an atom.

Answer

(a) 4.6 fm

(b) 0.61 to 1

31-11 §31.3 source ↗

What is the ratio of the velocity of a β particle to that of an α particle, if they have the same nonrelativistic kinetic energy?

Answer

85 . 4 to 1

31-13 §31.3 source ↗

The detail observable using a probe is limited by its wavelength. Calculate the energy of a γ-ray photon that has a wavelength of 1×10−16m, small enough to detect details about one-tenth the size of a nucleon. Note that a photon having this energy is difficult to produce and interacts poorly with the nucleus, limiting the practicability of this probe.

Answer

12.4 GeV

31-15 §31.3 source ↗

What is the ratio of the velocity of a 5.00-MeV β ray to that of an α particle with the same kinetic energy? This should confirm that βs travel much faster than αs even when relativity is taken into consideration. (See also Exercise 31.11.)

Answer

19.3 to 1

31-17 §31.4 source ↗

In the following eight problems, write the complete decay equation for the given nuclide in the complete ZAXN notation. Refer to the periodic table for values of Z.

β− decay of 3H (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 → 2 3 He 1 + β − + ν ¯ e

31-19 §31.4 source ↗

In the following eight problems, write the complete decay equation for the given nuclide in the complete ZAXN notation. Refer to the periodic table for values of Z.

β+ decay of 50Mn.

Answer

25 50 M 25 → 24 50 Cr 26 + β + + ν e

31-21 §31.4 source ↗

In the following eight problems, write the complete decay equation for the given nuclide in the complete ZAXN notation. Refer to the periodic table for values of Z.

Electron capture by 7Be.

Answer

4 7 Be 3 + e − → 3 7 Li 4 + ν e

31-23 §31.4 source ↗

In the following eight problems, write the complete decay equation for the given nuclide in the complete ZAXN notation. Refer to the periodic table for values of Z.

α decay of 210Po, the isotope of polonium in the decay series of 238U that was discovered by the Curies. A favorite isotope in physics labs, since it has a short half-life and decays to a stable nuclide.

Answer

84 210 Po 126 → 82 206 Pb 124 + 2 4 He 2

31-25 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

β− decay producing 137Ba. The parent nuclide is a major waste product of reactors and has chemistry similar to potassium and sodium, resulting in its concentration in your cells if ingested.

Answer

55 137 Cs 82 → 56 137 Ba 81 + β − + ν ¯ e

31-27 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

α decay producing 228Ra. The parent nuclide is nearly 100% of the natural element and is found in gas lantern mantles and in metal alloys used in jets (228Ra is also radioactive).

Answer

90 232 Th 142 → 88 228 Ra 140 + 2 4 He 2

31-29 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

When an electron and positron annihilate, both their masses are destroyed, creating two equal energy photons to preserve momentum. (a) Confirm that the annihilation equation e++e−→γ+γ conserves charge, electron family number, and total number of nucleons. To do this, identify the values of each before and after the annihilation. (b) Find the energy of each γ ray, assuming the electron and positron are initially nearly at rest. (c) Explain why the two γ rays travel in exactly opposite directions if the center of mass of the electron-positron system is initially at rest.

Answer

(a) charge: +1+ −1= 0; electron family number:+1+ −1= 0;A: 0+ 0= 0

(b) 0.511 MeV

(c) The two γ rays must travel in exactly opposite directions in order to conserve momentum, since initially there is zero momentum if the center of mass is initially at rest.

31-31 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for β− decay given in the equation ZAXN→ Z+1AY N−1+ β−+ ν¯e. To do this, identify the values of each before and after the decay.

Answer

Z = Z + 1 − 1; A = A ; efn : 0 = + 1 + − 1

31-33 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

Confirm that charge, electron family number, and the total number of nucleons are all conserved by the rule for electron capture given in the equation ZAXN+ e−→ Z−1AYN+1+ νe. To do this, identify the values of each before and after the capture.

Answer

Z - 1 = Z − 1; A = A; efn : + 1 = + 1

31-35 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

(a) Write the complete α decay equation for 226Ra.

(b) Find the energy released in the decay.

Answer

(a) 88226Ra138→ 86222Rn136+ 24He2

(b) 4.87 MeV

31-37 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

(a) Write the complete β− decay equation for the neutron. (b) Find the energy released in the decay.

Answer

(a) n→p+β−+ν¯e

(b) ) 0.783 MeV

31-39 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

Calculate the energy released in the β+ decay of 22Na, the equation for which is given in the text. The masses of 22Na and 22Ne are 21.994434 and 21.991383 u, respectively.

Answer

1.82 MeV

31-41 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

(a) Calculate the energy released in the α decay of 238U.

(b) What fraction of the mass of a single 238U is destroyed in the decay? The mass of 234Th 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) 4.274 MeV

(b) 1.927×10−5

(c) Since U-238 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.

31-43 §31.4 source ↗

In the following four problems, identify the parent nuclide and write the complete decay equation in the ZAXN notation. Refer to the periodic table for values of Z.

(a) Write the complete reaction equation for electron capture by 15O.

(b) Calculate the energy released.

Answer

(a) 815O7+ e−→ 715N8+ νe

(b) 2.754 MeV

31-44 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

An old campfire is uncovered during an archaeological dig. Its charcoal is found to contain less than 1/1000 the normal amount of 14C. Estimate the minimum age of the charcoal, noting that 210=1024.

Answer

57,300 y

31-46 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

(a) Calculate the activity R in curies of 1.00 g of 226Ra. (b) Discuss 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 226Ra is now better known.

31-48 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% 232Th, with a half-life of 1.405×1010y. If an average lantern mantle contains 300 mg of thorium, what is its activity?

Answer

1.22 × 10 3 Bq

31-50 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

(a) Natural potassium contains 40K, which has a half-life of 1.277×109 y. What mass of 40K in a person would have a decay rate of 4140 Bq? (b) What is the fraction of 40K in natural potassium, given that the person has 140 g in his body? (These numbers are typical for a 70-kg adult.)

Answer

(a) 16.0 mg

(b) 0.0114%

31-52 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

50V has one of the longest known radioactive half-lives. In a difficult experiment, a researcher found that the activity of 1.00 kg of 50V is 1.75 Bq. What is the half-life in years?

Answer

1.48 × 10 17 y

31-54 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

A tree falls in a forest. How many years must pass before the 14C activity in 1.00 g of the tree’s carbon drops to 1.00 decay per hour?

Answer

5.6 × 10 4 y

31-56 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

A 5000-Ci 60Co source used for cancer therapy is considered too weak to be useful when its activity falls to 3500 Ci. How long after its manufacture does this happen?

Answer

2.71 y

31-58 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

The β− particles emitted in the decay of 3H (tritium) interact with matter to create light in a glow-in-the-dark exit sign. At the time of manufacture, such a sign contains 15.0 Ci of 3H. (a) What is the mass of the tritium? (b) What is its activity 5.00 y after manufacture?

Answer

(a) 1.56 mg

(b) 11.3 Ci

31-60 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

(a) The 210Po source used in a physics laboratory is labeled as having an activity of 1.0μCi on the date it was prepared. A student measures the radioactivity of this source with a Geiger counter and observes 1500 counts per minute. She notices that the source was prepared 120 days before her lab. What fraction of the decays is she observing with her apparatus? (b) Identify some of the reasons that only a fraction of the α s emitted are observed by the detector.

Answer

(a) 1.23×10−3

(b) Only part of the emitted radiation goes in the direction of the detector. Only a fraction of that causes a response in the detector. Some of the emitted radiation (mostly α particles) is observed within the source. Some is absorbed within the source, some is absorbed by the detector, and some does not penetrate the detector.

31-62 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

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

Answer

(a) 1.68 × 10 – 5 Ci

(b) 8.65 × 10 10 J

(c) $ 2.9 × 10 3

31-64 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

The Galileo space probe was launched on its long journey past several planets 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 becquerel? (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

(a) 6.97 × 10 15 Bq

(b) 6.24 kW

(c) 5.67 kW

31-68 §31.5 source ↗

Data from the appendices and the periodic table may be needed for these problems.

Unreasonable Results

The manufacturer of a smoke alarm decides that the smallest current of α radiation he can detect is 1.00 μA . (a) Find the activity in curies of an α emitter that produces a 1.00 μA current of α particles. (b) What is unreasonable about this result? (c) What assumption is responsible?

Answer

(a) 84.5 Ci

(b) An extremely large activity, many orders of magnitude greater than permitted for home use.

(c) The assumption of 1.00 μA is unreasonably large. Other methods can detect much smaller decay rates.

31-69 §31.6 source ↗

2H is a loosely bound isotope of hydrogen. Called deuterium or heavy hydrogen, it is stable but relatively rare—it is 0.015% of natural hydrogen. Note that deuterium has Z=N, which should tend to make it more tightly bound, but both are odd numbers. Calculate BE/A, the binding energy per nucleon, for 2H and compare it with the approximate value obtained from the graph in Figure 31.24.
Figure 31.24 in the book ↗
A graph of average binding energy per nucleon, BE/A, for stable nuclei. The most tightly bound nuclei are those with A near 60, where the attractive nuclear force has its greatest effect. At higher A s, the Coulomb repulsion progressively reduces the binding energy per nucleon, because the nuclear force is short ranged. The spikes on the curve are very tightly bound nuclides and indicate shell closures.

Answer

1.112 MeV, consistent with graph

31-71 §31.6 source ↗

209Bi is the heaviest stable nuclide, and its BE/A is low compared with medium-mass nuclides. Calculate BE/A, the binding energy per nucleon, for 209Bi and compare it with the approximate value obtained from the graph in Figure 31.24.
Figure 31.24 in the book ↗
A graph of average binding energy per nucleon, BE/A, for stable nuclei. The most tightly bound nuclei are those with A near 60, where the attractive nuclear force has its greatest effect. At higher A s, the Coulomb repulsion progressively reduces the binding energy per nucleon, because the nuclear force is short ranged. The spikes on the curve are very tightly bound nuclides and indicate shell closures.

Answer

7.848 MeV, consistent with graph

31-73 §31.6 source ↗

(a) Calculate BE/A for 12C. Stable and relatively tightly bound, this nuclide is most of natural carbon. (b) Calculate BE/A for 14C. Is the difference in BE/A between 12C and 14C significant? One is stable and common, and the other is unstable and rare.

Answer

(a) 7.680 MeV, consistent with graph

(b) 7.520 MeV, consistent with graph. Not significantly different from value for 12C, but sufficiently lower to allow decay into another nuclide that is more tightly bound.

31-75 §31.6 source ↗

The purpose of this problem is to show in three ways that the binding energy of the electron in a hydrogen atom is negligible compared with the masses of the proton and electron. (a) Calculate the mass equivalent in u of the 13.6-eV binding energy of an electron in a hydrogen atom, and compare this with the mass of the hydrogen atom obtained from Appendix A. (b) Subtract the mass of the proton given in Table 31.2 from the mass of the hydrogen atom given in Appendix A. You will find the difference is equal to the electron’s mass to three digits, implying the binding energy is small in comparison. (c) Take the ratio of the binding energy of the electron (13.6 eV) to the energy equivalent of the electron’s mass (0.511 MeV). (d) Discuss how your answers confirm the stated purpose of this problem.
Table 31.2 Masses of the Proton, Neutron, and Electron in the book ↗
Particle Symbol kg u MeV/c2
Proton p 1 . 67262 × 10 − 27 1.007276 938.27
Neutron n 1 . 67493 × 10 − 27 1.008665 939.57
Electron e 9 . 1094 × 10 − 31 0.00054858 0.511

Answer

(a) 1.46×10−8u vs. 1.007825 u for 1H

(b) 0.000549 u

(c) 2.66×10−5

31-76 §31.6 source ↗

Unreasonable Results

A particle physicist discovers a neutral particle with a mass of 2.02733 u that he assumes is two neutrons bound together. (a) Find the binding energy. (b) What is unreasonable about this result? (c) What assumptions are unreasonable or inconsistent?

Answer

(a) –9.315 MeV

(b) The negative binding energy implies an unbound system.

(c) This assumption that it is two bound neutrons is incorrect.

31-78 §31.7 source ↗

Integrated Concepts

A 2.00-T magnetic field is applied perpendicular to the path of charged particles in a bubble chamber. What is the radius of curvature of the path of a 10 MeV proton in this field? Neglect any slowing along its path.

Answer

22.8 cm

31-79 §31.7 source ↗

(a) Write the decay equation for the α decay of 235 U . (b) What energy is released in this decay? The mass of the daughter nuclide is 231.036298 u. (c) Assuming the residual nucleus is formed in its ground state, how much energy goes to the α particle?

Answer

(a) 92235 U 143 → 90 231 Th 141 + 2 4 He 2

(b) 4.679 MeV

(c) 4.599 MeV

31-81 §31.7 source ↗

Unreasonable Results

A physicist scatters γ rays from a substance and sees evidence of a nucleus 7.5×10–13m in radius. (a) Find the atomic mass of such a nucleus. (b) What is unreasonable about this result? (c) What is unreasonable about the assumption?

Answer

a) 2.4 × 10 8 u

(b) The greatest known atomic masses are about 260. This result found in (a) is extremely large.

(c) The assumed radius is much too large to be reasonable.

31-82 §31.7 source ↗

Unreasonable Results

A frazzled theoretical physicist reckons that all conservation laws are obeyed in the decay of a proton into a neutron, positron, and neutrino (as in β+ decay of a nucleus) and sends a paper to a journal to announce the reaction as a possible end of the universe due to the spontaneous decay of protons. (a) What energy is released in this decay? (b) What is unreasonable about this result? (c) What assumption is responsible?

Answer

(a) –1.805 MeV

(b) Negative energy implies energy input is necessary and the reaction cannot be spontaneous.

(c) Although all conversation laws are obeyed, energy must be supplied, so the assumption of spontaneous decay is incorrect.

31-84 §31.7 source ↗

Critical Thinking (a) What is the density of 235U in u/fm3? (b) If 235U splits into 141Ba and 92Kr+3 neutrons, how does the density of 235U compare with the density of each element it splits into?

Answer

(a) r= r 0 A 1/3 =1.2 (235) 1/3 =7.4 fm   V= 4π r 3 3 =1700  fm 3   ρ= m v =0.14  u/fm 3

(b) For barium: r= r 0 A 1/3 = 1.2 (142) 1/3 = 6.3 fm
V= 4π r 3 3 =1047  fm 3 ρ= m v =0.14  u/fm 3
For krypton: r= r 0 A 1/3 = 1.2 (92) 1/3 = 5.4 fm
V= 4π r 3 3 =660  fm 3 ρ= m v =0.14  u/fm 3
To two significant figures, they are all alike.

AP Test Prep 31.1 §31.1 source ↗

A nucleus is observed to emit a γ ray with a frequency of 6.3× 10 19  Hz. What must happen to the nucleus as a consequence?
  1. a.
    The nucleus must gain 0.26 MeV.
  2. b.
    The nucleus must also emit an α particle of energy 0.26 MeV in the opposite direction.
  3. c.
    The nucleus must lose 0.26 MeV.
  4. d.
    The nucleus must also emit a β particle of energy 0.26 MeV in the opposite direction.

Answer

C. The nucleus must lose 0.26 MeV.

AP Test Prep 31.3 §31.3 source ↗

A typical carbon nucleus contains 6 neutrons and 6 protons. The 6 protons are all positively charged and in very close proximity, with separations on the order of 10-15 meters, which should result in an enormous repulsive force. What prevents the nucleus from dismantling itself due to the repulsion of the electric force?
  1. a.
    The attractive nature of the strong nuclear force overpowers the electric force.
  2. b.
    The weak nuclear force barely offsets the electric force.
  3. c.
    Magnetic forces generated by the orbiting electrons create a stable minimum in which the nuclear charged particles reside.
  4. d.
    The attractive electric force of the surrounding electrons is equal in all directions and cancels out, leaving no net electric force.

Answer

A. The attractive nature of the strong nuclear force overpowers the electric force.

AP Test Prep 31.5 §31.4 source ↗

95 241 Am is commonly used in smoke detectors because its α decay process provides a useful tool for detecting the presence of smoke particles. When 95 241 Am undergoes α decay, what is the resulting nucleus? If 95 241 Am were to undergo β decay, what would be the resulting nucleus? Explain each answer.

Answer

When 95 241 Am undergoes α decay, it loses 2 neutrons and 2 protons. The resulting nucleus is therefore 93 237 Np .

AP Test Prep 31.7 §31.4 source ↗

Explain why the overall charge of the nucleus is increased by +1 during the β decay process.

Answer

During this process, the nucleus emits a particle with -1 charge. In order for the overall charge of the system to remain constant, the charge of the nucleus must therefore increase by +1.

AP Test Prep 31.9 §31.4 source ↗

Are the following reactions possible? For each, explain why or why not.

(a) U 92 238 → R 88 234 a+ H 2 4 e

(b) R 88 223 a→ P 82 209 b+ C 6 14

(c) C 6 14 → N 7 14 + e − + v ¯ e

(d) M 12 24 g→ N 11 23 a+ e + + v e

Answer

(a) No. Nucleon number is conserved (238 = 234 + 4), but the atomic number or charge is NOT conserved (92 ≠ 88+2).

(b) Yes. Nucleon number is conserved (223 = 209 + 14), and atomic number is conserved (88 = 82 + 6).

(c) Yes. Nucleon number is conserved (14 = 14), and charge is conserved if the electron’s charge is properly counted (6 = 7 + (-1)).

(d) No. Nucleon number is not conserved (24 ≠ 23). The positron released counts as a charge to conserve charge, but it doesn’t count as a nucleon.

AP Test Prep 31.11 §31.5 source ↗

When P 84 215 o decays, the product is P 82 211 b. The half-life of this decay process is 1.78 ms. If the initial sample contains 3.4 x 1017 parent nuclei, how many are remaining after 35 ms have elapsed? What kind of decay process is this (alpha, beta, or gamma)?

Answer

This must be alpha decay since 4 nucleons (2 positive charges) are lost from the parent nucleus. The number remaining is found from:

N( t )= N 0 e( −0.693t t 1 2 )=3.4× 10 17 e( −( 0.693 )( 0.035 ) 0.00173 )

N(t)= 4.1× 10 11 nuclei

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