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2 Nuclear Reactions: Spontaneous Decay Of A Nucleus

2 Nuclear Reactions: Spontaneous Decay Of A Nucleus

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of nuclear physics. The Curies and others found that some nuclei are unstable; that

is, certain nuclei decay spontaneously into two or more particles. This process is

called radioactive decay, and the term radioactivity refers to the process in which

a nucleus spontaneously emits either particles or radiation.

When a nucleus decays, it can emit particles with mass such as electrons or neutrons, or it can emit photons (electromagnetic radiation). At fi rst, physicists did not

know the identities of the particles emitted in radioactive decay, so these unknown

decay products were initially given the names alpha, beta, and gamma. Eventually, the

nature of these particles was deduced from experiments, but these somewhat mysterious names are still widely used. Even though their identity was initially unknown,

these three types of decay products could be distinguished according to their properties, including their electric charge and their ability to penetrate matter.

Loosely speaking, we sometimes refer to alpha, beta, and gamma radiation emitted in radioactive decay. The term radiation is reminiscent of electromagnetic radiation (Chapter 23). We now know that only gamma radiation, also called “gamma

rays,” is actually a form of electromagnetic radiation, while alpha and beta “radiation” consists of particles with mass.

Alpha particles. An alpha particle is composed of two protons and two neutrons.

It is thus the nucleus of a He atom and is also denoted 42He. An alpha particle

produced by nuclear decay usually does not have any bound electrons, so alpha

particles almost always have a charge of ϩ2e (the charge of two protons). A typical

radioactive decay involving an alpha particle is that of the radium nucleus 22886Ra.

This process is written in a form very similar to a chemical reaction:

parent nucleus

226 Ra

88



daughter nucleus



S



alpha particle



ϩ



222 Rn

86



4 He

2



(30.12)



This decay reaction (Fig. 30.6A) begins with a parent nucleus of 22886Ra containing

88 protons and N ϭ A Ϫ Z ϭ 226 Ϫ 88 ϭ 138 neutrons. This nucleus spontaneously decays into two particles called decay products. One decay product is an

isotope of radon, 22862Rn, which contains N ϭ A Ϫ Z ϭ 222 Ϫ 86 ϭ 136 neutrons.

In this reaction, 22862Rn is called the daughter nucleus. The other decay product in

Equation 30.12 is an alpha particle, 42He. The total number of nucleons is conserved

(does not change) in the reaction. Since the reaction products contain 222 nucleons (22862Rn) plus 4 nucleons (42He), there are a total of 226 nucleons (protons plus

neutrons) at the start of the reaction and the same number at the end. This particular reaction also conserves the numbers of protons and neutrons separately. That

is, there are 88 protons at the start (in 22886Ra) and 88 at the end (86 in the radon

daughter nucleus and 2 in the alpha particle). You can view this reaction as simply

ejecting two protons and two neutrons (the alpha particle) from the original radium

nucleus. In his scattering experiments, Rutherford obtained his alpha particles from

the alpha decay of nuclei such as 21844Po.



ALPHA DECAY

226

88Ra



222

86Rn



86 protons

ϩ

136 neutrons



88 protons

ϩ

138 neutrons



BETA DECAY



4

2He



p n

n p

Alpha

particle



14

6C



Beta particle

(an electron)





Parent

nucleus



Parent

n

nucleus

Antineutrino



Daughter

nucleus

A



14

7N



Daughter

nucleus



Alpha decay: emission of an alpha

particle from a nucleus



Figure 30.6



A Example of

alpha decay: the original parent

nucleus (22886 Ra) splits to form

222 Rn and an alpha particle (4 He).

86

2

B Example of beta decay: the

original 146C nucleus produces 147 N

plus an electron (a beta particle)

and an antineutrino (n).



ϭ proton

ϭ neutron



B



30.2 | NUCLEAR REACTIONS: SPONTANEOUS DECAY OF A NUCLEUS



1027



The particular alpha decay reaction in Equation 30.12 has important medical

consequences. It contributes to the (ill) health of miners and can also lead to a

health hazard in homes. For more details, see Section 30.4.

Beta particles. There are two varieties of beta particles. One variety is negatively

charged and is simply an electron. The other type is identical to an electron except it

has a positive charge of precisely ϩe. This positive beta particle is called a positron

(denoted eϩ, with the superscript indicating its positive charge) and is an example

of antimatter (Chapter 31). 2 Electrons and positrons have the same mass and are

both point charges.

A typical radioactive decay producing a beta particle is

parent

nucleus



Beta decay: emission of a beta

particle (an electron or positron)

from a nucleus, along with an

antineutrino or a neutrino



14C

6



daughter

nucleus



S



14 N

7



beta particle

(electron)







ϩ



antineutrino



ϩ



n



(30.13)



In this reaction (shown schematically in Figure 30.6B), the emitted beta particle

is an electron denoted by eϪ; the superscript negative sign indicates that the particle is negatively charged. This reaction also produces a particle called an antineutrino, denoted by n. The antineutrino is another example of antimatter and is

thus a “cousin” of the positron. We’ll discuss neutrinos and antimatter further in

Chapter 31.

As with alpha decay, beta decay also conserves the total number of nucleons. It

is useful to rewrite Equation 30.13 to indicate explicitly both the number of protons

and neutrons and the electric charge at the start and end of the reaction:

number of protons: 6 S 7

number of neutrons: 8 S 7

S 147 N ϩ eϪ ϩ n

16e S 17e 2 e 1 0

14C

6



charge:



(30.14)



Hence, this reaction converts a neutron in the parent 146C nucleus into a proton,

conserving the total number of nucleons (neutrons plus protons). At the same time,

electric charge is conserved since the total charge is ϩ6e at the beginning and at

the end of the reaction. The positive charge of the new proton is balanced by the

creation of an electron.

The beta decay reaction in Equation 30.13 is widely used in archeology in the

technique called carbon dating. We’ll discuss this application of radioactivity in

Section 30.5.

Gamma decay. Gamma decay produces gamma “rays,” which are photons, quanta

of electromagnetic radiation denoted by the Greek letter g. One example of a nuclear

decay that produces gamma rays is

parent nucleus

(in excited state)



Gamma decay: emission of a

gamma ray photon from an excited

nucleus



14 N*

7



daughter

nucleus



S



14 N

7



gamma

ray



ϩ



g



(30.15)



The asterisk on the left in this equation denotes that the nucleus is in an excited

state. In Chapter 29, we saw that an atom in excited state can undergo a transition to a state of lower energy and emit a photon in the process. In the same way,

a nucleus that is initially in an excited state can undergo a transition to a state of

lower energy and emit a gamma ray photon. A particular nucleus is often placed in

an excited state as a result of alpha or beta decay. For example, a 147 N* nucleus in an

excited state (as on the left in Eq. 30.15) can be produced by beta decay (although

we did not indicate it in Eq. 30.13). A highly simplified energy-level diagram for the

reaction in Equation 30.15 is shown in Figure 30.7. This diagram shows only the

initial and fi nal states in Equation 30.15. The decay from the excited to the ground

2 Here and throughout this chapter we follow the standard notation of nuclear physics and denote

particles such as an electron by a “non-italic” letter.



1028



CHAPTER 30 | NUCLEAR PHYSICS



state in Figure 30.7 produces a gamma ray with an energy equal to the difference in

the energies of the two levels.

A more realistic energy-level diagram, this time for the excited states of 6280Ni, is

shown in Figure 30.8. A 6280Ni* nucleus can emit gamma rays with many different

energies, depending on which excited and fi nal states are involved. For the states

shown here, the emitted gamma rays have energies of a few MeV.

Gamma ray energies are much higher than those of visible light or X-ray photons. The energy of a gamma ray photon depends on the particular nuclear decay

that produces it, but typical gamma rays have energies from 10 keV (104 eV) to

100 MeV (108 eV) or even higher. Recall that the spectral emission of atoms (Chapter 29) involved photons with energies of typically 10 eV. This comparison shows

again that the overall energy scale of nuclear decay—and of nuclear processes in

general—is much greater than a typical atomic scale energy.

CO N C E P T C H E C K 3 0. 2 | Wavelength of a Gamma Ray Photon

According to the energy-level diagram for 6280Ni in Figure 30.8, this nucleus

can emit a gamma ray photon with an energy of 1.33 MeV. What is the wavelength of this photon?

(a) 9.3 ϫ 10Ϫ13 m ϭ 9.3 ϫ 10Ϫ4 nm

(b) 2.7 ϫ 10Ϫ31 m ϭ 2.7 ϫ 10Ϫ23 nm

(c) 2.3 ϫ 102 m

How does the wavelength of the gamma ray compare with the wavelength of

visible light?



Conservation Rules in Nuclear Reactions



14 *

7N



Gamma ray

photon



14

7N



excited state (147 N*) can decay to

its ground state (147 N) by emitting

a gamma ray photon.



ENERGY LEVELS OF 60

28Ni

E(MeV)

2.51

2.16

g

E ϭ 1.18 MeV

1.33



Conservation of mass–energy. The principle of conservation of energy applies to

all physical processes, including nuclear reactions. The total energy at the start of

a reaction such as the gamma decay in Equation 30.15 must equal the total energy

at the end. Indeed, this principle allows us to use an energy-level diagram (as in

Figs. 30.7 and 30.8) to calculate the energy of the gamma ray produced in such a

reaction. In addition, we must account for the results of special relativity and allow

for the conversion of mass to energy and vice versa. We’ll discuss some specific

examples later in this section.

Conservation of momentum. All nuclear reactions must conserve momentum. The

application of momentum conservation in nuclear physics is similar to our work on

collisions in Chapter 7.

Conservation of electric charge. All nuclear reactions conserve electric charge,

but the total number of charged particles may change in a reaction. For example,

the beta decay in Equation 30.14 produces a proton and an electron. The total

charge produced is thus zero (so the total charge does not change), but the number

of charged particles increases.

Conservation of nucleon number. The number of nucleons—that is, the number of

protons plus neutrons—does not change. For example, in the beta decay reaction

in Equation 30.14, a neutron is converted into a proton, but the total number of

neutrons plus protons is unchanged by the reaction.



Radioactive Decay Series

When a nucleus undergoes radioactive decay through the emission of an alpha particle or a beta particle, it is converted into another type of nucleus. An important

example is the alpha decay of 23928U:

S 23904Th ϩ 42He



Ground state



Figure 30.7 A nucleus in an



g

E ϭ 2.16 MeV



The radioactive processes of alpha, beta, and gamma decay can be understood in

terms of the following conservation principles.



238U

92



Excited state



(30.16)



g

E ϭ 1.33 MeV

0



Figure 30.8 Energy-level

diagram showing a few of the

energy levels of the nucleus 6280Ni.

This nucleus can emit gamma ray

photons with several different

energies, depending on the initial

and fi nal states.



Insight 30.2

GAMMA RAYS AND X-RAYS

The terms gamma ray and X-ray both

refer to portions of the electromagnetic spectrum. Typical gamma ray

energies can be as low as 10 keV and

as large as 100 MeV or even higher.

X-ray energies are typically in the

range of 100 eV to a few hundred keV.

Hence, the gamma and X-ray energy

ranges overlap. By convention, the

distinction between them is made by

the origin of the radiation. Gamma

rays are produced in nuclear reactions

(such as radioactive decay), while Xrays are generated by process involving atomic electrons as described in

Chapter 29.



30.2 | NUCLEAR REACTIONS: SPONTANEOUS DECAY OF A NUCLEUS



1029



Figure 30.9 Sequence of

nuclear decays starting with the

parent nucleus 23928U. The horizontal arrows denote beta decay, and

the diagonal arrows show alpha

decay.



238

92U



A ϭ Number of protons ϩ neutrons



238

b decay



234



234

91 Pa



234

90Th



a decay

230

226

88Ra



226

222

86Rn



222

218

84 Po



218

214

210

206

80



206

82 Pb



82



88

84

86

Z ϭ Number of protons



90



92



This decay converts uranium (U) into thorium (Th). The ability to convert one element into another was a goal of many chemists and alchemists throughout the Middle Ages. This conversion is precisely what happens in alpha and beta decay, but it is

not an economical approach for producing any element on a commercial scale (and

certainly not for producing the gold so eagerly sought by medieval alchemists).

Although Equation 30.16 describes a complete nuclear reaction, the decay reaction often continues. That is, after 23928U decays into 23904Th, this thorium nucleus

decays further. Figure 30.9 shows a sequence of nuclear decays starting from 23928U,

with each decay indicated by an arrow. A horizontal arrow pointing to the right

denotes a beta decay similar to the one in Equation 30.13, in which the number of

protons, Z, increases by one. For example, the second decay in Figure 30.9 is the

reaction

23 4Th

90



S 23914Pa 1 e2 1 n



(30.17)



in which thorium (Th) is converted into protactinium (Pa) plus an electron and an

antineutrino. Each beta decay (horizontal arrow) in Figure 30.9 follows the same

pattern. The diagonal arrows in the figure denote alpha decay; for example, the

diagonal arrow at the upper right links 23928U to 23904Th in the reaction in Equation

30.16. Each alpha decay reduces the proton number by two and the neutron number

by two. Some nuclei, such as 21848Po, can undergo either beta decay or alpha decay, so

both a horizontal arrow and a diagonal arrow originate at these nuclei.

In the radioactive decay series in Figure 30.9, 23928U is the original parent nucleus

while many unstable daughter nuclei are produced, and the decay proceeds until

reaching the nucleus 20826 Pb. This nucleus is a stable fi nal product of the decay of

238U. This particular decay series is especially important; among the daughter nuclei

92

are 22886Ra (radium), which is one of the radioactive nuclei discovered by Pierre and

Marie Curie, and 22862Rn (radon), which is a significant health hazard. We’ll investigate other radioactive nuclei produced by other decay series in the end-of-chapter

questions.

CO N C E P T C H E C K 3 0. 3 Alpha Decay of Radium

Which of the following reaction equations correctly describes the alpha decay of

223Ra?

88

(a) 22883Ra S 22907 Th ϩ 42He

(b) 22907 Th S 22883Ra ϩ 42He

(c) 22883Ra S 21869Rn ϩ 42He

1030



CHAPTER 30 | NUCLEAR PHYSICS



CO N C E P T C H E C K 3 0. 4 Beta Decay

Consider a nucleus that undergoes beta decay and emits an electron. Which of

the following statements is true?

(a) The atomic number Z decreases by one.

(b) The atomic number Z decreases by two.

(c) The atomic number Z increases by one.

(d) The atomic number Z increases by two.

(e) Z and A are unchanged.



Calculating the Binding Energy of a Nucleus

We have mentioned several times that the mass–energy relation of special relativity

is important for understanding nuclear binding energies. Recall from Chapter 27

that the rest energy of a particle (such as a nucleus) with rest mass m is

(30.18)

E ϭ mc 2

Let’s now consider carefully how this relation applies to the nucleus 24He, an alpha

particle. With an understanding of the mass and binding energy in this case, we will

be ready to discuss the energy released in nuclear reactions and how to calculate it.

An alpha particle consists of two protons and two neutrons, and from Table 30.1

we can add up the masses of two isolated protons plus two isolated neutrons to get



m(2 protons ϩ 2 neutrons) ϭ 2(1.0073 u) ϩ 2(1.0087 u) ϭ 4.0320 u



(30.19)



Here we have included many more significant figures than usual because small mass

differences will be extremely important in our fi nal result. The result in Equation

30.19 is slightly larger than the mass of an alpha particle, which is 4.0015 u. (This

is the mass of just a helium nucleus, a helium atom without the two electrons.) The

difference is due to the binding energy of the alpha particle, Ebinding. The mass difference is

Dm 5 4.0015 2 4.0320 u 5 20.0305 u

(30.20)

Applying the relation between mass and energy from special relativity (Eq. 30.18),

the mass difference in Equation 30.20 corresponds to an energy

Ebinding 5 1 Dm 2 c 2



To evaluate E binding, we note that 1 atomic mass unit (1 u) corresponds to

931.5 MeV/c 2 (Eq. 30.11). Using this conversion factor with the value of ⌬m in

Equation 30.20 gives

Ebinding 5 1 20.0305 u 2 a



931.5 MeV/c 2 2

bc 5 228.4 MeV

u



(30.21)



This binding energy is negative, indicating that an alpha particle has a lower energy

than a collection of separated protons and neutrons. An alpha particle is therefore

more stable than a collection of separate protons and neutrons and so will not decay

spontaneously into protons and neutrons.



Finding the Energy Released in a Nuclear Reaction

Let’s now apply our understanding of nuclear binding energy as illustrated for an

alpha particle in Equations 30.19 through 30.21 to calculate the energy released in

a nuclear decay process. The alpha decay reaction of radium was studied by Marie

and Pierre Curie and plays an important role in many decay processes like the one

diagramed in Figure 30.9:

226 Ra

88



S



222 Rn

86



ϩ



4 He

2



(30.22)



m 5 226.0254 u S 222.0176 u 1 4.0026 u

total mass ϭ 226.0254 u S

226.0202 u

(Note that the masses here and in other examples below include the electrons for

each atom, unlike Eq. 30.20 where the mass value for the alpha particle was for the

30.2 | NUCLEAR REACTIONS: SPONTANEOUS DECAY OF A NUCLEUS



1031



INITIAL (before decay)

226

88Ra



88 p

ϩ 138 n

A

FINAL (after decay)

222

86Rn



86 p

ϩ 136 n



4

2He



p n

n p



nucleus only.) For simplicity, Figure 30.10A shows the parent nucleus 22886Ra as being

at rest. Will the reaction products 22862Rn and 24He also be at rest after the reaction? The

answer is no; a nucleus such as 22886Ra will decay spontaneously only if the total rest

energy of the final products is less than the rest energy of the original parent nucleus.

Total energy must be conserved, so the reaction in Equation 30.22 will release energy,

usually in the form of kinetic energy of the decay products. (A portion of the energy

is sometimes also used to put a daughter nucleus into an excited state, similar to the

excited 147N* nucleus produced by beta decay; Eqs. 30.13 and 30.15.) That is why Figure 30.10B shows the 22862Rn and 24He nuclei traveling away after the reaction.

To calculate the energy released in the decay of 22886Ra, we must consider the

masses of the nuclei in Equation 30.22. These masses are listed in Table A.4 and are

given in Equation 30.22. The mass at the start of the reaction is mstart ϭ m(22886Ra)

ϭ 226.0254 u, whereas the total mass at the end is m end ϭ m(22862Rn) ϩ m(42He) ϭ

226.0202 u. The change in mass in the reaction is thus



B



Dm 5 mend 2 mstart 5 226.0202 2 226.0254 u 5 20.0052 u



Figure 30.10 When a 22886Ra

nucleus undergoes alpha decay,

it emits an alpha particle (42He)

and a 22862Rn nucleus. This decay

leaves these two particles with

some kinetic energy. To conserve

momentum, the reaction products

are emitted in opposite directions.



(30.23)



2



which corresponds (from Eq. 30.18) to an energy Ereaction 5 Dmc . Using the defi nition of the mass unit u, we get

Ereaction 5 Dmc 2 5 2 1 0.0052 u 2 a



931.5 MeV/c 2 2

bc 5 24.8 MeV

1u



(30.24)



This energy is negative, indicating that, as expected, the nuclei of the reaction

products have a lower rest energy than the parent nucleus. This reaction releases

4.8 MeV, typically as the kinetic energy of the products. We’ll explore this issue in

the next example.



E X AMPLE 30.2



Speed of an Alpha Particle



The alpha decay of 22886Ra releases 4.8 MeV (Eq. 30.24). If one-tenth of this energy

goes into kinetic energy of the emitted alpha particle, what is the speed of the alpha

particle? Use the classical relation for the kinetic energy (KE 5 12mv 2).

RECOGNIZE T HE PRINCIPLE



The kinetic energy of the alpha particle is related to its speed by KE 5 12mv 2, so given

the kinetic energy we can fi nd v.

SK E TCH T HE PROBLEM



Figure 30.10B describes the problem.

IDENT IF Y T HE REL AT IONSHIPS



We are given that the alpha particle has a kinetic energy of

1

1 4.8 MeV 2 5 4.8 3 105 eV

KE 5 12mav 2 5 10



(1)



We must convert this energy to SI units (joules) so that we can obtain v in meters per

second. The mass of the alpha particle is given in atomic mass units, and we must also

convert this to SI units (kilograms) using Equation 30.11.

SOLV E



Rearranging Equation (1) to solve for v gives



v5



2KE

Å ma



Converting KE to joules gives



KE 5 1 4.8 3 105 eV 2 a



1032



CHAPTER 30 | NUCLEAR PHYSICS



1.60 3 10219 J

b 5 7.7 3 10214 J

1 eV



(2)



Converting m a to kilograms, we have



ma 5 1 4.00 u 2 a



1.66 3 10227 kg

b 5 6.6 3 10227 kg

1u



Substituting these values in Equation (2), we fi nd



v5



2 1 7.7 3 10214 J 2

2KE

5 4.8 3 106 m/s

5

Å ma

Å 6.6 3 10227 kg



What does it mean?

While this speed is only about 2% of the speed of light, the alpha particle still

carries a lot of energy compared to the binding energy of electrons in an atom or

molecule. For this reason, nuclear decay products can do significant damage when

they travel through human tissue (Section 30.4).



Half-life

An important aspect of spontaneous nuclear decay is the decay time; that is, if you

are given a 22886Ra nucleus, how long will you have to wait for it to decay by the

process in Equation 30.22? Experiments show that individual nuclei decay one at a

time, at random times. It is not possible to predict when a particular 22886Ra nucleus

will decay. This randomness is a feature of quantum theory; there is no classical

analogy.

Although one cannot predict when a particular nucleus will decay, one can give

the probability for decay. This probability is specified in terms of the half-life of the

nucleus, T1/2 . Suppose a total of N 0 nuclei are present at time t ϭ 0. Half of these

nuclei will decay during a time equal to one half-life t ϭ T1/2 . Hence, half of the

original nuclei will remain at t ϭ T1/2 as shown graphically in Figure 30.11. If we

then wait until another half-life has passed—that is, until t ϭ 2T1/2 —the number

of original, undecayed, nuclei will fall to N 0/4. Half of the nuclei decay during each

time interval of length ⌬t ϭ T1/2 .

The decay curve in Figure 30.11 is described by the exponential function. We

defi ne the decay constant l in such a way that

N 5 N0e2lt



(30.25)



The value of the decay constant for a particular isotope can be determined through

experimental measurements of N as a function of time and comparing the results to

the exponential function in Equation 30.25. The decay constant l is closely related

to the half-life. From the defi nition of T1/2 , the number of nuclei is N ϭ N 0/2 at t ϭ

T1/2 . Inserting these values into Equation 30.25 leads to

N0

5 N0e2lT1/2

2

e2lT1/2 5 12



(30.26)



Taking the natural logarithm of both sides and doing some rearranging leads to

ln 1 e



2lT1/2



2 5 2lT1/2 5 ln 1 1/2 2 5 2ln 2

lT1/2 ϭ ln 2



N0



1

2



N0



1

4



N0



1

8



N0

0



ln 2

0.693

<

(30.27)

l

l

using ln 2 Ϸ 0.693. Equation 30.27 thus relates the value of the decay constant to

the half-life. This relation is an inverse one: a large decay constant means a short

half-life and vice versa.

Values of T1/2 vary widely; a half-life can be as short as a tiny fraction of a second or

longer than the age of the Earth. Values for a few important nuclei are given in Table

T1/2 5



N



N ϭ N0eϪlt



T1/2 2T1/2 3T1/2 4T1/2



t



Figure 30.11 Radioactive decay

is described by an exponential

decay curve. The curve shows the

number N of nuclei remaining

after a time t. If there are N 0 nuclei

at t ϭ 0, only half are left at time

t ϭ T1/2 .



30.2 | NUCLEAR REACTIONS: SPONTANEOUS DECAY OF A NUCLEUS



1033



Half-lives

of Nucleons and Some

Important Nuclei



Ta b l e 3 0 . 2



Nucleus



Half-life



n (a free

neutron)



10.4 min



1H

1



(a proton)



Stable



2H

1



(deuterium)



Stable



3H

1



(tritium)



12.33 yr



14C

6



5730 yr



60 Co

27



5.27 yr



12 3I

53



13.2 h



222 Rn

86



3.82 days



226 Ra

88



1600 yr



235 U

92



7.04 ϫ 108 yr



238U

92



4.47 ϫ 109 yr



30.2, and a more complete listing is found in Table A.4. The value of T1/2 is important

in many effects and applications of radioactivity (Sections 30.4 and 30.5).



PET Scans and the Half-Life of 209F

CO N C E P T C H E C K 3 0. 5 |

20

The isotope 9F is used in a medical procedure called positron emission tomography (PET). The half-life of 209F is approximately 110 minutes. If your doctor has

a sample with 16 g of pure 209F at t ϭ 0, how much 209F will he have 330 minutes

later, (a) 12 g, (b) 8.0 g, (c) 4.0 g, (d) 2.0 g, or (e) 1.0 g?

E X AMPLE 30.3



Half-life and the Radioactive

Decay of Tritium



Tritium is an isotope of hydrogen containing two neutrons and is denoted by 31H . It

is a component in some fusion bombs (i.e., “hydrogen bombs”) as we’ll discuss in the

next section. Tritium is radioactive and decays into an isotope of helium (32He) with a

half-life of T1/2 ϭ 12.33 yr. Suppose a fusion bomb contains 10 kg of tritium when it

is fi rst assembled. How much tritium (in kilograms) will it contain after 30 years on

the shelf?

RECOGNIZE T HE PRINCIPLE



If there are N 0 tritium nuclei at t ϭ 0, the number remaining after a time t is



N 5 N0e2lt



(1)



We weren’t given the value of l for tritium, but we can fi nd it from the given half-life

using the relation between l and T1/2 in Equation 30.27.

SK E TCH T HE PROBLEM



Figure 30.11 describes the problem. After a time T1/2 passes, half of the radioactive

nuclei have decayed.

IDENT IF Y T HE PRINCIPLE S



We fi rst rearrange Equation 30.27 to fi nd the decay constant for tritium in terms of

the half-life:



l5



ln2

0.693

5

5 0.056 yr21

T1/2

12.33 yr



SOLV E



Using this value of l in Equation (1), we fi nd that the fraction N/N 0 of tritium nuclei

that remain after t ϭ 30 yr is



N

21

5 e2lt 5 e210.056 yr 2130 yr2 5 0.19

N0

Hence, only 19% of the original tritium nuclei will remain. The initial mass was

10 kg, so the fi nal mass of tritium is 0.19 ϫ 10 kg ϭ 1.9 kg .



What does it mean?

Because 32He is not useful for making bombs, the decay of tritium will cause a

fusion bomb to stop working after a certain period of time. As a result, fusion

bombs containing tritium must be periodically “reloaded.”



E X AMPLE 30.4



Isotope Abundance and 23925U



The isotope 23925 U has a half-life for spontaneous fission of 7.0 ϫ 108 yr. In deposits of

naturally occurring uranium, this isotope is only about 0.72% of the total fraction of



1034



CHAPTER 30 | NUCLEAR PHYSICS



uranium nuclei. Hence, if you had a sample containing 100 g of pure uranium, only

0.72 g would be 23925 U. The Earth is believed to be about 4.5 billion ϭ 4.5 ϫ 109 years

old, or about (4.5 ϫ 109)/(7.0 ϫ 108) < 6 half-lives of 23925 U. How many grams of 23925 U

were in your sample when the Earth was formed 4.5 ϫ 109 yr ago?

RECOGNIZE T HE PRINCIPLE



As we go back in time, the amount of 23925 U increases by a factor of two for every halflife. Hence, if we go back in time by one half-life, the amount of 23925 U is doubled; if we

go back in time another half-life, the amount of 23925 U doubles again, and so on.

SK E TCH T HE PROBLEM



Figure 30.12 shows how the amount of 23925 U in our sample grows as we go back in

time by six half-lives.

MASS OF

BACK IN TIME



235

92U



46 g



23.0 g



11.5 g



5.76 g



2.88 g



1.44 g



6T1/2



5T1/2



4T1/2



3T1/2



2T1/2



T1/2



0.72 g



PRESENT

4.2 ϫ 109 yr



7.0 ϫ 108 yr



Figure 30.12 Example 30.4. As we go back in time, the amount of 23925U doubles every T1/2 ϭ



7.0 ϫ 108 yr.



IDENT IF Y T HE REL AT IONSHIPS AND SOLV E



© Cengage Learning/David Rogers



According to Figure 30.12, the total mass of 23925 U after going back in time six halflives is approximately 46 g .

We can be a bit more mathematical by noting that after one half-life, the mass

increases by a factor of two, after two half-lives it increases by 22 ϭ 4, and after six

half-lives it increases by 26 ϭ 64, which again gives



mstart 5 26mend 5 26 1 0.72 g 2 5 46 g



What does it mean?

Geophysicists continue to investigate the composition of the Earth when it fi rst

formed. At that time, the uranium was more equally apportioned between 23928U

and 23925U. At present (about six half-lives later), most of the original 23925U has

decayed. Notice that the half-life of 23928U is about 4.5 ϫ 109 yr, so only about

half of the original 23928U has decayed since the formation of the Earth.



A

Alpha particle





؉



؊

؉



How Do We “Measure” Radioactivity?

The half-life of a nucleus tells how quickly (on average) the nucleus will undergo

radioactive decay. Each decay can produce potentially harmful products such as

the alpha particle produced in the decay of 22886Ra (Eq. 30.22). A nucleus with a

short half-life is more likely to decay, so it is potentially more dangerous than

a nucleus with a long half-life. Since a sample with more radioactive nuclei will be

more dangerous than one with a small number of such nuclei, to fully assess such

dangers we must also know how many nuclei are present in a sample. For this reason, the “strength” of a radioactive sample is measured using a property called the

activity, which can be measured with a Geiger counter (Fig. 30.13). The activity of

a sample is proportional to the number of nuclei that decay in 1 second. If all other

factors are similar, a sample with a high activity is more dangerous than a sample

with a low activity. We often say that a sample with a high activity level is “hot.”



Gas molecules

B



Figure 30.13 A Activity can be

measured with a Geiger counter.

B Charged particles passing

through the detector chamber ionize gas molecules, causing a short

pulse of current to flow. Some

Geiger counters give an audible

“click” to indicate each current

pulse.



30.2 | NUCLEAR REACTIONS: SPONTANEOUS DECAY OF A NUCLEUS



1035



One common unit of activity is the curie (Ci), which is defi ned as



Insight 30.3

GEIGER COUNTERS

A Geiger counter (Fig. 30.13) is a useful tool to study and monitor radioactive materials. Invented by Hans

Geiger (a student of Rutherford), the

Geiger counter detects the passage of

a fast-moving particle (typically produced by radioactive decay) through a

gas. Such a particle ionizes gas atoms

and molecules, allowing a current

between the central wire and the wall

of the container. Typically, this current comes as short pulses, giving the

familiar “clicks” of a Geiger counter

(Fig. 30.13B). Since each click corresponds to a single nuclear decay, a

Geiger counter measures activity.



1 Ci 5 3.7 3 1010 decays/s



(30.28)



Hence, a sample of radioactive material has an activity of 1 Ci if it exhibits

3.7 ϫ 1010 decays each second; 1 g of naturally occurring radium has an activity of

approximately 1 Ci. In practice, a sample with an activity of 1 Ci is extremely dangerous due to the large number of alpha and beta particles and gamma rays that it

emits. Most studies or medical procedures with radioactive substances involve samples with activities of millicuries (mCi ϭ 10Ϫ3 Ci) or microcuries (mCi ϭ 10Ϫ6 Ci).

The official SI unit of activity is the becquerel (Bq), named in honor of Henri

Becquerel (1852–1908), who shared the Nobel Prize in Physics with Marie and

Pierre Curie for his discovery of spontaneous radioactivity. The becquerel is

defi ned by

1 Bq 5 1 decay/s



(30.29)



As time passes and nuclei in a sample decay, the number of remaining unstable

nuclei drops. Hence, the activity of a sample decreases with time. We thus say

(loosely speaking) that a “hot” radioactive sample (such as the nuclear reactor at

Chernobyl in Ukraine) becomes “cooler” with time and therefore safer to deal with

or clean up.

While a sample with a high activity level is generally more dangerous than one

with low activity, there are complicating factors due to the variable amount of

energy carried by decay products as we’ll discuss in Section 30.4.

CO N C E P T C H E C K 3 0.6 | How Does the Activity Level of a Radioactive



Sample Change with Time?

Consider a radioactive material that decays with a half-life of T1/2 . How long

does it take for the activity of the sample to decrease by a factor of four, (a) T1/2 ,

(b) 2T1/2 , (c) 3T1/2 , or (d) 4T1/2?



E X AMPLE 30.5



Activity of Tritium



You are given a sample containing exactly 1 mole of tritium (31H). The tritium decays

with a half-life of 12.33 yr. What is the activity of the sample? Express your answer

in units of decays per second. Hint: Inserting t ϭ 1 s in the exponential decay law

N 5 N0e2lt gives the number of nuclei remaining after 1 s. The activity is the number

of nuclei that decay in the same 1-s time period.

RECOGNIZE T HE PRINCIPLE



If a sample initially contains N 0 nuclei, the number remaining after t ϭ 1 s is



Nremaining 5 N0e2lt

Hence, the number that decay during this time is



Ndecays 5 N0 2 N0e2lt 5 N0 1 1 2 e2lt 2



N



The activity equals the number of nuclei that decay per second.



N0

Number that have

decayed in 1 s.

Number remaining

after 1 s.



1s



t



Figure 30.14 Example 30.5.

Not to scale.



1036



(1)



SK E TCH T HE PROBLEM



Figure 30.14 shows how the number of nuclei remaining decreases with time as well as

the number of nuclei that decay in t ϭ 1 s.

IDENT IF Y T HE REL AT IONSHIPS



To evaluate Equation (1) for our given tritium sample, we must fi nd both N 0 and

the decay constant l. The sample originally contains exactly 1 mole of tritium, so

N 0 ϭ 6.02 ϫ 1023 (Avogadro’s number). From Equation 30.27, the decay constant is

related to the half-life through l 5 0.693/T1/2.



CHAPTER 30 | NUCLEAR PHYSICS



Inserting the known half-life of tritium (12.33 yr) gives



l5



0.693

0.693

0.693

5 1.8 3 1029 s21

5

5

T1/2

12.33 yr

3.9 3 108 s



SOLV E



Inserting our values of N 0 and l as well as t ϭ 1 s into Equation (1) leads to



4



activity ϭ Ndecays ϭ 1.1 ϫ 1015 per second ϭ 1.1 ϫ 1015 Bq

What does it mean?

Nuclei with a shorter half-life decay more rapidly (and have a larger value of l)

than those with a long half-life. Hence, when comparing two samples, the one

with a short half-life will have a higher activity (and hence be more dangerous)

than the one with a long half-life, assuming the number of nuclei in the two

samples is the same. 3



N ϭ neutron number



The exponential factor is very close to one, so we have to keep extra significant figures

in performing this calculation. We fi nd



130

120

110

100

90

80

70

60

50

40

30

He

20

10

0



3 0. 3



|



Pb



Nuclei on this

line have equal

numbers of

neutrons and

protons.



Z



s21 21 1 s 2



ϭ



29



N



Ndecays 5 N0 1 1 2 e2lt 2 5 1 6.02 3 1023 2 3 1 2 e211.8310



10 20 30 40 50 60 70 80 90



Atomic number Z ϭ number of

protons



S TA B I L I T Y O F T H E N U C L E U S :

FISSION AND FUSION



Figure 30.15 Stable nuclei

To be stable, a nucleus containing two or more protons must contain neutrons.

Neutrons are thus necessary for the stability of essentially all matter. Figure 30.15

shows a plot of the neutron number N as a function of the number of protons Z (i.e.,

the atomic number) for all known stable nuclei. The dashed line shows the function

N ϭ Z; nuclei that lie on this line have equal numbers of neutrons and protons,

while nuclei above this line contain an excess of neutrons. Low-mass nuclei such as

He and C tend to have equal numbers of protons and neutrons and are thus on or

near the dashed line corresponding to N ϭ Z in Figure 30.15. Heavy nuclei, however, have many more neutrons than protons; for example, Pb (lead) nuclei contain

around 40 more neutrons than protons.

The reason these additional neutrons are needed can be seen by considering the

forces between protons and neutrons. Figure 30.16A shows a highly schematic nucleus

with two protons and two neutrons, representing 24He. This nucleus is extremely

stable, so this arrangement of protons and neutrons is particularly favored. We now

add another proton to make lithium (Li). To bind this additional proton and achieve

a stable arrangement of protons and neutrons similar to 24He, we need two more

neutrons as shown in Figure 30.16B. This highly simplified picture leaves out the

possibility that the protons in Li might rearrange to better “share” the attraction of

the neutrons, just as two atoms share electrons when they form a chemical bond.

In reality, 36Li and 37Li are both stable isotopes. However, this simple picture does

explain why nuclei with large numbers of protons (high Z) require greater numbers

of neutrons to be stable, accounting for the trend that N Ͼ Z for heavy nuclei in

Figure 30.15. This trend is essential for understanding nuclear fission.



Nuclear Binding Energy

We have discussed how the mass of a nucleus is related to its binding energy, and as

an example we calculated the binding energy of an alpha particle (Eq. 30.21). We

3Notice also that our calculation gives the average activity during a 1-s interval, but because the halflife here is much greater than 1 s, the instantaneous activity will have almost exactly the same value.



usually have a greater number of

neutrons than protons, so N Ͼ Z

for all but the lightest nuclei. This

trend is especially noticeable at

high Z.



n

4

2He



p



p

n

A

n



n

p



p

n



p



7

3Li



n



B



Figure 30.16



A The isotope

contains two protons and

two neutrons. This nucleus is

stable because the neutrons can

(roughly speaking) sit between

the protons, holding the nucleus

together. (Compare with Fig.

30.5.) B If a third proton is added,

more neutrons are needed to keep

the protons apart and provide the

needed binding due to the strong

force. As the number of protons

grows, extra neutrons are needed.

For this reason, 73Li (pictured here)

is more stable than 63Li.



4 He

2



30.3 | STABILITY OF THE NUCLEUS: FISSION AND FUSION



1037



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