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)
eϪ
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)
eϪ
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 energylevel 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 “nonitalic” 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 energylevel 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 Xray 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 energylevel 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 energylevel 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 Energylevel
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 XRAYS
The terms gamma ray and Xray 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.
Xray energies are typically in the
range of 100 eV to a few hundred keV.
Hence, the gamma and Xray 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 endofchapter
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 onetenth 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).
Halflife
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 halflife 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 halflife 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 halflife 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 halflife. 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 halflife. This relation is an inverse one: a large decay constant means a short
halflife and vice versa.
Values of T1/2 vary widely; a halflife 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
Halflives
of Nucleons and Some
Important Nuclei
Ta b l e 3 0 . 2
Nucleus
Halflife
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 HalfLife 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 halflife 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
Halflife 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
halflife 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 halflife
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 halflife:
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 halflife 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 halflives 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 halflife, the amount of 23925 U is doubled; if we
go back in time another halflife, 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 halflives.
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 halflife, the mass
increases by a factor of two, after two halflives it increases by 22 ϭ 4, and after six
halflives 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 halflives later), most of the original 23925U has
decayed. Notice that the halflife 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
eϪ
؉
؊
؉
How Do We “Measure” Radioactivity?
The halflife 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 halflife is more likely to decay, so it is potentially more dangerous than
a nucleus with a long halflife. 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 fastmoving 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 halflife 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 halflife 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 1s 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 halflife through l 5 0.693/T1/2.
CHAPTER 30  NUCLEAR PHYSICS
Inserting the known halflife 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 halflife decay more rapidly (and have a larger value of l)
than those with a long halflife. Hence, when comparing two samples, the one
with a short halflife will have a higher activity (and hence be more dangerous)
than the one with a long halflife, 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. Lowmass 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 1s 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