20.5

Strategy:

In balancing nuclear equations, note that the sum of
atomic numbers and that of mass numbers must match on both sides of the
equation.

Solution:

a.

On the left side of this
equation the atomic number sum is 13 (12 + 1) and the mass number sum
is 27 (26 + 1). These sums must
be the same on the right side.
Remember that the atomic and mass numbers of an alpha particle
are 2 and 4, respectively. The
atomic number of X is therefore 11 (13 
2) and the mass number is 23 (27 
4). X is sodium23 (Na).

b.

On the left side of this
equation the atomic number sum is 28 (27 + 1) and the mass number sum
is 61 (59 + 2). These sums must
be the same on the right side.
The atomic number of X is therefore 1 (28  27) and the mass number
is also 1 (61  60). X is a proton (p
or H).

c.

On the left side of this equation the atomic number sum
is 92 (92 + 0) and the mass number sum is 236 (235 + 1). These sums must be the same on the
right side. The atomic number of
X is therefore 0 and the
mass number is 1 . X is
a neutron (n).

d.

On the left side of this equation the atomic number sum
is 26 (24 + 2) and the mass number sum is 57 (53 + 4). These sums must be the same on the
right side. The atomic number of
X is therefore 26 (26  0) and the mass number
is 56 (57  1). X is iron56
(Fe).

e.

On the left side of this equation the atomic number
sum is 8 and the mass number sum is 20.
The sums must be the same on the right side. The atomic number of X is therefore 1 (8  9) and the mass number
is 0 (20  20). X is a b
particle (b).




20.6

a.

The sum of the mass numbers must be conserved. Thus, the unknown product will have a
mass
number of 0.
The atomic number must be conserved. Thus, the nuclear charge of
the unknown product must be 1. The particle is a b
particle.

b.

Balancing the mass numbers
first, we find that the unknown product must have a mass of 40. Balancing the nuclear charges, we find
that the atomic number of the unknown must be 20. Element number 20 is calcium (Ca).

c.

Balancing the mass numbers, we
find that the unknown product must have a mass of 4. Balancing the nuclear charges, we find
that the nuclear charge of the unknown must be 2. The unknown particle is an alpha (a)
particle.

d.

Balancing the mass numbers, we
find that the unknown products must have a combined mass of 2. Balancing the nuclear charges, we find
that the combined nuclear charge of the two unknown particles must be 0. The unknown particles are neutrons.



20.13

Strategy:

We first convert the mass in amu to grams. Then, assuming the nucleus to be
spherical, we calculate its volume.
Dividing mass by volume gives density.

Solution:

The mass is:
The volume is, V = 4/3πr^{3}.
The density is:
2.72 × 10^{14}
g/cm^{3}



20.14

The principal
factor for determining the stability of a nucleus is the neutrontoproton ratio (n/p). For stable elements of low atomic number,
the n/p ratio is close to 1.
As the atomic number increases, the n/p ratios of
stable nuclei become greater than 1.
The following rules are useful in predicting nuclear stability.
1)
Nuclei that contain 2, 8, 20, 50, 82, or 126
protons or neutrons are generally more stable than nuclei that do not
possess these numbers. These numbers
are called magic numbers.
2) Nuclei with even numbers of both
protons and neutrons are generally more stable than those with odd numbers
of these particles (see Table 20.2 of the text).
a.

Lithium9 should be less
stable. The neutrontoproton
ratio is too high. For small
atoms, the n/p ratio will be close to 1:1.

b.

Sodium25 is less stable. Its neutrontoproton ratio is probably
too high.

c.

Scandium48 is less
stable because of odd numbers of protons and neutrons. We would not expect calcium48 to be
stable even though it has a magic number of protons. Its n/p ratio is too
high.



20.15

Nickel (Ni), selenium (Se), and cadmium (Cd)
have more stable isotopes. All three
have even atomic numbers (see Table 20.2 of the text).


20.16

a.

Neon17 should be radioactive. It falls below the belt of stability
(low n/p ratio).

b.

Calcium45 should be radioactive. It falls above the belt of stability
(high n/p ratio).

c.

All technetium isotopes are radioactive.

d.

Mercury195 should be radioactive. Mercury196 has an even number of both
neutrons and protons.

e.

All curium isotopes are unstable.



20.17

Strategy:

We can use Equation 20.1, DE = (Dm)c^{2}, to solve the problem. Recall the following conversion factor:

Solution:

The mass change is:

Think About It:

Is this mass measurable with ordinary laboratory
analytical balances?



20.18

The energy
loss in one second is:
Therefore the rate of mass loss is 6 ´
10^{9} kg/s.


20.19

Strategy:

To calculate the nuclear binding energy, we first
determine the difference between the mass of the nucleus and the mass of
all the protons and neutrons, which gives us the mass defect. Next, we apply Einstein's massenergy
relationship [DE = (Dm)c^{2}], (Equation 20.1); and use the procedure shown
in Sample Problem 20.2.

Solution:

a.

There are 4 neutrons
and 3 protons in a Li7 nucleus. The predicted mass is:
(3)(mass of proton) + (4)(mass of neutron) = (3)(1.007825 amu) + (4)(1.008665
amu)
predicted mass = 7.058135 amu
The mass defect, that is the difference between the
predicted mass and the measured mass is:
Dm =
7.01600 amu  7.058135 amu =
0.042135 amu
The mass that is converted
in energy, that is the energy released is:
6.30 × 10^{12} kg·m^{2}/s^{2} = 6.30
× 10^{12} J
Note that we report the binding energy without
referring to its sign.
Therefore, the nuclear binding energy is 6.30 × 10^{12} J. When comparing the stability of any
two nuclei we must account for the fact that they have different
numbers of nucleons. For this
reason, it is more meaningful to use the nuclear binding energy per nucleon, defined as
The binding energy per
nucleon for ^{7}Li is:

b.

Using the same
procedure as in (a), using 1.007825 amu for and 1.008665
amu for, we can show that for chlorine35:
Nuclear binding energy =
4.78 ´
10^{11} J
Nuclear binding energy per
nucleon = 1.37 ´
10^{12} J/nucleon




20.20

We use the procedure shown in Sample Problem 20.2 of the
text.
a.

The binding energy is
the energy required for the process
There are 2 protons and 2 neutrons in the helium
nucleus. The mass of 2 protons is
(2)(1.007825
amu) = 2.015650 amu
and the mass of 2
neutrons is
(2)(1.008665
amu) = 2.017330 amu
Therefore, the predicted
mass of is
2.015650 + 2.017330 = 4.032980 amu, and the mass defect is
Dm
= 4.032980 amu 
4.002603 amu = 0.030377 amu 5.044×10^{29}
kg
The energy change (DE) for the process is
DE
= (Dm)c^{2}
= (5.044×10^{29}
kg)(3.00 ´
10^{8} m/s)^{2}
=
4.54 × 10^{12}
kg·m^{2}/s^{2} = 4.54
× 10^{12}
J
This is the nuclear binding energy. It’s the energy required to break up
one helium4 nucleus into 2 protons and 2 neutrons.
For the helium4 nucleus,

b.

The binding energy is
the energy required for the process
There are 74 protons and 110 neutrons in the tungsten
nucleus. The mass of 74 protons is
(74)(1.007825
amu) = 74.57905 amu
and the mass of 110
neutrons is
(110)(1.008665
amu) = 110.9532 amu
Therefore, the predicted
mass of is
74.57905 + 110.9532 = 185.5323 amu, and the mass defect is
Dm
= 185.5323 amu 
183.950928 amu = 1.5814 amu 2.626 × 10^{27}
kg
The energy change (DE) for the process is
DE
= (Dm)c^{2}
= (2.626 × 10^{27}
kg)(3.00 ´
10^{8} m/s)^{2}
=
2.36 × 10^{10}
kg·m^{2}/s^{2} = 2.36
× 10^{10}
J



20.21

Strategy:

The mass of one ^{48}Cr atom is the sum of the
masses of its constituents (24 protons, 24 neutrons, and 24 electrons)
plus the mass defect. We can use the given nuclear binding energy and
Equation 20.1 to compute the mass defect. Remember that 1 J = 1 kg·m/s^{2}.

Solution:

Calculate the mass defect:
Calculate the mass of the ^{48}Cr
atom (ignore the mass of the electrons):
mass ^{48}Cr atom = (24
protons)( 1.007825 amu/proton) + (24 neutrons)(1.008665 amu)
– 7.32
× 10^{–28} kg
= 48.39576 amu –
7.32 × 10^{–28} kg
= (48.39576 amu)(1 kg/6.022 × 10^{26}
amu) – 7.32 × 10^{–28} kg
= 8.036 × 10^{–26} kg –
7.32 × 10^{–28} kg
= 7.963
× 10^{–26} kg



20.22

Strategy:

The mass of one ^{192}Ir atom is the sum of
the masses of its constituents (77 protons, 115 neutrons, and 77
electrons) plus the mass defect. We can use the given nuclear binding
energy and Equation 20.1 to compute the mass defect. Remember that 1 J =
1 kg·m/s^{2}.

Solution:

Calculate the mass defect:
Calculate the mass of the ^{192}Ir
atom (ignore the mass of the electrons):
mass ^{192}Ir atom = (77
protons)( 1.007825 amu/proton) + (115 neutrons)(1.008665 amu)
– 2.71 × 10^{–27} kg
= 193.599 amu –
7.32 × 10^{–28} kg
= (193.599 amu)(1 kg/6.022 × 10^{26}
amu) – 2.71 × 10^{–27} kg
= 3.215 × 10^{–25} kg –
2.71 × 10^{–27} kg
= 3.188
× 10^{–25} kg



20.25

Strategy:

Alpha emission decreases the atomic number by two and
the mass number by four. Beta
emission increases the atomic number by one and has no effect on the mass
number.

Solution:




20.26

According to
Section 20.3, the number of radioactive nuclei at time zero (N_{0}) and time t (N_{t}) is
and the corresponding halflife of the reaction is given
by:
(Section
14.3)
Using the information given in the problem and the first
equation above, we can calculate the rate constant, k. Then, the halflife
can be calculated from the rate constant.
We can use the following equation to calculate the rate
constant k for each point.
From day 0 to day 1, we have
k =
0.251 d^{1}
Following the same procedure for the other days,
t
(d)

mass (g)

k
(d^{1})

0

500


1

389

0.251

2

303

0.250

3

236

0.250

4

184

0.249

5

143

0.252

6

112

0.244

The average value of k is 0.249 d^{1}.
We use the average value of k to calculate the halflife.
2.78 d


20.27

The number of atoms decreases by half for each
halflife. For ten halflives we
have:


20.28

Since all radioactive decay processes have firstorder
rate laws, the decay rate is proportional to the amount of radioisotope at
any time. The halflife is given by
the following equation:
(1)
There is also an equation that relates the number of
nuclei at time zero (N_{0})
and time t (N_{t}).
We can use this equation to solve for the rate constant,
k. Then, we can substitute k into Equation (1) to calculate the
halflife.
The time interval is:
(2:15 p.m., 12/17/10)  (1:00 p.m., 12/3/10) =
14 d + 1 hr + 15 min = 20,235 min
k =
1.8 ´
10^{4}
min^{1}
Substitute k
into equation (1) to calculate the halflife.
3.9 × 10^{3}
min or 2.7 d


20.29

Strategy:

We use the equation from Section 20.3 (a variation of
Equation 14.3) to solve for t,
the age of the artifact. Recall
that the ratio in this equation can be expressed as the number of
radioactive nuclei or as the activity (disintegrations per unit time) of
the sample.
The problem
gives the halflife. We must first
solve for k using the equation

Solution:

Rearranging the equation
to solve for t gives
t3.09 × 10^{3} yr

Think About It:

Note that because the activity is more than half the original activity, we expect the age of
the artifact to be less than a
halflife.



20.30

The equation for the overall process is:
The final product isotope must be.


20.31

Let’s consider the decay of A first.
Let’s convert k
to units of day^{1}.
Next, use the firstorder rate equation to calculate the
amount of A left after 30 days.
Let x be the
amount of A left after 30 days.
x
» 0
Thus, no A remains.
For B: As
calculated above, all of A is converted to B in less than 30 days. In fact, essentially all of A is gone in
less than 1 day! This means that at
the beginning of the 30 day period, there is 1.00 mol of B present. The half life of B is 15 days, so that
after two halflives (30 days), there should be
0.25 mole of B left.
For C: As
in the case of A, the halflife of C is also very short. Therefore, at the end of the 30day
period, no C is left.
For D: D is
not radioactive. 0.75 mol of B
reacted in 30 days; therefore, due to a 1:1 mole ratio between B and D,
there should be 0.75 mole of D present after 30 days.


20.32

We use the integrated rate law to solve for t, the age of the charcoal. Recall that the ratio in this equation
can be expressed as the number of radioactive nuclei or as the activity
(disintegrations per unit time) of the sample.
In Problem 20.29 we used the halflife of carbon14 to
calculate the rate constant, k.
Rearranging the equation to
solve for t gives
t4.05 × 10^{3} yr


20.33

Strategy:

We use the integrated
rate law to solve for the activity at time t, given that time t
= 8.4 × 10^{3} yr.

Solution:

Taking the inverse ln of both sides of the equation, we get
=e^{1.019}
= 0.361
N_{t} = N_{0}(0.361) = (15.3
dpm)(0.361) = 5.5 dpm



20.34

We first convert the mass of ^{206}Pb to the mass of ^{238}U
that decayed to produce it:
We find the mass of ^{238}U at t = 0 by adding this mass to the mass of ^{238}U in the
rock now.
Mass of ^{238}U at t = 0 = 1.09 g + 0.09 g = 1.18 g
We use the halflife given in the problem to solve for the rate
constant, k.
Then we use the integrated rate law to solve for t, the age of the rock.
The age of the rock, t = 5.2 × 10^{8} yr


20.35

Strategy:

We begin by using the
integrated rate law and the rate constant, k (determined in Problem 20.34), to determine the ratio of ^{238}U
at t = 0 to ^{238}U at t = 1.7 × 10^{8} yr.

Solution:

= 
0.02618
It simplifies the
problem to put the larger mass in the numerator. Doing this simply changes the sign of
the result:
= 0.02618
Taking the inverse natural log of both sides gives
The mass ratio of ^{238}U
at t = 0 to ^{238}U now
(at t = 1.7 × 10^{8}
yr) is 1.027. Since an infinite
number of mass combinations that would give this ratio, we must assume a
value for one of the masses. It
makes the mass easier if we assume that the mass of ^{238}U now
is 1.000 g. This gives a mass of
1.027 g ^{238}U at t =
0. The difference between these
masses is the mass of ^{238}U that has decayed to ^{206}Pb. However, because a nucleus loses mass
as it decays, the mass of ^{206}Pb will be less than the mass of ^{238}U
that decayed. We determine the
mass of ^{206}Pb as follows:
Finally, if there
currently a mass of 1.000 g ^{238}U and a mass of 0.023 g ^{206}Pb,
the ratio of ^{238}U
to ^{206}Pb is
43:1



20.38

In the shorthand notation for nuclear reactions, the first symbol
inside the parentheses is the bombarding particle (reactant) and the second
symbol is the emitted particle (product).


20.39


20.40


20.41

Remember that it is
unnecessary to include the subscripted atomic number. The elemental symbol and atomic mass is
sufficient to identify an isotope unambiguously.


20.42

All you need is a highintensity
alpha particle emitter. Any heavy
element like plutonium or curium will do.
Place the bismuth209
sample next to the alpha emitter and wait.
The reaction is:


20.43

Upon
bombardment with neutrons, mercury198 is first converted
to mercury199,
which then emits a proton. The
reaction is:


20.54

The easiest
experiment would be to add a small amount of aqueous iodide containing some
radioactive iodine to a saturated solution of lead(II) iodide. If the equilibrium is dynamic,
radioactive iodine will eventually be detected in the solid lead(II) iodide.
Could this
technique be used to investigate the forward and reverse rates of this
reaction?


20.55

The fact that the radioisotope appears
only in the I_{2} shows that the IO is formed
only from the
IO. Does
this result rule out the possibility that I_{2} could be formed
from IO as well? Can you suggest an experiment to answer
the question?


20.56

On paper, this is a simple experiment. If
one were to dope part of a crystal with a radioactive tracer, one could
demonstrate diffusion in the solid state by detecting the tracer in a
different part of the crystal at a later time. This actually happens with many
substances. In fact, in some
compounds one type of ion migrates easily while the other remains in fixed
position!


20.57

Add iron59 to
the person’s diet, and allow a few days for the iron59 isotope to be
incorporated into the person’s body.
Isolate red blood cells from a blood sample and monitor
radioactivity from the hemoglobin molecules present in the red blood cells.


20.58

Strategy:

Neutron activation has no effect on the atomic number
and increases the mass number by one.
Electron capture decreases the atomic number by one and has no
effect on the mass number.
To identify the product of the electroncapture decay
of ^{125}I, determine the atomic number for the unknown species,
X, by decreasing the atomic number by one. Use the atomic number to determine the
identity of the unknown species.

Setup:

The atomic number of xenon is 54.

Solution:

The balanced equation for the neutron activation of ^{124}Xe
is:
The balanced equation for the electron capture of ^{125}Xe
is:
Mass number of X = (125 + 0) = 125. The atomic number of X = (53 – 1) = 52.
The balanced equation for the electron capture of ^{125}I is:
The product of
the electroncapture decay of is.



20.59

Strategy:

Use Equation 14.5 to solve for k.
Then use Equation 14.3 to solve for time.

Setup:

At 5.00 percent of the original value, we assume [A]_{o}
= 100 and [A]_{t} = 5.

Solution:

Solving Equation 14.5 for k gives
Solving Equation14.3 for t
gives
t = 257 days



20.64

The design and
operation of a Geiger counter are discussed in Figure 20.19 of the text.


20.65

We start with the integrated firstorder rate law (Equation
14.3):
We can calculate
the rate constant, k, from the
halflife using Equation 14.5 of the text, and then substitute into
Equation 14.3 to solve for the time.
_{}
Substituting:
t = 65.3 yr


20.66

Apparently there
is a sort of Pauli exclusion principle for nucleons as well as for
electrons. When neutrons pair with
neutrons and when protons pair with protons, their spins cancel. Eveneven
nuclei are the only ones with no net spin.


20.67

a.

The balanced equation
is:

b.

The number of tritium
(T) atoms in 1.00 kg of water is:
= 6.68 ´
10^{8} T atoms
The number of disintegrations
per minute (dpm) will be:
rate
= k (number of T atoms) = kN =
rate =
70.5 T atoms/min = 70.5
dpm



20.68

a.

One millicurie represents 3.70 ´
10^{7} disintegrations/s.
The rate of decay of the isotope is given by the rate law: rate = kN, where N is the
number of atoms in the sample. We
find the value of k in units of
s^{1}:
k =
The number of atoms (N) in a 0.500 g sample of
neptunium237
is:
rate
of decay = kN
= (9.99 ´
10^{15}
s^{1})(1.27
´
10^{21} atoms) = 1.27 ´
10^{7} atoms/s
We can also say that:
rate
of decay = 1.27 ´
10^{7} disintegrations/s
The activity in millicuries
is:

b.

The decay equation is:



20.69


20.70

We use the same procedure as in Problem 20.20.
Isotope Atomic Mass Nuclear
Binding Energy
(amu) (J/nucleon)
a.

^{ 10}B 10.0129 1.040
´
10^{12}

b.

^{ 11}B 11.009305 1.111
´
10^{12}

c.

^{ 14}N 14.003074 1.199
´
10^{12}

d.

^{ 56}Fe 55.93494 1.410
´
10^{12}



20.71

The balanced
nuclear equations are:


20.72

When an isotope is above the belt of
stability, the neutron/proton ratio is too high. The only mechanism to correct this
situation is beta emission; the process turns a neutron into a proton. Direct neutron emission does not occur.
Oxygen18 is a stable isotope.


20.73

Because both Ca
and Sr belong to Group 2A, radioactive strontium that has been ingested
into the human body becomes concentrated in bones (replacing Ca) and can
damage blood cell production.
Since radioactive decay kinetics are firstorder, the
decay rate is given by R = kN, where k is the decay constant and N
is the number or radioactive nuclei.
A curie (1 Ci) is defined as 3.70 × 10^{10}
disintegrations per second. So, the activity in curies is:


20.74

The age of the
fossil can be determined by radioactively dating the age of the deposit
that contains the fossil.


20.75

Normally the
human body concentrates iodine in the thyroid gland. The purpose of the large doses of KI is
to displace radioactive iodine from the thyroid and allow its excretion
from the body.


20.76

a.

The nuclear equation is:

b.

Xray analysis
only detects shapes, particularly of metal objects. Bombs can be made in a variety of
shapes and sizes and can be constructed of "plastic"
explosives. Thermal neutron
analysis is much more specific than Xray analysis. However, articles that are high in
nitrogen other than explosives (such as silk, wool, and polyurethane)
will give "false positive" test results.



20.77

a.



b.





20.78

Because of the
relative masses, the force of gravity on the sun is much greater than it is
on Earth. Thus the nuclear particles
on the sun are already held much closer together than the equivalent
nuclear particles on the earth. Less
energy (lower temperature) is required on the sun to force fusion
collisions between the nuclear particles.


20.79

Step 1: The halflife of carbon14 is 5715 years. From the halflife, we can calculate the
rate constant, k.
Step 2: The age of the object can now be
calculated using the following equation.
N =
the number of radioactive nuclei.
In the problem, we are given disintegrations per second per
gram. The number of disintegrations
is directly proportional to the number of radioactive nuclei. We can write,
t = 2.77 ´
10^{3} yr


20.80

The
neutrontoproton ratio for tritium equals 2 and is thus outside the belt
of stability. In a more
elaborate analysis, it can be shown that the decay of tritium to ^{3}He
is exothermic; thus, the total energy of the products is less than the
reactant.


20.81

x
= 1422
Percent of C14 left 0.070%


20.82

a.

The balanced equation
is:

b.

First, calculate the
rate constant k.
Then,
calculate the age of the rock by substituting k into the following equation. (N_{t} = 0.18N_{0})
t =
3.0 ´
10^{9} yr



20.83

a.

In the ^{90}Sr
decay, the mass defect is:
Dm
= (mass ^{90}Y +
mass e^{})

mass ^{90}Sr
= [(89.907152 amu + 5.4857 ´
10^{4}
amu) 
89.907738 amu] = −3.743 ´
10^{−5} amu
The energy change is given
by:
DE = (Dm)c^{2}
= (6.216
´
10^{32}
kg)(3.00 ´
10^{8} m/s)^{2}
= 5.59
´
10^{15}
kg m^{2}/s^{2}
= 5.59
´
10^{15}
J
Similarly, for the ^{90}Y
decay, we have
Dm
= (mass ^{90}Zr + mass e^{})

mass ^{90}Y
= [(89.904703 amu + 5.4857 ´
10^{4}
amu) 
89.907152 amu] = −1.900 ´
10^{−3} amu
and the energy change is:
DE = (3.156
´
10^{30}
kg)(3.00 ´
10^{8} m/s)^{2}
= 2.84
´
10^{13}
J
The energy released in the
above two decays is 5.59 ´
10^{15}
J and 2.84 ´
10^{13}
J. The total amount of energy
released is:
(5.59
´
10^{15}
J) + (2.84 ´
10^{13}
J) = 2.90 ´
10^{13}
J.

b.

This calculation requires that
we know the rate constant for the decay.
From the halflife, we can calculate k.
k =
To calculate the number of
moles of ^{90}Sr decaying in a year, we apply the following
equation:
where x is the number of moles of ^{90}Sr nuclei left
over. Solving, we obtain:
x
= 0.9756 mol ^{90}Sr
Thus
the number of moles of nuclei which decay in a year is
(1.00

0.9756) mol = 0.0244 mol =
0.024 mol
This is a reasonable number
since it takes 28.1 years for 0.5 mole of ^{90}Sr to decay.

c.

Since the halflife
of ^{90}Y is much shorter than that of ^{90}Sr, we can
safely assume that all the ^{90}Y
formed from ^{90}Sr will be converted to ^{90}Zr. The energy changes calculated in part
(a) refer to the decay of individual nuclei. In 0.024 mole, the number of nuclei
that have decayed is:
Realize that there are two
decay processes occurring, so we need to add the energy released for each
process calculated in part (a).
Thus, the heat released from 1 mole of 90Sr waste in a year is given by:
This amount is roughly equivalent to the heat
generated by burning 50 tons of coal!
Although the heat is released slowly during the course of a year,
effective ways must be devised to prevent heat damage to the storage
containers and subsequent leakage of radioactive material to the
surroundings.



20.84

A radioactive
isotope with a shorter halflife because more radiation would be emitted
over a certain period of time.


20.85

One curie represents 3.70 ´ 10^{10}
disintegrations/s. The rate of decay
of the isotope is given by the rate law:
rate = kN, where N is the number of atoms in the
sample and k is the firstorder
rate constant. We find the value of k in units of s^{1}:
Now, we can calculate N, the number of Ra atoms in the sample.
rate
= kN
3.7
´
10^{10} disintegrations/s
= (1.37 ´
10^{11}
s^{1})N
N
= 2.70 ´
10^{21} Ra atoms
By definition, 1 curie corresponds to exactly 3.7 ´
10^{10} nuclear disintegrations per second which is the decay rate
equivalent to that of 1 g of radium. Thus, the mass of 2.70 ´
10^{21} Ra atoms is 1 g.
6.1 × 10^{23}
atoms/mol = N_{A}


20.86

First, let’s calculate the number of disintegrations/s
to which 7.4 mCi corresponds.
This is the rate of decay. We can now calculate the number of
iodine131 atoms to which this radioactivity corresponds. First, we calculate the halflife in
seconds:
rate = kN
Therefore,
2.74
´
10^{8} disintegrations/s
= (9.90 ´
10^{7}
s^{1})N
N =
2.8 ´
10^{14} iodine131 atoms


20.87

a.

Dm =
234.03596 + 4.002603  238.05078 =
0.01222
amu
DE = 2.029 × 10^{}^{29}
kg (3.00 × 10^{8} m/s)^{2} = 1.83 × 10^{}^{12} J

b.

The a
particle will move away faster because it is smaller.



20.88

There was
radioactive material inside the box.


20.89

U238,
and Th232,
They are still present because of their long halflives.


20.90

All except
gravitational have a nuclear origin.


20.91

This wavelength is clearly in the
gray
region of the electromagnetic spectrum.


20.92

The a particles emitted by ^{241}Am
ionize the air molecules between the plates. The voltage from the battery makes one
plate positive and the other negative, so each plate attracts ions of
opposite charge. This creates a
current in the circuit attached to the plates. The presence of smoke particles between
the plates reduces the current, because the ions that collide with smoke
particles (or steam) are usually absorbed (and neutralized) by the
particles. This drop in current
triggers the alarm.


20.93

Only ^{3}H has a suitable halflife. The other halflives are either too long
or too short to determine the time span of 6 years accurately.


20.94

a.

The nuclear
submarine can be submerged for a long period without refueling.

b.

Conventional diesel engines receive an input of oxygen. A nuclear reactor does not.



20.95

A small scale
chain reaction (fission of ^{235}U) took place. Copper played the crucial role of
reflecting neutrons from the splitting uranium235 atoms back into the
uranium sphere to trigger the chain reaction. Note that a sphere has the most
appropriate geometry for such a chain reaction. In fact, during the implosion process
prior to an atomic explosion, fragments of uranium235 are pressed roughly
into a sphere for the chain reaction to occur (see Section 20.5 of the
text).


20.96

From the halflife, we
can determine the rate constant, k. Next, using the firstorder integrated
rate law, we can calculate the amount of copper remaining. Finally, from the initial amount of Cu
and the amount remaining, we can calculate the amount of Zn produced.
Next, plug the amount of
copper, the time, and the rate constant into the firstorder integrated
rate law, to calculate the amount of copper remaining.
grams
Cu remaining = 31.0 g
The quantity of Zn
produced is:
g Zn =
initial g Cu  g Cu remaining =
84.0 g 
31.0 g = 53.0
g Zn


20.97

In this problem, we are asked
to calculate the molar mass of a radioactive isotope. Grams of sample are given in the problem,
so if we can find moles of sample we can calculate the molar mass. The rate constant can be calculated from
the halflife. Then, from the rate
of decay and the rate constant, the number of radioactive nuclei can be
calculated. The number of
radioactive nuclei can be converted to moles.
First, we convert the halflife
to units of minutes because the rate is given in dpm (disintegrations per
minute). Then, we calculate the rate
constant from the halflife.
Next, we calculate the number of radioactive nuclei from the rate and
the rate constant.
rate = kN
2.9 ´ 10^{4}
dpm = (1.0 ´ 10^{15}
min^{1})N
N = 2.9 ´ 10^{19}
nuclei
Convert to moles of nuclei, and then determine the molar mass.


20.98

a.

First, we calculate the rate
constant using Equation 14.5 of the text.
Next, we use Equation 14.3 of
the text to calculate the number of ^{27}Mg nuclei remaining
after
30.0 minutes.
N_{t} =
4.71 × 10^{11} Mg nuclei remain

b.

The activity is given
by
activity = kN
We first convert the
rate constant to units of s^{1}.
At t = 0,
activity = (1.22 × 10^{3} s^{1})(4.20 × 10^{12}
nuclei) = 5.12 × 10^{9} decays/s
At t = 30.0 minutes,
activity = (1.22 × 10^{3} s^{1})(4.71 × 10^{11}
nuclei) = 5.75 × 10^{8} decays/s

c.

The probability is just the
firstorder decay constant, 1.22 × 10^{3} s^{1}. This is valid if the
halflife is large compared to one second, which is true in this case.



20.99

a.

The volume of a sphere is
Volume is proportional to the number of nucleons. Therefore,
V µ A (mass number)
r^{3} µ A
r =
r_{0}A^{1/3}, where r_{0}
is a proportionality constant.

b.

We can calculate the volume of the ^{238}U nucleus by
substituting the equation derived in part (a) into the equation for the
volume of a sphere.



20.100

The ignition of a fission bomb requires an ample supply of
neutrons. In addition to the normal
neutron source placed in the bomb, the high temperature attained during the
chain reaction causes a small scale nuclear fusion between deuterium and
tritium.
The additional neutrons produced will enhance the efficiency of the
chain reaction and result in a more powerful bomb.


20.101

The halflife is the time required for the reactant
concentration to drop to half its original value. After
45 days, there will be 7.5 isotopes of X remaining. After an additional 45 days (total time =
90 days), there will be 3.75 isotopes of X remaining. After 105 days (an additional onethird
of the halflife), there will be approximately 3 isotopes of X remaining.
Diagram (b)
most closely represents the sample of X after 105 days.


20.102

a.


b.


c.

Assume that all the energy
corresponding to the mass defect ends up as kinetic energy of the alpha
particle.

d.

Calculate the number of Polonium atoms in 1µg:
Since the halflife of
polonium210 is 138 days, half of the original atoms will decay in 138
days (one halflife). The number of alpha particles produced during this
138day period is equal to the number of
^{210}Po atoms that disintegrated:
.
Using the result from part (c):
(1.5×10^{15} alpha
particles)(1.03×10^{–12} J/alpha particle) = 1.5×10^{3} J = 1.5 kJ



20.103

Strategy:

We are asked to find the volume
of helium formed at STP from the radioactive decay of radium226. The
volume is easily calculated using the ideal gas law:
V = n_{He}RT/P.
The overall decay
process produces 5 mol He for each mole of ^{226}Ra that decays:
mol He = 5(mol ^{226}Ra
decayed) = 5(initial mol ^{226}Ra – final
mol ^{226}Ra).
The initial amount of ^{226}Ra
can be found easily from the initial mass (1.00 g) and the molar mass
(226 g/mol) of radium226. But, calculating the final amount of ^{226}Ra
does not appear to be straightforward since the decay sequence from
radium226 to lead206 involves several intermediate radioisotopes, each
with its own halflife. Fortunately, though, the halflife of the first
step in the decay sequence (radium226 ®
radon222, halflife 1600 years) is significantly longer than any other
halflife in the sequence, so we can approximate the overall decay
process using firstorder kinetics and the rate constant of the slow
(first) step:
^{226}Ra ® ^{206}Pb +
5 ^{4}He t_{1/2} » 1600 yr
Plugging the known values into
the above equation, we can solve it for the final moles of ^{226}Ra,
which in turn gives us the moles of He produced and the volume of He
produced.

Solution:

Calculate the initial moles of ^{226}Ra:
.
Next, find the final moles of ^{226}Ra:
The number of moles of He produced is:
mol He = 5(0.00422 mol – 0.00419 mol) = 1.5×10^{–4} mol
He
Finally, the volume of He at STP is:



20.104

Since the new
particle’s mass exceeds the sum of the masses of the electron and positron,
the process violates the law of conservation of mass. But, it does not
violate Einstein’s more general law of massenergy conservation, DE = Dmc^{2}. The large mass of
the new particle reflects that fact that the process is extremely
endothermic.


20.105

a.

Strategy:

The species written first is a reactant. The species written last is a
product. Within the parenthesis,
the bombarding particle (a reactant) is written first, followed by the
emitted particle (a product).
Determine the atomic number for the unknown species,
X, by summing the atomic numbers on both sides of the reaction.
Σ
reactant atomic numbers = Σ product atomic numbers
Use the atomic number to determine the identity of
the unknown species.

Setup:

The bombarding and emitted particles are represented
by and γ, respectively.
The atomic number of Ir is 77.
Gamma emission has no effect on either the atomic
number or the mass number.
Neutron activation has no effect on the atomic number and
increases the mass number by one.
Therefore, The atomic number of X = (77 + 0) =77

Solution:

X =
The balanced nuclear equation is


b.

Strategy:

To calculate the nuclear binding energy, we first
determine the difference between the mass of the nucleus and the mass
of all the protons and neutrons, which yields the mass defect. Next, we must apply Einstein’s
massenergy relationship [DE = (Dm)c^{2}]. The nuclear binding energy per
nucleon is given by the following equation:

Solution:

There are 53 protons and 72 neutrons in the^{ 125}I
nucleus. The mass of 53 atoms is:
53 × 1.007825
= 53.41473 amu
and the mass of 72 neutrons is:
72 × 1.008665
amu = 72.62388 amu
Therefore, the predicted mass for is 53.41473 + 72.62388 = 126.03861 amu, and
the mass defect is:
Dm =
124.904624  126.03861 amu = 1.13399
amu
The energy released is:
=
−1.69 × 10^{−10}
Note that we report the binding energy without
referring to its sign.
Therefore, the nuclear binding energy is 1.69 × 10^{10} J.
The binding energy per
nucleon for ^{125}I is:



