Chapter 20

Nuclear Chemistry

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 sodium-23 (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 iron-56 (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 neutron-to-proton 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.	Lithium-9 should be less stable. The neutron-to-proton ratio is too high. For small atoms, the n/p ratio will be close to 1:1.
b.	Sodium-25 is less stable. Its neutron-to-proton ratio is probably too high.
c.	Scandium-48 is less stable because of odd numbers of protons and neutrons. We would not expect calcium-48 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.	Neon-17 should be radioactive. It falls below the belt of stability (low n/p ratio).
b.	Calcium-45 should be radioactive. It falls above the belt of stability (high n/p ratio).
c.	All technetium isotopes are radioactive.
d.	Mercury-195 should be radioactive. Mercury-196 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 ´ 10⁹ 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 mass-energy relationship [DE = (Dm)c²], (Equation 20.1); and use the procedure shown in Sample Problem 20.2.

Solution:

a.

There are 4 neutrons and 3 protons in a Li-7 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 ^7Li is:

b.

Using the same procedure as in (a), using 1.007825 amu for and 1.008665 amu for, we can show that for chlorine-35: 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 helium-4 nucleus into 2 protons and 2 neutrons. For the helium-4 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 ^48Cr 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 ^48Cr atom (ignore the mass of the electrons): mass ^48Cr 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 ^192Ir 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 ^192Ir atom (ignore the mass of the electrons): mass ^192Ir 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:

a.
b.
c.

20.26

According to Section 20.3, the number of radioactive nuclei at time zero (N₀) and time t (N_t) is

and the corresponding half-life 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 half-life 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 half-life.

2.78 d

20.27

The number of atoms decreases by half for each half-life. For ten half-lives we have:

20.28

Since all radioactive decay processes have first-order rate laws, the decay rate is proportional to the amount of radioisotope at any time. The half-life is given by the following equation:

(1)

There is also an equation that relates the number of nuclei at time zero (N₀) 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 half-life.

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 half-life.

3.9 × 10³ 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 half-life. 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 half-life.

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 first-order 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 half-lives (30 days), there should be
0.25 mole of B left.

For C: As in the case of A, the half-life of C is also very short. Therefore, at the end of the 30-day 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 half-life of carbon-14 to calculate the rate constant, k.

Rearranging the equation to solve for t gives

t4.05 × 10³ 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 Nt = N0(0.361) = (15.3 dpm)(0.361) = 5.5 dpm

20.34

We first convert the mass of ²⁰⁶Pb to the mass of ²³⁸U that decayed to produce it:

We find the mass of ²³⁸U at t = 0 by adding this mass to the mass of ²³⁸U in the rock now.

Mass of ²³⁸U at t = 0 = 1.09 g + 0.09 g = 1.18 g

We use the half-life 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 × 10⁸ 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 ^238U at t = 0 to ^238U 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 ^238U at t = 0 to ^238U 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 ^238U now is 1.000 g. This gives a mass of 1.027 g ^238U at t = 0. The difference between these masses is the mass of ^238U that has decayed to ^206Pb. However, because a nucleus loses mass as it decays, the mass of ^206Pb will be less than the mass of ^238U that decayed. We determine the mass of ^206Pb as follows: Finally, if there currently a mass of 1.000 g ^238U and a mass of 0.023 g ^206Pb, the ratio of ^238U to ^206Pb 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).

a.
b.
c.

20.39

a.	^14N(a,p)^17O

b.	^9Be(a,n)^12C

c.	^238U(d,2n)^238Np

20.40

a.

b.

c.

20.41

a.	^40Ca(d,p)^41Ca

b.	^32S(n,p)^32P

c.	^239Pu(a,n)^242Cm

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 high-intensity alpha particle emitter. Any heavy element like plutonium or curium will do. Place the bismuth-209 sample next to the alpha emitter and wait. The reaction is:

20.43

Upon bombardment with neutrons, mercury-198 is first converted to mercury-199, 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 I₂ shows that the IO is formed only from the

IO. Does this result rule out the possibility that I₂ 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 iron-59 to the person’s diet, and allow a few days for the iron-59 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 electron-capture decay of ^125I, 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 ^124Xe is: The balanced equation for the electron capture of ^125Xe is: Mass number of X = (125 + 0) = 125. The atomic number of X = (53 – 1) = 52. The balanced equation for the electron capture of ^125I is: The product of the electron-capture 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 first-order rate law (Equation 14.3):

We can calculate the rate constant, k, from the half-life 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. Even-even 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 neptunium-237 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

a.
b.
c.
d.

20.70

We use the same procedure as in Problem 20.20.

Isotope Atomic Mass Nuclear Binding Energy

(amu) (J/nucleon)

a.	^10B 10.0129 1.040 ´ 10^-12
b.	^11B 11.009305 1.111 ´ 10^-12
c.	^14N 14.003074 1.199 ´ 10^-12
d.	^56Fe 55.93494 1.410 ´ 10^-12

20.71

The balanced nuclear equations are:

a.
b.
c.
d.

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.

Oxygen-18 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 first-order, 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 × 10¹⁰ 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.	X-ray 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 X-ray 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 half-life of carbon-14 is 5715 years. From the half-life, 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 ´ 10³ yr

20.80

The neutron-to-proton 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 ³He is exothermic; thus, the total energy of the products is less than the reactant.

20.81

x = 1422

Percent of C-14 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. (Nt = 0.18N0) t = 3.0 ´ 10^9 yr

20.83

a.	In the ^90Sr decay, the mass defect is: Dm = (mass ^90Y + mass e^-) - mass ^90Sr = [(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 ^90Y decay, we have Dm = (mass ^90Zr + mass e^-) - mass ^90Y = [(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 half-life, we can calculate k. k = To calculate the number of moles of ^90Sr decaying in a year, we apply the following equation: where x is the number of moles of ^90Sr nuclei left over. Solving, we obtain: x = 0.9756 mol ^90Sr 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 ^90Sr to decay.
c.	Since the half-life of ^90Y is much shorter than that of ^90Sr, we can safely assume that all the ^90Y formed from ^90Sr will be converted to ^90Zr. 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 half-life because more radiation would be emitted over a certain period of time.

20.85

One curie represents 3.70 ´ 10¹⁰ 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 first-order 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 ´ 10¹⁰ disintegrations/s = (1.37 ´ 10^-11 s^-1)N

N = 2.70 ´ 10²¹ Ra atoms

By definition, 1 curie corresponds to exactly 3.7 ´ 10¹⁰ nuclear disintegrations per second which is the decay rate equivalent to that of 1 g of radium. Thus, the mass of 2.70 ´ 10²¹ Ra atoms is 1 g.

6.1 × 10²³ 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 iodine-131 atoms to which this radioactivity corresponds. First, we calculate the half-life in seconds:

rate = kN

Therefore,

2.74 ´ 10⁸ disintegrations/s = (9.90 ´ 10^-7 s^-1)N

N = 2.8 ´ 10¹⁴ iodine-131 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

U-238, and Th-232,

They are still present because of their long half-lives.

20.90

All except gravitational have a nuclear origin.

20.91

This wavelength is clearly in the g-ray region of the electromagnetic spectrum.

20.92

The a particles emitted by ²⁴¹Am 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 ³H has a suitable half-life. The other half-lives 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 ²³⁵U) took place. Copper played the crucial role of reflecting neutrons from the splitting uranium-235 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 uranium-235 are pressed roughly into a sphere for the chain reaction to occur (see Section 20.5 of the text).

20.96

From the half-life, we can determine the rate constant, k. Next, using the first-order 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 first-order 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 half-life. 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 half-life to units of minutes because the rate is given in dpm (disintegrations per minute). Then, we calculate the rate constant from the half-life.

Next, we calculate the number of radioactive nuclei from the rate and the rate constant.

rate = kN

2.9 ´ 10⁴ dpm = (1.0 ´ 10^-15 min^-1)N

N = 2.9 ´ 10¹⁹ 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 ^27Mg nuclei remaining after 30.0 minutes. Nt = 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 first-order decay constant, 1.22 × 10^-3 s^-1. This is valid if the half-life 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 = r0A^1/3, where r0 is a proportionality constant.
b.	We can calculate the volume of the ^238U 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 half-life 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 one-third of the half-life), 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 half-life of polonium-210 is 138 days, half of the original atoms will decay in 138 days (one half-life). The number of alpha particles produced during this 138-day period is equal to the number of ^210Po 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 radium-226. The volume is easily calculated using the ideal gas law: V = nHeRT/P. The overall decay process produces 5 mol He for each mole of ^226Ra that decays: mol He = 5(mol ^226Ra decayed) = 5(initial mol ^226Ra – final mol ^226Ra). The initial amount of ^226Ra can be found easily from the initial mass (1.00 g) and the molar mass (226 g/mol) of radium-226. But, calculating the final amount of ^226Ra does not appear to be straightforward since the decay sequence from radium-226 to lead-206 involves several intermediate radioisotopes, each with its own half-life. Fortunately, though, the half-life of the first step in the decay sequence (radium-226 ® radon-222, half-life 1600 years) is significantly longer than any other half-life in the sequence, so we can approximate the overall decay process using first-order kinetics and the rate constant of the slow (first) step: ^226Ra ® ^206Pb + 5 ^4He t1/2 » 1600 yr Plugging the known values into the above equation, we can solve it for the final moles of ^226Ra, which in turn gives us the moles of He produced and the volume of He produced.

Solution:

Calculate the initial moles of ^226Ra: . Next, find the final moles of ^226Ra: 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 mass-energy conservation, DE = Dmc². 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 mass-energy 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^125I 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 ^125I is: