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Chapter 101

Chapter 10 Nuclear Chemistry Solutions to In-Chapter Problems 10.1 Refer to Example 10.1 to answer the question.

The atomic number (Z) = the number of protons. The mass number (A) = the number of protons + the number of neutrons. Isotopes are written with the mass number to the upper left of the element symbol and the

atomic number to the lower left of the element symbol.

Atomic Number Mass Number Number of

Protons Number of Neutrons

Isotope Symbol

Cobalt-59 27 59 27 32 27Co59

Cobalt-60 27 60 27 33 27Co

60

10.2 Fill in the table as in Example 10.1, using the rules in Answer 10.1.

Atomic Number Mass Number Number of Protons Number of Neutrons

Sr3885a.

38 85 38 47

Ga3167b.

31 67 31 36 Selenium-75c. 34 75 34 41

10.3 An particle has no electrons around the nucleus and a +2 charge, whereas a helium atom has

two electrons around the nucleus and is neutral. 10.4 Use Table 10.1 to identify Q in each of the following symbols.

Q42Q01 Qa. b. c.0

+1

! particle " particle positron 10.5 Write a balanced nuclear equation as in Example 10.2.

[1] Write an incomplete equation with the original nucleus on the left and the particle emitted on the right. Radon-222 has an atomic number of 86.

222 He2

4 + ?86Rn [2] Calculate the mass number and the atomic number of the newly formed nucleus on the

right. Radon-222 emits an particle. Mass number: Subtract the mass of an particle (4) to obtain the mass of the new

nucleus; 222 4 = 218. Atomic number: Subtract the two protons of an particle to obtain the atomic number of

the new nucleus; 86 2 = 84.

[3] Use the atomic number to identify the new nucleus and complete the equation.

Nuclear Chemistry 102

From the periodic table, the element with an atomic number of 84 is polonium, Po. Write the mass number and the atomic number with the element symbol to complete the

equation.

2He + 84Po222 4 21886Rn

10.6 Work backwards to write the equation that produces radon-222.

?

22688Ra

2He + 86Rn2224

2He + 86Rn2224Mass number = 222 + 4 = 226

Atomic number = 86 + 2 = 88 10.7 Write the balanced nuclear equation for each isotope as in Example 10.2 or Answer 10.5.

84Po218 4

2He + 82Pb214a.

b. 90Th230

2He + 88Ra4 226

c. 99Es252

2He + 97Bk4 248

10.8 Write a balanced nuclear equation for the emission of iodine-131 as in Example 10.2.

[1] Write an incomplete equation with the original nucleus on the left and the particle emitted on the right. Use the identity of the element to determine the atomic number; iodine has an atomic

number of 53.

I53131 +e1

0 ?

[2] Calculate the mass number and the atomic number of the newly formed nucleus on the

right. Mass number: Since a particle has no mass, the masses of the new particle and the

original particle are the same, 131. Atomic number: Since emission converts a neutron into a proton, the new nucleus has

one more proton than the original nucleus; 53 = 1 + ?. Thus, the new nucleus has an atomic number of 54.

[3] Use the atomic number to identify the new nucleus and complete the equation.

From the periodic table, the element with an atomic number of 54 is xenon, Xe. Write the mass number and the atomic number with the element symbol to complete the

equation.

I53131 +e1

0 Xe54131

Chapter 103

10.9 Write a balanced nuclear equation for the emission of each isotope as in Example 10.2 and Answer 10.8.

F9

20 +e10 Ne

1020a.

Sr3892 +e1

0 Y3992b.

c. Cr2455 +e1

0 Mn2555

10.10 Write a balanced nuclear equation for positron emission as in Example 10.3. a. [1] Write an incomplete equation with the original nucleus on the left and the particle

emitted on the right. Use the identity of the element to determine the atomic number; arsenic has an atomic

number of 33.

As3374 + ?e+1

0

[2] Calculate the mass number and the atomic number of the newly formed nucleus on the

right. Mass number: Since a + particle has no mass, the masses of the new particle and the

original particle are the same, 74. Atomic number: Since + emission converts a proton into a neutron, the new nucleus has

one fewer proton than the original nucleus; 33 1 = 32. Thus, the new nucleus has an atomic number of 32.

[3] Use the atomic number to identify the new nucleus and complete the equation.

From the periodic table, the element with an atomic number of 32 is germanium, Ge. Write the mass number and the atomic number with the element symbol to complete the

equation.

As3374 +e+1

0 Ge3274

b. Use the same steps as in part (a).

O815 +e+1

0 N715

10.11 When and emission occur together, the atomic number increases by one (due to the

particle), and a ray is released.

Ir77192 + e1

0Pt78192 +

Both ! particles and " rays are emitted.

"

10.12 When emission occurs alone, there is no change in the atomic number or the mass number.

When and emission occur together, the atomic number increases by one (due to the particle), and a ray is released.

Nuclear Chemistry 104

Both ! particles and " rays are emitted.

40 + "K1940 + e1

0B511 + " 20Cab.B5

11

" emission aloneno change in atomic number or mass number

Atomic number increases, but no change in the mass number.

a.

10.13 To calculate the amount of radioisotope present after the given number of half-lives, multiply the

initial mass by for each half-life.

xinitial mass

12

x 12

1.00 g = 0.250 g of phosphorus-32 remains.

The mass is halved two times.

a.

xinitial mass

12

x 12

x 12

x 12

1.00 g = 0.0625 g of phosphorus-32 remains.

The mass is halved four times.

b.

xinitial mass

12

x 12

x 12

x 12

1.00 g

= 0.00391 g of phosphorus-32 remains.The mass is halved eight times.

c. x12

x 12

x 12

x 12

xinitial mass

12

x 12

x 12

x 12

1.00 g

= 9.54 x 107 g of phosphorus-32 remains.The mass is halved twenty times.

d. x12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

x 12

10.14 To calculate the amount of radioisotope present, first determine the number of half-lives that

occur in the given amount of time. Then multiply the initial mass by for each half-life to determine the amount present.

6.0 hours 1 half-life

6.0 hoursx = 1.0 half-life x

initial mass

12

160. mg = 80. mg of Tc-99ma.

18.0 hours 1 half-life6.0 hours

x = 3.0 half-lives xinitial mass

12

160. mg = 20.0 mg of Tc-99mb. x 12x 1

2

24.0 hours 1 half-life

6.0 hoursx = 4.0 half-lives x

initial mass

12

160. mg = 10.0 mg of Tc-99mc. x 12x 1

2x 1

2

2 days 1 half-life

6.0 hoursx = 8 half-lives

xinitial mass

12

160. mg = 0.6 mg of Tc-99m

d.

x 12

x 12

x 12

24 hours1 day

x

x 12

x 12

x 12

x 12

Chapter 105

10.15 If an artifact has 1/8 of the amount of C-14 compared to living organisms, it has decayed by three half-lives ( ).

1 half-life

5,730 yearsx =3 half-lives 17,200 years

10.16 Use the amount of radioactivity (mCi/mL) as a conversion factor to convert the dose of radioactivity from millicuries to a volume in milliliters.

x =110 mCi dose 1 mL25 mCi

Millicuries cancel.

4.4 mLAnswer

mCimL conversion factor

10.17 Use the conversion factor given to convert mrem to rem.

x =200 mrem 1 rem1000 mrem

0.2 rema. x =0.014 rem1 rem

1000 mrem 14 mremb.

larger dose 10.18 Calculate the number of half-lives that pass in nine days.

9 days 1 half-life3 days

x = 3 half-lives 12

=x 12

x 12

18

10.19

b. Sm62153 + e10Eu63

153

a. Sm62153

62 protons, 62 electrons153 62 = 91 neutrons

One ! particleis emitted.

Atomic number increases.

c. xinitial activity

12

150 mCi = 9.4 mCix 12x 1

2x 1

2

10.20 The emission of a positron decreases the atomic number by one, but the mass number stays the

same. Use Example 10.3.

N713 +e+1

0 C613

Nuclear Chemistry 106

10.21 Nuclear fission splits an atom into two lighter nuclei. Write the equation with the information given, and then balance the equation.

92U235 + 3 0n51Sb +

133 1+ 0n1

235 + 3 0n51Sb +133

41Nb100 1+ 0n

1

?

The atomic number must be 41 = niobium:92 = 51 + X

The mass number of niobium must be 100:235 + 1 = 133 + X + 3, X = 100

92U 10.22 Balance each reaction.

+ +12 H1

1 H e+10a. 1

1 H b. 1

1 H + 12 H 2

3 He c. 1

1 H + 23 He 2

4 He + e+10

Solutions to End-of-Chapter Problems 10.23 Refer to Example 10.1 to answer the question.

The atomic number (Z) = the number of protons. The mass number (A) = the number of protons + the number of neutrons. Isotopes are written with the mass number to the upper left of the element symbol and the

atomic number to the lower left of the element symbol