Chapter 20, Problem 11E

(a)

Interpretation Introduction

Interpretation:

The equation for the decay of $Am$ by alpha production needs to be determined.

Concept Introduction:

Radioactive decay of unstable nuclei leads to formation of smaller nuclei that are stable and during this course alpha, beta and gamma radiations are emitted. When alpha particles are released, the new nuclei will display a decrease of 4 units in mass number and decrease of 2 units in atomic number.

Answer to Problem 11E

The equation for the decay of $Am$ by alpha production is $Am\to Np+He$ .

Explanation of Solution

The decay of $Am$ with production of alpha particle is depicted by equation as $Am\to ?+He$ .

While computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 241 =? + 4

x = 237

Atomic number: 95 =? + 2

y= 93

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 237 and atomic number 93 is Np. Hence, the element is $Np$ . The release of alpha particle during decay results in mass number showing a decrease of 4 units and atomic number showing decrease of 2 units.

Hence, the decay of the decay of $Am$ by alpha production is $Am\to Np+He$ .

(b)

Interpretation Introduction

Interpretation:

The balanced equation for the complete decay of $Am$ needs to be determined, wherein the, a, β a, a, β a, a, a, β a and β are emitted successively.

Concept Introduction:

Alpha decay involves the release of alpha particle, namely, $He$ with a new nucleus which will have mass number reduced by 4 units and atomic number reduced by 2 units.

Beta decay involves the release of beta particle, namely, $e$ with a new nucleus which will have same mass number but atomic number decreased by 1 unit.

Answer to Problem 11E

The final product obtained complete decay of $Am$ is $Bi$ .

Explanation of Solution

The complete decay of $Am$ in which a, a,β, a, a,β, a ,a, a,β, a and β are emitted successively show that 8 a and 4 β are involved.

In other words, it indicates that $8\text{ x }He$ and $4\text{ x}e$ which implies that there should be

Mass number = 8 x 4 + 0 = 32 units

Atomic number = 16 + (-4) = 12 units

While computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 241 = 0 + 32

x = 209

Atomic number: 95 =? + 12

y= 83

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 209 and atomic number 83 is Bi. Hence, the element is $Bi$ .

The decay of $Am$ with successive emission of 8 a and 4 β gives the final product as $Bi$ .

(c)

Interpretation Introduction

Interpretation:

The intermediate products formed by complete decay of $Am$ needs to be determined in which a, a, β a, a, β a, a, a, β a and β are emitted successively.

Concept Introduction:

There is a release of alpha particle in alpha decay wherein the emitted is $He$ along with a new nucleus that will have a reduction of mass number by 4 units and atomic number by 2 units.

In Beta decay, there is of a beta particle which is $e$ along with a new nucleus that displays the same mass number with reduction in atomic number by 1 unit.

Answer to Problem 11E

The various intermediates formed by complete decay of $Am$ are depicted $Np$ , $Pa$ , $U$ , $Th$ , $Ra$ , $Ac$ , $F$ , $At$ , $Bi$ , $Po$ , $Pb$ , $Bi$ .

Explanation of Solution

-The alpha decay of $Am$ is depicted as $Am\to ?+He$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 241 =? + 4

x = 237

Atomic number: 95 =? + 2

y= 93

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 237 and atomic number 93 is Np. Hence, the element is $Np$ .

The alpha decay of $Am$ is depicted as $Am\to Np+He$

-The alpha decay of $Np$ is depicted as $Np\to ?+He$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 237 =? + 4

x = 230

Atomic number: 93 =? + 2

y= 91

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 233 and atomic number 91 is Pa. Hence, the element is $Pa$ .

The alpha decay of $Np$ is depicted as $Np\to Pa+He$

-The beta decay of $Pa$ is $Pa\to ?+e$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 233 =? + 0

x = 233

Atomic number: 91 =? + (-1)

y= 92

\Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 233 and atomic number 92 is U. Hence, the element is $U$ .

The beta decay of $Pa$ is depicted as $Pa\to U+e$

-The alpha decay of $U$ is depicted as $U\to ?+He$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 233 =? + 4

x = 229

Atomic number: 92 =? + 2

y= 90

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 229 and atomic number 90 is Th. Hence, the element is $Th$ .

The alpha decay of $U$ is depicted as $U\to Th+He$

-The alpha decay of $Th$ is depicted as $Th\to ?+He$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 229 =? + 4

x = 225

Atomic number: 90 =? + 2

y= 88

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 225 and atomic number 88 is Ra. Hence, the element is $Ra$ .

The alpha decay of $Th$ is depicted as $Th\to Ra+He$

-The beta decay of $Ra$ is $Ra\to ?+e$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 225 =? + 0

x = 225

Atomic number: 88 =? + (-1)

y= 89

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 225 and atomic number 89 is Ac. Hence, the element is $Ac$ .

The beta decay of $Ra$ is depicted as $Ra\to Ac+e$

-The alpha decay of $Ac$ is depicted as $Ac\to ?+He$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 225 =? + 4

x = 221

Atomic number: 89 =? + 2

y= 87

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 221 and atomic number 87 is Fr. Hence, the element is $Fr$ .

The alpha decay of $Ac$ is depicted as $Ac\to Fr+He$

-The alpha decay of $Fr$ is depicted as $Fr\to ?+He$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 221 =? + 4

x = 217

Atomic number: 87 =? + 2

y= 85

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 217 and atomic number 85 is at. Hence, the element is $At$

The alpha decay of $Fr$ is depicted as $Fr\to At+He$

-The alpha decay of $At$ is depicted as $At\to ?+He$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 217 =? + 4

x = 213

Atomic number: 85 =? + 2

y= 83

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 213 and atomic number 83 is Bi. Hence, the element is $Bi$ .

The alpha decay of $At$ is depicted as $At\to Bi+He$

-The beta decay of $Bi$ is $Bi\to ?+e$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 213 =? + 0

x = 213

Atomic number: 83 =? + (-1)

y= 84

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 213 and atomic number 84 is Po. Hence, the element is $Po$ .

The beta decay of $Bi$ is depicted as $Bi\to Po+e$

-The alpha decay of $Po$ is depicted as $Po\to ?+He$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 213 =? + 4

x = 209

Atomic number: 84 =? + 2

y= 82

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 209 and atomic number 82 is Pb. Hence, the element is $Pb$ .

The alpha decay of $Po$ is depicted as $Po\to Pb+He$

-The beta decay of $Pb$ is $Pb\to ?+e$

Computing the mass numbers and atomic mass for unknown element, it is found that

Mass number: 209 =? + 0

x = 209

Atomic number: 82 =? + (-1)

y= 83

Hence unknown element is $x$ . Looking through the periodic table, it is found that element with mass number 209 and atomic number 83 is Bi. Hence, the element is $Bi$ .

The beta decay of $Pb$ is depicted as $Pb\to Bi+e$ .