The equation for radioactive decay of G 68 a where electron capture takes place needs to be determined. Concept Introduction : Electron capture involves the absorption of electrons by the proton rich nucleus from the inner shells. During this process, the proton changes to neutron releasing an electron and atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

Question

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Answer 1

Question

Chapter 20, Problem 7E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of G68a where electron capture takes place needs to be determined.

Concept Introduction:

Electron capture involves the absorption of electrons by the proton rich nucleus from the inner shells. During this process, the proton changes to neutron releasing an electron and atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

(a)

Expert Solution

Answer to Problem 7E

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

Explanation of Solution

The radioactive decay of G68a where electron capture takes place is depicted as as below

G3168a+e−10→?

(As Ga has atomic number as 31)

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

Mass Number: 68 + 0 =?

x = 68

Atomic number: 31 + (-1) =?

y= 30

Hence unknown element is x3068 . Looking through the periodic table, it is found that element with mass number 68 and atomic number 30 is Zn. Hence, the element is Z3068n . The electron capture releases electron in which the atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of C62u which releases positron needs to be determined.

Concept Introduction:

Beta decay takes place when there is excess of protons or neutrons which gets transformed to other. In beta plus decay, proton gets converted to neutron releasing positron and neutrino. The emission of positron will decrease the atomic number by 1 with no changes in mass number.

(b)

Expert Solution

Answer to Problem 7E

The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Explanation of Solution

The radioactive decay of C2962u which produces positron is depicted as below

C2962u→e10+?

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

Mass number: 62 = 0 +?

x = 62

Atomic Number: 29 =? + (1)

y= 28

Hence unknown element is x2862 . Looking through the periodic table, it is found that element with mass number 62 and atomic number 28 is Ni. Hence, the element is N2862i . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Hence, the overall equation for the radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of F212r where there is alpha particle release needs to be determined.

Concept Introduction:

Radioactive decay takes place when an atomic nucleus is unstable such that it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma. Release of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

(c)

Expert Solution

Answer to Problem 7E

The radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

Explanation of Solution

The radioactive decay of F87212r where alpha particles are released and the equation is depicted as

F87212r→?+H24e

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

Mass number: 212 = ?+4

x = 208

Atomic number: 87=? + 2

y= 85

Hence unknown element is x85208 . Looking through the periodic table, it is found that element with mass number 208 and atomic number 85 is Are Hence, the element is A85208r . The emission of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

The radioactive decay of F87212r with release of alpha particles is depicted as F87212r→A85208r+H24e .

Hence, the overall equation for the radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of S129b which emits beta particles needs to be explained.

Concept Introduction:

Radioactive decay releases radiations like alpha, beta and gamma, when unstable nuclei get converted to smaller stable fragments. The emission of β particle will increase the atomic number by 1 with no changes in mass number.

(d)

Expert Solution

Answer to Problem 7E

The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Explanation of Solution

The radioactive decay of S129b which emits beta particle is depicted as

S51129b→?+e−10

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

Mass number: 129 = x+ 0

x = 129

Atomic number: 51 = x + (- 1)

x= 52

Hence unknown element is x52129 . Looking through the periodic table, it is found that element with mass number129 and atomic number 52 is Te. Hence, the element is T52129e . The emission of beta particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Hence, the overall equation for the radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

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Answer 2

Question

Chapter 20, Problem 7E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of G68a where electron capture takes place needs to be determined.

Concept Introduction:

Electron capture involves the absorption of electrons by the proton rich nucleus from the inner shells. During this process, the proton changes to neutron releasing an electron and atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

(a)

Expert Solution

Answer to Problem 7E

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

Explanation of Solution

The radioactive decay of G68a where electron capture takes place is depicted as as below

G3168a+e−10→?

(As Ga has atomic number as 31)

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

Mass Number: 68 + 0 =?

x = 68

Atomic number: 31 + (-1) =?

y= 30

Hence unknown element is x3068 . Looking through the periodic table, it is found that element with mass number 68 and atomic number 30 is Zn. Hence, the element is Z3068n . The electron capture releases electron in which the atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of C62u which releases positron needs to be determined.

Concept Introduction:

Beta decay takes place when there is excess of protons or neutrons which gets transformed to other. In beta plus decay, proton gets converted to neutron releasing positron and neutrino. The emission of positron will decrease the atomic number by 1 with no changes in mass number.

(b)

Expert Solution

Answer to Problem 7E

The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Explanation of Solution

The radioactive decay of C2962u which produces positron is depicted as below

C2962u→e10+?

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

Mass number: 62 = 0 +?

x = 62

Atomic Number: 29 =? + (1)

y= 28

Hence unknown element is x2862 . Looking through the periodic table, it is found that element with mass number 62 and atomic number 28 is Ni. Hence, the element is N2862i . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Hence, the overall equation for the radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of F212r where there is alpha particle release needs to be determined.

Concept Introduction:

Radioactive decay takes place when an atomic nucleus is unstable such that it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma. Release of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

(c)

Expert Solution

Answer to Problem 7E

The radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

Explanation of Solution

The radioactive decay of F87212r where alpha particles are released and the equation is depicted as

F87212r→?+H24e

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

Mass number: 212 = ?+4

x = 208

Atomic number: 87=? + 2

y= 85

Hence unknown element is x85208 . Looking through the periodic table, it is found that element with mass number 208 and atomic number 85 is Are Hence, the element is A85208r . The emission of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

The radioactive decay of F87212r with release of alpha particles is depicted as F87212r→A85208r+H24e .

Hence, the overall equation for the radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of S129b which emits beta particles needs to be explained.

Concept Introduction:

Radioactive decay releases radiations like alpha, beta and gamma, when unstable nuclei get converted to smaller stable fragments. The emission of β particle will increase the atomic number by 1 with no changes in mass number.

(d)

Expert Solution

Answer to Problem 7E

The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Explanation of Solution

The radioactive decay of S129b which emits beta particle is depicted as

S51129b→?+e−10

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

Mass number: 129 = x+ 0

x = 129

Atomic number: 51 = x + (- 1)

x= 52

Hence unknown element is x52129 . Looking through the periodic table, it is found that element with mass number129 and atomic number 52 is Te. Hence, the element is T52129e . The emission of beta particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Hence, the overall equation for the radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

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Answer 3

Question

Chapter 20, Problem 7E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of G68a where electron capture takes place needs to be determined.

Concept Introduction:

Electron capture involves the absorption of electrons by the proton rich nucleus from the inner shells. During this process, the proton changes to neutron releasing an electron and atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

(a)

Expert Solution

Answer to Problem 7E

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

Explanation of Solution

The radioactive decay of G68a where electron capture takes place is depicted as as below

G3168a+e−10→?

(As Ga has atomic number as 31)

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

Mass Number: 68 + 0 =?

x = 68

Atomic number: 31 + (-1) =?

y= 30

Hence unknown element is x3068 . Looking through the periodic table, it is found that element with mass number 68 and atomic number 30 is Zn. Hence, the element is Z3068n . The electron capture releases electron in which the atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of C62u which releases positron needs to be determined.

Concept Introduction:

Beta decay takes place when there is excess of protons or neutrons which gets transformed to other. In beta plus decay, proton gets converted to neutron releasing positron and neutrino. The emission of positron will decrease the atomic number by 1 with no changes in mass number.

(b)

Expert Solution

Answer to Problem 7E

The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Explanation of Solution

The radioactive decay of C2962u which produces positron is depicted as below

C2962u→e10+?

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

Mass number: 62 = 0 +?

x = 62

Atomic Number: 29 =? + (1)

y= 28

Hence unknown element is x2862 . Looking through the periodic table, it is found that element with mass number 62 and atomic number 28 is Ni. Hence, the element is N2862i . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Hence, the overall equation for the radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of F212r where there is alpha particle release needs to be determined.

Concept Introduction:

Radioactive decay takes place when an atomic nucleus is unstable such that it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma. Release of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

(c)

Expert Solution

Answer to Problem 7E

The radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

Explanation of Solution

The radioactive decay of F87212r where alpha particles are released and the equation is depicted as

F87212r→?+H24e

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

Mass number: 212 = ?+4

x = 208

Atomic number: 87=? + 2

y= 85

Hence unknown element is x85208 . Looking through the periodic table, it is found that element with mass number 208 and atomic number 85 is Are Hence, the element is A85208r . The emission of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

The radioactive decay of F87212r with release of alpha particles is depicted as F87212r→A85208r+H24e .

Hence, the overall equation for the radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of S129b which emits beta particles needs to be explained.

Concept Introduction:

Radioactive decay releases radiations like alpha, beta and gamma, when unstable nuclei get converted to smaller stable fragments. The emission of β particle will increase the atomic number by 1 with no changes in mass number.

(d)

Expert Solution

Answer to Problem 7E

The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Explanation of Solution

The radioactive decay of S129b which emits beta particle is depicted as

S51129b→?+e−10

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

Mass number: 129 = x+ 0

x = 129

Atomic number: 51 = x + (- 1)

x= 52

Hence unknown element is x52129 . Looking through the periodic table, it is found that element with mass number129 and atomic number 52 is Te. Hence, the element is T52129e . The emission of beta particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Hence, the overall equation for the radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

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Answer 4

Question

Chapter 20, Problem 7E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of G68a where electron capture takes place needs to be determined.

Concept Introduction:

Electron capture involves the absorption of electrons by the proton rich nucleus from the inner shells. During this process, the proton changes to neutron releasing an electron and atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

(a)

Expert Solution

Answer to Problem 7E

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

Explanation of Solution

The radioactive decay of G68a where electron capture takes place is depicted as as below

G3168a+e−10→?

(As Ga has atomic number as 31)

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

Mass Number: 68 + 0 =?

x = 68

Atomic number: 31 + (-1) =?

y= 30

Hence unknown element is x3068 . Looking through the periodic table, it is found that element with mass number 68 and atomic number 30 is Zn. Hence, the element is Z3068n . The electron capture releases electron in which the atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of C62u which releases positron needs to be determined.

Concept Introduction:

Beta decay takes place when there is excess of protons or neutrons which gets transformed to other. In beta plus decay, proton gets converted to neutron releasing positron and neutrino. The emission of positron will decrease the atomic number by 1 with no changes in mass number.

(b)

Expert Solution

Answer to Problem 7E

The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Explanation of Solution

The radioactive decay of C2962u which produces positron is depicted as below

C2962u→e10+?

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

Mass number: 62 = 0 +?

x = 62

Atomic Number: 29 =? + (1)

y= 28

Hence unknown element is x2862 . Looking through the periodic table, it is found that element with mass number 62 and atomic number 28 is Ni. Hence, the element is N2862i . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Hence, the overall equation for the radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of F212r where there is alpha particle release needs to be determined.

Concept Introduction:

Radioactive decay takes place when an atomic nucleus is unstable such that it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma. Release of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

(c)

Expert Solution

Answer to Problem 7E

The radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

Explanation of Solution

The radioactive decay of F87212r where alpha particles are released and the equation is depicted as

F87212r→?+H24e

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

Mass number: 212 = ?+4

x = 208

Atomic number: 87=? + 2

y= 85

Hence unknown element is x85208 . Looking through the periodic table, it is found that element with mass number 208 and atomic number 85 is Are Hence, the element is A85208r . The emission of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

The radioactive decay of F87212r with release of alpha particles is depicted as F87212r→A85208r+H24e .

Hence, the overall equation for the radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of S129b which emits beta particles needs to be explained.

Concept Introduction:

Radioactive decay releases radiations like alpha, beta and gamma, when unstable nuclei get converted to smaller stable fragments. The emission of β particle will increase the atomic number by 1 with no changes in mass number.

(d)

Expert Solution

Answer to Problem 7E

The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Explanation of Solution

The radioactive decay of S129b which emits beta particle is depicted as

S51129b→?+e−10

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

Mass number: 129 = x+ 0

x = 129

Atomic number: 51 = x + (- 1)

x= 52

Hence unknown element is x52129 . Looking through the periodic table, it is found that element with mass number129 and atomic number 52 is Te. Hence, the element is T52129e . The emission of beta particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Hence, the overall equation for the radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

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Answer 5

Chapter 20, Problem 7E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of G68a where electron capture takes place needs to be determined.

Concept Introduction:

Electron capture involves the absorption of electrons by the proton rich nucleus from the inner shells. During this process, the proton changes to neutron releasing an electron and atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

(a)

Expert Solution

Answer to Problem 7E

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

Explanation of Solution

The radioactive decay of G68a where electron capture takes place is depicted as as below

G3168a+e−10→?

(As Ga has atomic number as 31)

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

Mass Number: 68 + 0 =?

x = 68

Atomic number: 31 + (-1) =?

y= 30

Hence unknown element is x3068 . Looking through the periodic table, it is found that element with mass number 68 and atomic number 30 is Zn. Hence, the element is Z3068n . The electron capture releases electron in which the atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of C62u which releases positron needs to be determined.

Concept Introduction:

Beta decay takes place when there is excess of protons or neutrons which gets transformed to other. In beta plus decay, proton gets converted to neutron releasing positron and neutrino. The emission of positron will decrease the atomic number by 1 with no changes in mass number.

(b)

Expert Solution

Answer to Problem 7E

The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Explanation of Solution

The radioactive decay of C2962u which produces positron is depicted as below

C2962u→e10+?

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

Mass number: 62 = 0 +?

x = 62

Atomic Number: 29 =? + (1)

y= 28

Hence unknown element is x2862 . Looking through the periodic table, it is found that element with mass number 62 and atomic number 28 is Ni. Hence, the element is N2862i . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Hence, the overall equation for the radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of F212r where there is alpha particle release needs to be determined.

Concept Introduction:

Radioactive decay takes place when an atomic nucleus is unstable such that it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma. Release of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

(c)

Expert Solution

Answer to Problem 7E

The radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

Explanation of Solution

The radioactive decay of F87212r where alpha particles are released and the equation is depicted as

F87212r→?+H24e

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

Mass number: 212 = ?+4

x = 208

Atomic number: 87=? + 2

y= 85

Hence unknown element is x85208 . Looking through the periodic table, it is found that element with mass number 208 and atomic number 85 is Are Hence, the element is A85208r . The emission of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

The radioactive decay of F87212r with release of alpha particles is depicted as F87212r→A85208r+H24e .

Hence, the overall equation for the radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of S129b which emits beta particles needs to be explained.

Concept Introduction:

Radioactive decay releases radiations like alpha, beta and gamma, when unstable nuclei get converted to smaller stable fragments. The emission of β particle will increase the atomic number by 1 with no changes in mass number.

(d)

Expert Solution

Answer to Problem 7E

The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Explanation of Solution

The radioactive decay of S129b which emits beta particle is depicted as

S51129b→?+e−10

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

Mass number: 129 = x+ 0

x = 129

Atomic number: 51 = x + (- 1)

x= 52

Hence unknown element is x52129 . Looking through the periodic table, it is found that element with mass number129 and atomic number 52 is Te. Hence, the element is T52129e . The emission of beta particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Hence, the overall equation for the radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Answer 6

Chapter 20, Problem 7E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of G68a where electron capture takes place needs to be determined.

Concept Introduction:

Electron capture involves the absorption of electrons by the proton rich nucleus from the inner shells. During this process, the proton changes to neutron releasing an electron and atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

(a)

Expert Solution

Answer to Problem 7E

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

Explanation of Solution

The radioactive decay of G68a where electron capture takes place is depicted as as below

G3168a+e−10→?

(As Ga has atomic number as 31)

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

Mass Number: 68 + 0 =?

x = 68

Atomic number: 31 + (-1) =?

y= 30

Hence unknown element is x3068 . Looking through the periodic table, it is found that element with mass number 68 and atomic number 30 is Zn. Hence, the element is Z3068n . The electron capture releases electron in which the atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

Answer 7

Chapter 20, Problem 7E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of G68a where electron capture takes place needs to be determined.

Concept Introduction:

Electron capture involves the absorption of electrons by the proton rich nucleus from the inner shells. During this process, the proton changes to neutron releasing an electron and atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

Answer 8

(a)

Expert Solution

Answer to Problem 7E

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The radioactive decay of G68a where electron capture takes place is depicted as as below

G3168a+e−10→?

(As Ga has atomic number as 31)

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

Mass Number: 68 + 0 =?

x = 68

Atomic number: 31 + (-1) =?

y= 30

Hence unknown element is x3068 . Looking through the periodic table, it is found that element with mass number 68 and atomic number 30 is Zn. Hence, the element is Z3068n . The electron capture releases electron in which the atomic number of the nucleus gets reduced by 1 unit keeping the mass number the same.

The radioactive decay of G68a where electron capture takes place is depicted as G3168a+e−10→Z3068n .

Answer 13

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of C62u which releases positron needs to be determined.

Concept Introduction:

Beta decay takes place when there is excess of protons or neutrons which gets transformed to other. In beta plus decay, proton gets converted to neutron releasing positron and neutrino. The emission of positron will decrease the atomic number by 1 with no changes in mass number.

(b)

Expert Solution

Answer to Problem 7E

The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Explanation of Solution

The radioactive decay of C2962u which produces positron is depicted as below

C2962u→e10+?

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

Mass number: 62 = 0 +?

x = 62

Atomic Number: 29 =? + (1)

y= 28

Hence unknown element is x2862 . Looking through the periodic table, it is found that element with mass number 62 and atomic number 28 is Ni. Hence, the element is N2862i . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Hence, the overall equation for the radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Answer 14

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of C62u which releases positron needs to be determined.

Concept Introduction:

Beta decay takes place when there is excess of protons or neutrons which gets transformed to other. In beta plus decay, proton gets converted to neutron releasing positron and neutrino. The emission of positron will decrease the atomic number by 1 with no changes in mass number.

Answer 15

(b)

Expert Solution

Answer to Problem 7E

The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The radioactive decay of C2962u which produces positron is depicted as below

C2962u→e10+?

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

Mass number: 62 = 0 +?

x = 62

Atomic Number: 29 =? + (1)

y= 28

Hence unknown element is x2862 . Looking through the periodic table, it is found that element with mass number 62 and atomic number 28 is Ni. Hence, the element is N2862i . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Hence, the overall equation for the radioactive decay of C2962u which produces positron is depicted as C2962u→e10+N2862i .

Answer 20

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of F212r where there is alpha particle release needs to be determined.

Concept Introduction:

Radioactive decay takes place when an atomic nucleus is unstable such that it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma. Release of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

(c)

Expert Solution

Answer to Problem 7E

The radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

Explanation of Solution

The radioactive decay of F87212r where alpha particles are released and the equation is depicted as

F87212r→?+H24e

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

Mass number: 212 = ?+4

x = 208

Atomic number: 87=? + 2

y= 85

Hence unknown element is x85208 . Looking through the periodic table, it is found that element with mass number 208 and atomic number 85 is Are Hence, the element is A85208r . The emission of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

The radioactive decay of F87212r with release of alpha particles is depicted as F87212r→A85208r+H24e .

Hence, the overall equation for the radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

Answer 21

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of F212r where there is alpha particle release needs to be determined.

Concept Introduction:

Radioactive decay takes place when an atomic nucleus is unstable such that it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma. Release of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

Answer 22

(c)

Expert Solution

Answer to Problem 7E

The radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The radioactive decay of F87212r where alpha particles are released and the equation is depicted as

F87212r→?+H24e

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

Mass number: 212 = ?+4

x = 208

Atomic number: 87=? + 2

y= 85

Hence unknown element is x85208 . Looking through the periodic table, it is found that element with mass number 208 and atomic number 85 is Are Hence, the element is A85208r . The emission of alpha particle causes a decrease in mass number by 4 units and atomic number by 2 units.

The radioactive decay of F87212r with release of alpha particles is depicted as F87212r→A85208r+H24e .

Hence, the overall equation for the radioactive decay of F87212r with release of alpha particles has equation as F87212r→A85208r+H24e .

Answer 27

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of S129b which emits beta particles needs to be explained.

Concept Introduction:

Radioactive decay releases radiations like alpha, beta and gamma, when unstable nuclei get converted to smaller stable fragments. The emission of β particle will increase the atomic number by 1 with no changes in mass number.

(d)

Expert Solution

Answer to Problem 7E

The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Explanation of Solution

The radioactive decay of S129b which emits beta particle is depicted as

S51129b→?+e−10

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

Mass number: 129 = x+ 0

x = 129

Atomic number: 51 = x + (- 1)

x= 52

Hence unknown element is x52129 . Looking through the periodic table, it is found that element with mass number129 and atomic number 52 is Te. Hence, the element is T52129e . The emission of beta particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Hence, the overall equation for the radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Answer 28

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of S129b which emits beta particles needs to be explained.

Concept Introduction:

Radioactive decay releases radiations like alpha, beta and gamma, when unstable nuclei get converted to smaller stable fragments. The emission of β particle will increase the atomic number by 1 with no changes in mass number.

Answer 29

(d)

Expert Solution

Answer to Problem 7E

The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The radioactive decay of S129b which emits beta particle is depicted as

S51129b→?+e−10

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

Mass number: 129 = x+ 0

x = 129

Atomic number: 51 = x + (- 1)

x= 52

Hence unknown element is x52129 . Looking through the periodic table, it is found that element with mass number129 and atomic number 52 is Te. Hence, the element is T52129e . The emission of beta particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of S51129b which emits beta particles has equation as S51129b→T52129e+e−10 .