The equation for radioactive decay of H 1 3 which produces β needs to be determined. Concept Introduction : Radioactive decay is process to convert unstable atomic nucleus to smaller stable nuclei and this is possible only when there is imbalance between the number of protons and neutrons. During this process radiations like alpha, beta, and gamma are emitted.

Question

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Answer 1

Question

Chapter 20, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of H13 which produces β needs to be determined.

Concept Introduction:

Radioactive decay is process to convert unstable atomic nucleus to smaller stable nuclei and this is possible only when there is imbalance between the number of protons and neutrons. During this process radiations like alpha, beta, and gamma are emitted.

(a)

Expert Solution

Answer to Problem 4E

The radioactive decay of H13 that produces β is depicted as H13→H23e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of H13 which produces β is depicted as below

H13→?+e−10

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

Mass number: 3 = 0 +?

x = 3

Atomic number: 1 =? + (-1)

y= 1

Hence unknown element is x23 . Looking through the periodic table, it is found that element with mass number 3 is He. Hence, the element is H23e . The emission of β particle will increase the atomic number by 1 with no changes in mass number.

The radioactive decay of H13 that emits β is depicted by equation H13→H23e+e−10

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of L38i which produces β and then later on a needs to be determined.

Concept Introduction:

When there is unstable atomic nucleus, then it gets converted to smaller stable nuclei through radioactive decay process. This happens when there is imbalance between the number of protons and neutrons. Radiations like alpha, beta, and gamma are emitted are by-products.

(b)

Expert Solution

Answer to Problem 4E

The radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of L38i which produces β is depicted as below

L38i→?+e−10

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

Mass number: 8 = 0 +?

x = 8

Atomic number: 3 =? + (-1)

y= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 is being. Hence, the element is B48e . The emission of β particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of L38i which produces β is depicted by equation L38i→B48e+e−10 ............…. (1)

Then there is emission of alpha particles indicating that there will be decrease in mass number by 4 units and atomic number by 2 units. This is depicted as below

B48e→H24e+H24e ............. (2)

Adding equation (1) and (2) we get

L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10

(As the B48e on both sides will cancel each other)

Hence, the overall equation for the radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B47e where there is electron capture 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. This happens only when the number of protons and neutrons are not balanced.

(c)

Expert Solution

Answer to Problem 4E

The radioactive decay of B47e where there is electron capture has equation as B47e+e−10→L37i

Explanation of Solution

The radioactive decay of B47e where there is electron capture and the equation is depicted as

B47e+e−10→?

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

Mass number: 7 + 0 = ?

x = 7

Atomic number: 4 + (-1) =?

y= 3

Hence unknown element is x37 . Looking through the periodic table, it is found that element with mass number 7 and atomic number 3 is Li. Hence, the element is L37i . The electron capture will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B47e where electron capture is depicted as B47e+e−10→L37i

Hence, the overall equation for the radioactive decay of B47e where there is electron capture is depicted as B47e+e−10→L37i

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B58 which emits positron needs to be determined.

Concept Introduction:

When an atomic nucleus is unstable then it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma and this process is called as radioactive decay. This happens only when there is imbalance in number of protons and neutrons.

(d)

Expert Solution

Answer to Problem 4E

The radioactive decay of B58 which emits positron has equation as B58→B48e+e10

Explanation of Solution

The radioactive decay of B58 which emits positron is depicted as

B58→x+e10

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

Mass number: 8 = x+ 0

x = 8

Atomic number: 5 = x + 1

x= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 and atomic number 4 is Be. Hence, the element is B48e . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B58 which emits positron is depicted as B58→B48e+e10

Hence, the overall equation for the radioactive decay B58 which emits positron has equation as B58→B48e+e10

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Answer 2

Question

Chapter 20, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of H13 which produces β needs to be determined.

Concept Introduction:

Radioactive decay is process to convert unstable atomic nucleus to smaller stable nuclei and this is possible only when there is imbalance between the number of protons and neutrons. During this process radiations like alpha, beta, and gamma are emitted.

(a)

Expert Solution

Answer to Problem 4E

The radioactive decay of H13 that produces β is depicted as H13→H23e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of H13 which produces β is depicted as below

H13→?+e−10

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

Mass number: 3 = 0 +?

x = 3

Atomic number: 1 =? + (-1)

y= 1

Hence unknown element is x23 . Looking through the periodic table, it is found that element with mass number 3 is He. Hence, the element is H23e . The emission of β particle will increase the atomic number by 1 with no changes in mass number.

The radioactive decay of H13 that emits β is depicted by equation H13→H23e+e−10

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of L38i which produces β and then later on a needs to be determined.

Concept Introduction:

When there is unstable atomic nucleus, then it gets converted to smaller stable nuclei through radioactive decay process. This happens when there is imbalance between the number of protons and neutrons. Radiations like alpha, beta, and gamma are emitted are by-products.

(b)

Expert Solution

Answer to Problem 4E

The radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of L38i which produces β is depicted as below

L38i→?+e−10

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

Mass number: 8 = 0 +?

x = 8

Atomic number: 3 =? + (-1)

y= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 is being. Hence, the element is B48e . The emission of β particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of L38i which produces β is depicted by equation L38i→B48e+e−10 ............…. (1)

Then there is emission of alpha particles indicating that there will be decrease in mass number by 4 units and atomic number by 2 units. This is depicted as below

B48e→H24e+H24e ............. (2)

Adding equation (1) and (2) we get

L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10

(As the B48e on both sides will cancel each other)

Hence, the overall equation for the radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B47e where there is electron capture 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. This happens only when the number of protons and neutrons are not balanced.

(c)

Expert Solution

Answer to Problem 4E

The radioactive decay of B47e where there is electron capture has equation as B47e+e−10→L37i

Explanation of Solution

The radioactive decay of B47e where there is electron capture and the equation is depicted as

B47e+e−10→?

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

Mass number: 7 + 0 = ?

x = 7

Atomic number: 4 + (-1) =?

y= 3

Hence unknown element is x37 . Looking through the periodic table, it is found that element with mass number 7 and atomic number 3 is Li. Hence, the element is L37i . The electron capture will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B47e where electron capture is depicted as B47e+e−10→L37i

Hence, the overall equation for the radioactive decay of B47e where there is electron capture is depicted as B47e+e−10→L37i

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B58 which emits positron needs to be determined.

Concept Introduction:

When an atomic nucleus is unstable then it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma and this process is called as radioactive decay. This happens only when there is imbalance in number of protons and neutrons.

(d)

Expert Solution

Answer to Problem 4E

The radioactive decay of B58 which emits positron has equation as B58→B48e+e10

Explanation of Solution

The radioactive decay of B58 which emits positron is depicted as

B58→x+e10

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

Mass number: 8 = x+ 0

x = 8

Atomic number: 5 = x + 1

x= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 and atomic number 4 is Be. Hence, the element is B48e . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B58 which emits positron is depicted as B58→B48e+e10

Hence, the overall equation for the radioactive decay B58 which emits positron has equation as B58→B48e+e10

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Answer 3

Question

Chapter 20, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of H13 which produces β needs to be determined.

Concept Introduction:

Radioactive decay is process to convert unstable atomic nucleus to smaller stable nuclei and this is possible only when there is imbalance between the number of protons and neutrons. During this process radiations like alpha, beta, and gamma are emitted.

(a)

Expert Solution

Answer to Problem 4E

The radioactive decay of H13 that produces β is depicted as H13→H23e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of H13 which produces β is depicted as below

H13→?+e−10

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

Mass number: 3 = 0 +?

x = 3

Atomic number: 1 =? + (-1)

y= 1

Hence unknown element is x23 . Looking through the periodic table, it is found that element with mass number 3 is He. Hence, the element is H23e . The emission of β particle will increase the atomic number by 1 with no changes in mass number.

The radioactive decay of H13 that emits β is depicted by equation H13→H23e+e−10

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of L38i which produces β and then later on a needs to be determined.

Concept Introduction:

When there is unstable atomic nucleus, then it gets converted to smaller stable nuclei through radioactive decay process. This happens when there is imbalance between the number of protons and neutrons. Radiations like alpha, beta, and gamma are emitted are by-products.

(b)

Expert Solution

Answer to Problem 4E

The radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of L38i which produces β is depicted as below

L38i→?+e−10

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

Mass number: 8 = 0 +?

x = 8

Atomic number: 3 =? + (-1)

y= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 is being. Hence, the element is B48e . The emission of β particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of L38i which produces β is depicted by equation L38i→B48e+e−10 ............…. (1)

Then there is emission of alpha particles indicating that there will be decrease in mass number by 4 units and atomic number by 2 units. This is depicted as below

B48e→H24e+H24e ............. (2)

Adding equation (1) and (2) we get

L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10

(As the B48e on both sides will cancel each other)

Hence, the overall equation for the radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B47e where there is electron capture 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. This happens only when the number of protons and neutrons are not balanced.

(c)

Expert Solution

Answer to Problem 4E

The radioactive decay of B47e where there is electron capture has equation as B47e+e−10→L37i

Explanation of Solution

The radioactive decay of B47e where there is electron capture and the equation is depicted as

B47e+e−10→?

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

Mass number: 7 + 0 = ?

x = 7

Atomic number: 4 + (-1) =?

y= 3

Hence unknown element is x37 . Looking through the periodic table, it is found that element with mass number 7 and atomic number 3 is Li. Hence, the element is L37i . The electron capture will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B47e where electron capture is depicted as B47e+e−10→L37i

Hence, the overall equation for the radioactive decay of B47e where there is electron capture is depicted as B47e+e−10→L37i

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B58 which emits positron needs to be determined.

Concept Introduction:

When an atomic nucleus is unstable then it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma and this process is called as radioactive decay. This happens only when there is imbalance in number of protons and neutrons.

(d)

Expert Solution

Answer to Problem 4E

The radioactive decay of B58 which emits positron has equation as B58→B48e+e10

Explanation of Solution

The radioactive decay of B58 which emits positron is depicted as

B58→x+e10

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

Mass number: 8 = x+ 0

x = 8

Atomic number: 5 = x + 1

x= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 and atomic number 4 is Be. Hence, the element is B48e . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B58 which emits positron is depicted as B58→B48e+e10

Hence, the overall equation for the radioactive decay B58 which emits positron has equation as B58→B48e+e10

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Answer 4

Question

Chapter 20, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of H13 which produces β needs to be determined.

Concept Introduction:

Radioactive decay is process to convert unstable atomic nucleus to smaller stable nuclei and this is possible only when there is imbalance between the number of protons and neutrons. During this process radiations like alpha, beta, and gamma are emitted.

(a)

Expert Solution

Answer to Problem 4E

The radioactive decay of H13 that produces β is depicted as H13→H23e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of H13 which produces β is depicted as below

H13→?+e−10

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

Mass number: 3 = 0 +?

x = 3

Atomic number: 1 =? + (-1)

y= 1

Hence unknown element is x23 . Looking through the periodic table, it is found that element with mass number 3 is He. Hence, the element is H23e . The emission of β particle will increase the atomic number by 1 with no changes in mass number.

The radioactive decay of H13 that emits β is depicted by equation H13→H23e+e−10

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of L38i which produces β and then later on a needs to be determined.

Concept Introduction:

When there is unstable atomic nucleus, then it gets converted to smaller stable nuclei through radioactive decay process. This happens when there is imbalance between the number of protons and neutrons. Radiations like alpha, beta, and gamma are emitted are by-products.

(b)

Expert Solution

Answer to Problem 4E

The radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of L38i which produces β is depicted as below

L38i→?+e−10

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

Mass number: 8 = 0 +?

x = 8

Atomic number: 3 =? + (-1)

y= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 is being. Hence, the element is B48e . The emission of β particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of L38i which produces β is depicted by equation L38i→B48e+e−10 ............…. (1)

Then there is emission of alpha particles indicating that there will be decrease in mass number by 4 units and atomic number by 2 units. This is depicted as below

B48e→H24e+H24e ............. (2)

Adding equation (1) and (2) we get

L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10

(As the B48e on both sides will cancel each other)

Hence, the overall equation for the radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B47e where there is electron capture 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. This happens only when the number of protons and neutrons are not balanced.

(c)

Expert Solution

Answer to Problem 4E

The radioactive decay of B47e where there is electron capture has equation as B47e+e−10→L37i

Explanation of Solution

The radioactive decay of B47e where there is electron capture and the equation is depicted as

B47e+e−10→?

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

Mass number: 7 + 0 = ?

x = 7

Atomic number: 4 + (-1) =?

y= 3

Hence unknown element is x37 . Looking through the periodic table, it is found that element with mass number 7 and atomic number 3 is Li. Hence, the element is L37i . The electron capture will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B47e where electron capture is depicted as B47e+e−10→L37i

Hence, the overall equation for the radioactive decay of B47e where there is electron capture is depicted as B47e+e−10→L37i

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B58 which emits positron needs to be determined.

Concept Introduction:

When an atomic nucleus is unstable then it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma and this process is called as radioactive decay. This happens only when there is imbalance in number of protons and neutrons.

(d)

Expert Solution

Answer to Problem 4E

The radioactive decay of B58 which emits positron has equation as B58→B48e+e10

Explanation of Solution

The radioactive decay of B58 which emits positron is depicted as

B58→x+e10

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

Mass number: 8 = x+ 0

x = 8

Atomic number: 5 = x + 1

x= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 and atomic number 4 is Be. Hence, the element is B48e . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B58 which emits positron is depicted as B58→B48e+e10

Hence, the overall equation for the radioactive decay B58 which emits positron has equation as B58→B48e+e10

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Answer 5

Chapter 20, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of H13 which produces β needs to be determined.

Concept Introduction:

Radioactive decay is process to convert unstable atomic nucleus to smaller stable nuclei and this is possible only when there is imbalance between the number of protons and neutrons. During this process radiations like alpha, beta, and gamma are emitted.

(a)

Expert Solution

Answer to Problem 4E

The radioactive decay of H13 that produces β is depicted as H13→H23e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of H13 which produces β is depicted as below

H13→?+e−10

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

Mass number: 3 = 0 +?

x = 3

Atomic number: 1 =? + (-1)

y= 1

Hence unknown element is x23 . Looking through the periodic table, it is found that element with mass number 3 is He. Hence, the element is H23e . The emission of β particle will increase the atomic number by 1 with no changes in mass number.

The radioactive decay of H13 that emits β is depicted by equation H13→H23e+e−10

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of L38i which produces β and then later on a needs to be determined.

Concept Introduction:

When there is unstable atomic nucleus, then it gets converted to smaller stable nuclei through radioactive decay process. This happens when there is imbalance between the number of protons and neutrons. Radiations like alpha, beta, and gamma are emitted are by-products.

(b)

Expert Solution

Answer to Problem 4E

The radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of L38i which produces β is depicted as below

L38i→?+e−10

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

Mass number: 8 = 0 +?

x = 8

Atomic number: 3 =? + (-1)

y= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 is being. Hence, the element is B48e . The emission of β particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of L38i which produces β is depicted by equation L38i→B48e+e−10 ............…. (1)

Then there is emission of alpha particles indicating that there will be decrease in mass number by 4 units and atomic number by 2 units. This is depicted as below

B48e→H24e+H24e ............. (2)

Adding equation (1) and (2) we get

L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10

(As the B48e on both sides will cancel each other)

Hence, the overall equation for the radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B47e where there is electron capture 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. This happens only when the number of protons and neutrons are not balanced.

(c)

Expert Solution

Answer to Problem 4E

The radioactive decay of B47e where there is electron capture has equation as B47e+e−10→L37i

Explanation of Solution

The radioactive decay of B47e where there is electron capture and the equation is depicted as

B47e+e−10→?

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

Mass number: 7 + 0 = ?

x = 7

Atomic number: 4 + (-1) =?

y= 3

Hence unknown element is x37 . Looking through the periodic table, it is found that element with mass number 7 and atomic number 3 is Li. Hence, the element is L37i . The electron capture will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B47e where electron capture is depicted as B47e+e−10→L37i

Hence, the overall equation for the radioactive decay of B47e where there is electron capture is depicted as B47e+e−10→L37i

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B58 which emits positron needs to be determined.

Concept Introduction:

When an atomic nucleus is unstable then it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma and this process is called as radioactive decay. This happens only when there is imbalance in number of protons and neutrons.

(d)

Expert Solution

Answer to Problem 4E

The radioactive decay of B58 which emits positron has equation as B58→B48e+e10

Explanation of Solution

The radioactive decay of B58 which emits positron is depicted as

B58→x+e10

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

Mass number: 8 = x+ 0

x = 8

Atomic number: 5 = x + 1

x= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 and atomic number 4 is Be. Hence, the element is B48e . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B58 which emits positron is depicted as B58→B48e+e10

Hence, the overall equation for the radioactive decay B58 which emits positron has equation as B58→B48e+e10

Answer 6

Chapter 20, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of H13 which produces β needs to be determined.

Concept Introduction:

Radioactive decay is process to convert unstable atomic nucleus to smaller stable nuclei and this is possible only when there is imbalance between the number of protons and neutrons. During this process radiations like alpha, beta, and gamma are emitted.

(a)

Expert Solution

Answer to Problem 4E

The radioactive decay of H13 that produces β is depicted as H13→H23e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of H13 which produces β is depicted as below

H13→?+e−10

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

Mass number: 3 = 0 +?

x = 3

Atomic number: 1 =? + (-1)

y= 1

Hence unknown element is x23 . Looking through the periodic table, it is found that element with mass number 3 is He. Hence, the element is H23e . The emission of β particle will increase the atomic number by 1 with no changes in mass number.

The radioactive decay of H13 that emits β is depicted by equation H13→H23e+e−10

Answer 7

Chapter 20, Problem 4E

(a)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of H13 which produces β needs to be determined.

Concept Introduction:

Radioactive decay is process to convert unstable atomic nucleus to smaller stable nuclei and this is possible only when there is imbalance between the number of protons and neutrons. During this process radiations like alpha, beta, and gamma are emitted.

Answer 8

(a)

Expert Solution

Answer to Problem 4E

The radioactive decay of H13 that produces β is depicted as H13→H23e+e−10 , as β is e−10

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The radioactive decay of H13 which produces β is depicted as below

H13→?+e−10

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

Mass number: 3 = 0 +?

x = 3

Atomic number: 1 =? + (-1)

y= 1

Hence unknown element is x23 . Looking through the periodic table, it is found that element with mass number 3 is He. Hence, the element is H23e . The emission of β particle will increase the atomic number by 1 with no changes in mass number.

The radioactive decay of H13 that emits β is depicted by equation H13→H23e+e−10

Answer 13

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of L38i which produces β and then later on a needs to be determined.

Concept Introduction:

When there is unstable atomic nucleus, then it gets converted to smaller stable nuclei through radioactive decay process. This happens when there is imbalance between the number of protons and neutrons. Radiations like alpha, beta, and gamma are emitted are by-products.

(b)

Expert Solution

Answer to Problem 4E

The radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10 , as β is e−10

Explanation of Solution

The radioactive decay of L38i which produces β is depicted as below

L38i→?+e−10

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

Mass number: 8 = 0 +?

x = 8

Atomic number: 3 =? + (-1)

y= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 is being. Hence, the element is B48e . The emission of β particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of L38i which produces β is depicted by equation L38i→B48e+e−10 ............…. (1)

Then there is emission of alpha particles indicating that there will be decrease in mass number by 4 units and atomic number by 2 units. This is depicted as below

B48e→H24e+H24e ............. (2)

Adding equation (1) and (2) we get

L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10

(As the B48e on both sides will cancel each other)

Hence, the overall equation for the radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10

Answer 14

(b)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of L38i which produces β and then later on a needs to be determined.

Concept Introduction:

When there is unstable atomic nucleus, then it gets converted to smaller stable nuclei through radioactive decay process. This happens when there is imbalance between the number of protons and neutrons. Radiations like alpha, beta, and gamma are emitted are by-products.

Answer 15

(b)

Expert Solution

Answer to Problem 4E

The radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10 , as β is e−10

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

The radioactive decay of L38i which produces β is depicted as below

L38i→?+e−10

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

Mass number: 8 = 0 +?

x = 8

Atomic number: 3 =? + (-1)

y= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 is being. Hence, the element is B48e . The emission of β particle will increase the atomic number by 1 with no changes in mass number. The radioactive decay of L38i which produces β is depicted by equation L38i→B48e+e−10 ............…. (1)

Then there is emission of alpha particles indicating that there will be decrease in mass number by 4 units and atomic number by 2 units. This is depicted as below

B48e→H24e+H24e ............. (2)

Adding equation (1) and (2) we get

L38i→B48e+e−10B48e→H24e+H24eL38i→2H24e+e−10

(As the B48e on both sides will cancel each other)

Hence, the overall equation for the radioactive decay of L38i which produces β and then later on a. s depicted as L38i→2H24e+e−10

Answer 20

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B47e where there is electron capture 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. This happens only when the number of protons and neutrons are not balanced.

(c)

Expert Solution

Answer to Problem 4E

The radioactive decay of B47e where there is electron capture has equation as B47e+e−10→L37i

Explanation of Solution

The radioactive decay of B47e where there is electron capture and the equation is depicted as

B47e+e−10→?

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

Mass number: 7 + 0 = ?

x = 7

Atomic number: 4 + (-1) =?

y= 3

Hence unknown element is x37 . Looking through the periodic table, it is found that element with mass number 7 and atomic number 3 is Li. Hence, the element is L37i . The electron capture will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B47e where electron capture is depicted as B47e+e−10→L37i

Hence, the overall equation for the radioactive decay of B47e where there is electron capture is depicted as B47e+e−10→L37i

Answer 21

(c)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B47e where there is electron capture 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. This happens only when the number of protons and neutrons are not balanced.

Answer 22

(c)

Expert Solution

Answer to Problem 4E

The radioactive decay of B47e where there is electron capture has equation as B47e+e−10→L37i

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The radioactive decay of B47e where there is electron capture and the equation is depicted as

B47e+e−10→?

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

Mass number: 7 + 0 = ?

x = 7

Atomic number: 4 + (-1) =?

y= 3

Hence unknown element is x37 . Looking through the periodic table, it is found that element with mass number 7 and atomic number 3 is Li. Hence, the element is L37i . The electron capture will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B47e where electron capture is depicted as B47e+e−10→L37i

Hence, the overall equation for the radioactive decay of B47e where there is electron capture is depicted as B47e+e−10→L37i

Answer 27

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B58 which emits positron needs to be determined.

Concept Introduction:

When an atomic nucleus is unstable then it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma and this process is called as radioactive decay. This happens only when there is imbalance in number of protons and neutrons.

(d)

Expert Solution

Answer to Problem 4E

The radioactive decay of B58 which emits positron has equation as B58→B48e+e10

Explanation of Solution

The radioactive decay of B58 which emits positron is depicted as

B58→x+e10

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

Mass number: 8 = x+ 0

x = 8

Atomic number: 5 = x + 1

x= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 and atomic number 4 is Be. Hence, the element is B48e . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B58 which emits positron is depicted as B58→B48e+e10

Hence, the overall equation for the radioactive decay B58 which emits positron has equation as B58→B48e+e10

Answer 28

(d)

Interpretation Introduction

Interpretation:

The equation for radioactive decay of B58 which emits positron needs to be determined.

Concept Introduction:

When an atomic nucleus is unstable then it gets converted to smaller stable nuclei emitting radiations like alpha, beta, and gamma and this process is called as radioactive decay. This happens only when there is imbalance in number of protons and neutrons.

Answer 29

(d)

Expert Solution

Answer to Problem 4E

The radioactive decay of B58 which emits positron has equation as B58→B48e+e10

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

The radioactive decay of B58 which emits positron is depicted as

B58→x+e10

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

Mass number: 8 = x+ 0

x = 8

Atomic number: 5 = x + 1

x= 4

Hence unknown element is x48 . Looking through the periodic table, it is found that element with mass number 8 and atomic number 4 is Be. Hence, the element is B48e . The emission of positron will decrease the atomic number by 1 with no changes in mass number. The radioactive decay of B58 which emits positron is depicted as B58→B48e+e10