(a) Interpretation: A nuclear equation for the decay of U − 238 through alpha emission is to be stated. Concept introduction: Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Question

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Answer 1

Question

Chapter 18, Problem 11ST

Interpretation Introduction

(a)

Interpretation:

A nuclear equation for the decay of U−238 through alpha emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Alpha emission of U−238 releases Th90234 and He24 as shown below.

92238U→90234Th+24He.

Explanation of Solution

Alpha particle emission releases one alpha particle whose mass number decreases by 4 and atomic number decreasesby 2. The radioactive decay of U−238 by alpha emission releases one alpha particle He24 and gives Th90234 as shown in Figure 1.

Figure 1

Conclusion

Radioactive alpha decay of U−238 is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

A nuclear equation for the decay of U−239 through beta emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Beta emission of U−239 releases 93239Np and e−10 as shown below.

92239U→93239Np+−10e

Explanation of Solution

Beta emission decays into a proton and releases an electron or the atomic number increases by 1 and mass number remains the same. When U−239 beta emission takes place, it releases one electron and gives 1 increased atomic number element that is 93239Np as shpown in Figure 2.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 18, Problem 11ST , additional homework tip 2

Figure 2

Conclusion

Radioactive beta decay of U−239 is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

A nuclear equation for the decay of K−40 through positron emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Positron emission of K−40 releases 1840Ar and e10 as shown below.

1940K→1840Ar++10e

Explanation of Solution

In positron emission, an electron is released and its charge is +1 rather than −1. It can also be defined as the atomic number decreases by 1 and mass number remains the same in Positron emission. The positron emission of K−40 is shown in Figure 3.

Figure 3

Conclusion

Radioactive positron electron emission of K−40 is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

A nuclear equation for the decay of K−40 through electron capture is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Electron capture of K−40 releases 1840Ar as shown below.

1940K+−10e→1840Ar

Explanation of Solution

In electron capture decay, element accepts electron and results in atomic number decreased by 1 and mass number remains same. The electron capture of K−40 is shown in Figure 4.

Figure 4

Conclusion

Radioactive alpha decay of K−40 is shown in Figure 4.

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Students have asked these similar questions

Using reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NH3 (g) = N2 (g) +3H₂ —N2 (g) AGº = 34. kJ Now suppose a reaction vessel is filled with 4.19 atm of ammonia (NH3) and 9.94 atm of nitrogen (N2) at 378. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of NH 3 tend to rise or fall? ☐ x10 fall Х Is it possible to reverse this tendency by adding H₂? In other words, if you said the pressure of NH 3 will tend to rise, can that be changed to a tendency to fall by adding H₂? Similarly, if you said the pressure of NH3 will tend to fall, can that be changed to a tendency to rise by adding H₂? If you said the tendency can be reversed in the second question, calculate the minimum pressure of H₂ needed to reverse it. Round your answer to 2 significant digits. yes no atm 00. 18 Ar 무ㅎ ?

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Answer 2

Question

Chapter 18, Problem 11ST

Interpretation Introduction

(a)

Interpretation:

A nuclear equation for the decay of U−238 through alpha emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Alpha emission of U−238 releases Th90234 and He24 as shown below.

92238U→90234Th+24He.

Explanation of Solution

Alpha particle emission releases one alpha particle whose mass number decreases by 4 and atomic number decreasesby 2. The radioactive decay of U−238 by alpha emission releases one alpha particle He24 and gives Th90234 as shown in Figure 1.

Figure 1

Conclusion

Radioactive alpha decay of U−238 is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

A nuclear equation for the decay of U−239 through beta emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Beta emission of U−239 releases 93239Np and e−10 as shown below.

92239U→93239Np+−10e

Explanation of Solution

Beta emission decays into a proton and releases an electron or the atomic number increases by 1 and mass number remains the same. When U−239 beta emission takes place, it releases one electron and gives 1 increased atomic number element that is 93239Np as shpown in Figure 2.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 18, Problem 11ST , additional homework tip 2

Figure 2

Conclusion

Radioactive beta decay of U−239 is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

A nuclear equation for the decay of K−40 through positron emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Positron emission of K−40 releases 1840Ar and e10 as shown below.

1940K→1840Ar++10e

Explanation of Solution

In positron emission, an electron is released and its charge is +1 rather than −1. It can also be defined as the atomic number decreases by 1 and mass number remains the same in Positron emission. The positron emission of K−40 is shown in Figure 3.

Figure 3

Conclusion

Radioactive positron electron emission of K−40 is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

A nuclear equation for the decay of K−40 through electron capture is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Electron capture of K−40 releases 1840Ar as shown below.

1940K+−10e→1840Ar

Explanation of Solution

In electron capture decay, element accepts electron and results in atomic number decreased by 1 and mass number remains same. The electron capture of K−40 is shown in Figure 4.

Figure 4

Conclusion

Radioactive alpha decay of K−40 is shown in Figure 4.

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Answer 3

Question

Chapter 18, Problem 11ST

Interpretation Introduction

(a)

Interpretation:

A nuclear equation for the decay of U−238 through alpha emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Alpha emission of U−238 releases Th90234 and He24 as shown below.

92238U→90234Th+24He.

Explanation of Solution

Alpha particle emission releases one alpha particle whose mass number decreases by 4 and atomic number decreasesby 2. The radioactive decay of U−238 by alpha emission releases one alpha particle He24 and gives Th90234 as shown in Figure 1.

Figure 1

Conclusion

Radioactive alpha decay of U−238 is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

A nuclear equation for the decay of U−239 through beta emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Beta emission of U−239 releases 93239Np and e−10 as shown below.

92239U→93239Np+−10e

Explanation of Solution

Beta emission decays into a proton and releases an electron or the atomic number increases by 1 and mass number remains the same. When U−239 beta emission takes place, it releases one electron and gives 1 increased atomic number element that is 93239Np as shpown in Figure 2.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 18, Problem 11ST , additional homework tip 2

Figure 2

Conclusion

Radioactive beta decay of U−239 is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

A nuclear equation for the decay of K−40 through positron emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Positron emission of K−40 releases 1840Ar and e10 as shown below.

1940K→1840Ar++10e

Explanation of Solution

In positron emission, an electron is released and its charge is +1 rather than −1. It can also be defined as the atomic number decreases by 1 and mass number remains the same in Positron emission. The positron emission of K−40 is shown in Figure 3.

Figure 3

Conclusion

Radioactive positron electron emission of K−40 is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

A nuclear equation for the decay of K−40 through electron capture is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Electron capture of K−40 releases 1840Ar as shown below.

1940K+−10e→1840Ar

Explanation of Solution

In electron capture decay, element accepts electron and results in atomic number decreased by 1 and mass number remains same. The electron capture of K−40 is shown in Figure 4.

Figure 4

Conclusion

Radioactive alpha decay of K−40 is shown in Figure 4.

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Answer 4

Question

Chapter 18, Problem 11ST

Interpretation Introduction

(a)

Interpretation:

A nuclear equation for the decay of U−238 through alpha emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Alpha emission of U−238 releases Th90234 and He24 as shown below.

92238U→90234Th+24He.

Explanation of Solution

Alpha particle emission releases one alpha particle whose mass number decreases by 4 and atomic number decreasesby 2. The radioactive decay of U−238 by alpha emission releases one alpha particle He24 and gives Th90234 as shown in Figure 1.

Figure 1

Conclusion

Radioactive alpha decay of U−238 is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

A nuclear equation for the decay of U−239 through beta emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Beta emission of U−239 releases 93239Np and e−10 as shown below.

92239U→93239Np+−10e

Explanation of Solution

Beta emission decays into a proton and releases an electron or the atomic number increases by 1 and mass number remains the same. When U−239 beta emission takes place, it releases one electron and gives 1 increased atomic number element that is 93239Np as shpown in Figure 2.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 18, Problem 11ST , additional homework tip 2

Figure 2

Conclusion

Radioactive beta decay of U−239 is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

A nuclear equation for the decay of K−40 through positron emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Positron emission of K−40 releases 1840Ar and e10 as shown below.

1940K→1840Ar++10e

Explanation of Solution

In positron emission, an electron is released and its charge is +1 rather than −1. It can also be defined as the atomic number decreases by 1 and mass number remains the same in Positron emission. The positron emission of K−40 is shown in Figure 3.

Figure 3

Conclusion

Radioactive positron electron emission of K−40 is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

A nuclear equation for the decay of K−40 through electron capture is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Electron capture of K−40 releases 1840Ar as shown below.

1940K+−10e→1840Ar

Explanation of Solution

In electron capture decay, element accepts electron and results in atomic number decreased by 1 and mass number remains same. The electron capture of K−40 is shown in Figure 4.

Figure 4

Conclusion

Radioactive alpha decay of K−40 is shown in Figure 4.

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Answer 5

Chapter 18, Problem 11ST

Interpretation Introduction

(a)

Interpretation:

A nuclear equation for the decay of U−238 through alpha emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Alpha emission of U−238 releases Th90234 and He24 as shown below.

92238U→90234Th+24He.

Explanation of Solution

Alpha particle emission releases one alpha particle whose mass number decreases by 4 and atomic number decreasesby 2. The radioactive decay of U−238 by alpha emission releases one alpha particle He24 and gives Th90234 as shown in Figure 1.

Figure 1

Conclusion

Radioactive alpha decay of U−238 is shown in Figure 1.

Interpretation Introduction

(b)

Interpretation:

A nuclear equation for the decay of U−239 through beta emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Beta emission of U−239 releases 93239Np and e−10 as shown below.

92239U→93239Np+−10e

Explanation of Solution

Beta emission decays into a proton and releases an electron or the atomic number increases by 1 and mass number remains the same. When U−239 beta emission takes place, it releases one electron and gives 1 increased atomic number element that is 93239Np as shpown in Figure 2.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 18, Problem 11ST , additional homework tip 2

Figure 2

Conclusion

Radioactive beta decay of U−239 is shown in Figure 2.

Interpretation Introduction

(c)

Interpretation:

A nuclear equation for the decay of K−40 through positron emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Positron emission of K−40 releases 1840Ar and e10 as shown below.

1940K→1840Ar++10e

Explanation of Solution

In positron emission, an electron is released and its charge is +1 rather than −1. It can also be defined as the atomic number decreases by 1 and mass number remains the same in Positron emission. The positron emission of K−40 is shown in Figure 3.

Figure 3

Conclusion

Radioactive positron electron emission of K−40 is shown in Figure 3.

Interpretation Introduction

(d)

Interpretation:

A nuclear equation for the decay of K−40 through electron capture is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Electron capture of K−40 releases 1840Ar as shown below.

1940K+−10e→1840Ar

Explanation of Solution

In electron capture decay, element accepts electron and results in atomic number decreased by 1 and mass number remains same. The electron capture of K−40 is shown in Figure 4.

Figure 4

Conclusion

Radioactive alpha decay of K−40 is shown in Figure 4.

Answer 6

Chapter 18, Problem 11ST

Interpretation Introduction

(a)

Interpretation:

A nuclear equation for the decay of U−238 through alpha emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Alpha emission of U−238 releases Th90234 and He24 as shown below.

92238U→90234Th+24He.

Explanation of Solution

Alpha particle emission releases one alpha particle whose mass number decreases by 4 and atomic number decreasesby 2. The radioactive decay of U−238 by alpha emission releases one alpha particle He24 and gives Th90234 as shown in Figure 1.

Figure 1

Conclusion

Radioactive alpha decay of U−238 is shown in Figure 1.

Answer 7

Chapter 18, Problem 11ST

Interpretation Introduction

(a)

Interpretation:

A nuclear equation for the decay of U−238 through alpha emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Answer 8

Expert Solution

Answer to Problem 11ST

Alpha emission of U−238 releases Th90234 and He24 as shown below.

92238U→90234Th+24He.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Alpha particle emission releases one alpha particle whose mass number decreases by 4 and atomic number decreasesby 2. The radioactive decay of U−238 by alpha emission releases one alpha particle He24 and gives Th90234 as shown in Figure 1.

Figure 1

Answer 13

Conclusion

Radioactive alpha decay of U−238 is shown in Figure 1.

Answer 14

Interpretation Introduction

(b)

Interpretation:

A nuclear equation for the decay of U−239 through beta emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Beta emission of U−239 releases 93239Np and e−10 as shown below.

92239U→93239Np+−10e

Explanation of Solution

Beta emission decays into a proton and releases an electron or the atomic number increases by 1 and mass number remains the same. When U−239 beta emission takes place, it releases one electron and gives 1 increased atomic number element that is 93239Np as shpown in Figure 2.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 18, Problem 11ST , additional homework tip 2

Figure 2

Conclusion

Radioactive beta decay of U−239 is shown in Figure 2.

Answer 15

Interpretation Introduction

(b)

Interpretation:

A nuclear equation for the decay of U−239 through beta emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Answer 16

Expert Solution

Answer to Problem 11ST

Beta emission of U−239 releases 93239Np and e−10 as shown below.

92239U→93239Np+−10e

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

Beta emission decays into a proton and releases an electron or the atomic number increases by 1 and mass number remains the same. When U−239 beta emission takes place, it releases one electron and gives 1 increased atomic number element that is 93239Np as shpown in Figure 2.

Introductory Chemistry: Concepts and Critical Thinking (8th Edition), Chapter 18, Problem 11ST , additional homework tip 2

Figure 2

Answer 21

Conclusion

Radioactive beta decay of U−239 is shown in Figure 2.

Answer 22

Interpretation Introduction

(c)

Interpretation:

A nuclear equation for the decay of K−40 through positron emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Positron emission of K−40 releases 1840Ar and e10 as shown below.

1940K→1840Ar++10e

Explanation of Solution

In positron emission, an electron is released and its charge is +1 rather than −1. It can also be defined as the atomic number decreases by 1 and mass number remains the same in Positron emission. The positron emission of K−40 is shown in Figure 3.

Figure 3

Conclusion

Radioactive positron electron emission of K−40 is shown in Figure 3.

Answer 23

Interpretation Introduction

(c)

Interpretation:

A nuclear equation for the decay of K−40 through positron emission is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Answer 24

Expert Solution

Answer to Problem 11ST

Positron emission of K−40 releases 1840Ar and e10 as shown below.

1940K→1840Ar++10e

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

In positron emission, an electron is released and its charge is +1 rather than −1. It can also be defined as the atomic number decreases by 1 and mass number remains the same in Positron emission. The positron emission of K−40 is shown in Figure 3.

Figure 3

Answer 29

Conclusion

Radioactive positron electron emission of K−40 is shown in Figure 3.

Answer 30

Interpretation Introduction

(d)

Interpretation:

A nuclear equation for the decay of K−40 through electron capture is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Expert Solution

Answer to Problem 11ST

Electron capture of K−40 releases 1840Ar as shown below.

1940K+−10e→1840Ar

Explanation of Solution

In electron capture decay, element accepts electron and results in atomic number decreased by 1 and mass number remains same. The electron capture of K−40 is shown in Figure 4.

Figure 4

Conclusion

Radioactive alpha decay of K−40 is shown in Figure 4.

Answer 31

Interpretation Introduction

(d)

Interpretation:

A nuclear equation for the decay of K−40 through electron capture is to be stated.

Concept introduction:

Radioactive decay is the process that involves the emission of radiation by an unstable atomic nucleus. The atomic nucleus loses its energy. The process is spontaneous. It is also known as nuclear radiation. The decay is accompanied by the emission of alpha particles, beta particles, and gamma rays and so on. All the elements which have an atomic number greater than 83 are radioactive.

Answer 32

Expert Solution

Answer to Problem 11ST

Electron capture of K−40 releases 1840Ar as shown below.

1940K+−10e→1840Ar

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

In electron capture decay, element accepts electron and results in atomic number decreased by 1 and mass number remains same. The electron capture of K−40 is shown in Figure 4.

Figure 4