(a) Interpretation: The equation for W − 160 decay by alpha emission is to be stated. Concept introduction: Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Question

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Answer 1

Question

Chapter 18, Problem 10E

Interpretation Introduction

(a)

Interpretation:

The equation for W−160 decay by alpha emission is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for W−160 decay by alpha emission is shown below.

W74160→H72156f+H24e

Explanation of Solution

It is given that W−160 undergoes decay by alpha emission. An alpha particle is H24e. Therefore, in the case of alpha decay the atomic number of an element decreases by 2 and the mass number decreases by 4. The atomic number of tungsten is 74 which decreases by 2 that means resulting element has atomic number 72. The element which has atomic number 70 is Hafnium.

The resulting equation is shown below.

W74160→H72156f+H24e

Conclusion

The equation of W−160 decay is stated above.

Interpretation Introduction

(b)

Interpretation:

The equation for P−32 decay by beta emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for P−32 decay by beta emission is shown below.

P1532→S1632+e−10

Explanation of Solution

It is given that P−32 undergoes decay by beta emission. A beta particle is identical to an electron that is e−10. Therefore, in the case of beta decay the atomic number of an element increases by 1 and there is no change in mass number. The atomic number of phosphorus is 15 which increases by 1 that means resulting element has atomic number 16. The element which has an atomic number 16 is sulfur.

The resulting equation is shown below.

P1532→S1632+e−10

Conclusion

The equation of P−32 decay is stated above.

Interpretation Introduction

(c)

Interpretation:

The equation for Co−55 decay by positron emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Co−55 decay by positron emission is shown below.

C2755o→F2655e+e+10

Explanation of Solution

It is given that Co−55 undergoes decay by positron emission. A positron is opposite of an electron and represented by e+10. Therefore, in the case of positron the atomic number of an element decreases by 1 and there is no change in mass number. The atomic number of cobalt is 27 which decreases by 1 that means resulting element has atomic number 26. The element which has an atomic number 26 is iron.

The resulting equation is shown below.

C2755o→F2655e+e+10

Conclusion

The equation of Co−55 decay is stated above.

Interpretation Introduction

(d)

Interpretation:

The equation for Ti−44 decay by electron capture is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Ti−44 decay by electron capture is shown below.

T2244i+e−10→S2144c

Explanation of Solution

It is given that Ti−44 undergoes decay by electron capture. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by 1, and there is no change in mass number. The atomic number of titanium is 22 which decreases by 1 that means resulting element has atomic number 21. The element which has an atomic number 21 is scandium.

The resulting equation is shown below.

T2244i+e−10→S2144c

Conclusion

The equation of Ti−44 decay is stated above.

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Answer 2

Question

Chapter 18, Problem 10E

Interpretation Introduction

(a)

Interpretation:

The equation for W−160 decay by alpha emission is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for W−160 decay by alpha emission is shown below.

W74160→H72156f+H24e

Explanation of Solution

It is given that W−160 undergoes decay by alpha emission. An alpha particle is H24e. Therefore, in the case of alpha decay the atomic number of an element decreases by 2 and the mass number decreases by 4. The atomic number of tungsten is 74 which decreases by 2 that means resulting element has atomic number 72. The element which has atomic number 70 is Hafnium.

The resulting equation is shown below.

W74160→H72156f+H24e

Conclusion

The equation of W−160 decay is stated above.

Interpretation Introduction

(b)

Interpretation:

The equation for P−32 decay by beta emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for P−32 decay by beta emission is shown below.

P1532→S1632+e−10

Explanation of Solution

It is given that P−32 undergoes decay by beta emission. A beta particle is identical to an electron that is e−10. Therefore, in the case of beta decay the atomic number of an element increases by 1 and there is no change in mass number. The atomic number of phosphorus is 15 which increases by 1 that means resulting element has atomic number 16. The element which has an atomic number 16 is sulfur.

The resulting equation is shown below.

P1532→S1632+e−10

Conclusion

The equation of P−32 decay is stated above.

Interpretation Introduction

(c)

Interpretation:

The equation for Co−55 decay by positron emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Co−55 decay by positron emission is shown below.

C2755o→F2655e+e+10

Explanation of Solution

It is given that Co−55 undergoes decay by positron emission. A positron is opposite of an electron and represented by e+10. Therefore, in the case of positron the atomic number of an element decreases by 1 and there is no change in mass number. The atomic number of cobalt is 27 which decreases by 1 that means resulting element has atomic number 26. The element which has an atomic number 26 is iron.

The resulting equation is shown below.

C2755o→F2655e+e+10

Conclusion

The equation of Co−55 decay is stated above.

Interpretation Introduction

(d)

Interpretation:

The equation for Ti−44 decay by electron capture is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Ti−44 decay by electron capture is shown below.

T2244i+e−10→S2144c

Explanation of Solution

It is given that Ti−44 undergoes decay by electron capture. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by 1, and there is no change in mass number. The atomic number of titanium is 22 which decreases by 1 that means resulting element has atomic number 21. The element which has an atomic number 21 is scandium.

The resulting equation is shown below.

T2244i+e−10→S2144c

Conclusion

The equation of Ti−44 decay is stated above.

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Answer 3

Question

Chapter 18, Problem 10E

Interpretation Introduction

(a)

Interpretation:

The equation for W−160 decay by alpha emission is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for W−160 decay by alpha emission is shown below.

W74160→H72156f+H24e

Explanation of Solution

It is given that W−160 undergoes decay by alpha emission. An alpha particle is H24e. Therefore, in the case of alpha decay the atomic number of an element decreases by 2 and the mass number decreases by 4. The atomic number of tungsten is 74 which decreases by 2 that means resulting element has atomic number 72. The element which has atomic number 70 is Hafnium.

The resulting equation is shown below.

W74160→H72156f+H24e

Conclusion

The equation of W−160 decay is stated above.

Interpretation Introduction

(b)

Interpretation:

The equation for P−32 decay by beta emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for P−32 decay by beta emission is shown below.

P1532→S1632+e−10

Explanation of Solution

It is given that P−32 undergoes decay by beta emission. A beta particle is identical to an electron that is e−10. Therefore, in the case of beta decay the atomic number of an element increases by 1 and there is no change in mass number. The atomic number of phosphorus is 15 which increases by 1 that means resulting element has atomic number 16. The element which has an atomic number 16 is sulfur.

The resulting equation is shown below.

P1532→S1632+e−10

Conclusion

The equation of P−32 decay is stated above.

Interpretation Introduction

(c)

Interpretation:

The equation for Co−55 decay by positron emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Co−55 decay by positron emission is shown below.

C2755o→F2655e+e+10

Explanation of Solution

It is given that Co−55 undergoes decay by positron emission. A positron is opposite of an electron and represented by e+10. Therefore, in the case of positron the atomic number of an element decreases by 1 and there is no change in mass number. The atomic number of cobalt is 27 which decreases by 1 that means resulting element has atomic number 26. The element which has an atomic number 26 is iron.

The resulting equation is shown below.

C2755o→F2655e+e+10

Conclusion

The equation of Co−55 decay is stated above.

Interpretation Introduction

(d)

Interpretation:

The equation for Ti−44 decay by electron capture is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Ti−44 decay by electron capture is shown below.

T2244i+e−10→S2144c

Explanation of Solution

It is given that Ti−44 undergoes decay by electron capture. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by 1, and there is no change in mass number. The atomic number of titanium is 22 which decreases by 1 that means resulting element has atomic number 21. The element which has an atomic number 21 is scandium.

The resulting equation is shown below.

T2244i+e−10→S2144c

Conclusion

The equation of Ti−44 decay is stated above.

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Answer 4

Question

Chapter 18, Problem 10E

Interpretation Introduction

(a)

Interpretation:

The equation for W−160 decay by alpha emission is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for W−160 decay by alpha emission is shown below.

W74160→H72156f+H24e

Explanation of Solution

It is given that W−160 undergoes decay by alpha emission. An alpha particle is H24e. Therefore, in the case of alpha decay the atomic number of an element decreases by 2 and the mass number decreases by 4. The atomic number of tungsten is 74 which decreases by 2 that means resulting element has atomic number 72. The element which has atomic number 70 is Hafnium.

The resulting equation is shown below.

W74160→H72156f+H24e

Conclusion

The equation of W−160 decay is stated above.

Interpretation Introduction

(b)

Interpretation:

The equation for P−32 decay by beta emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for P−32 decay by beta emission is shown below.

P1532→S1632+e−10

Explanation of Solution

It is given that P−32 undergoes decay by beta emission. A beta particle is identical to an electron that is e−10. Therefore, in the case of beta decay the atomic number of an element increases by 1 and there is no change in mass number. The atomic number of phosphorus is 15 which increases by 1 that means resulting element has atomic number 16. The element which has an atomic number 16 is sulfur.

The resulting equation is shown below.

P1532→S1632+e−10

Conclusion

The equation of P−32 decay is stated above.

Interpretation Introduction

(c)

Interpretation:

The equation for Co−55 decay by positron emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Co−55 decay by positron emission is shown below.

C2755o→F2655e+e+10

Explanation of Solution

It is given that Co−55 undergoes decay by positron emission. A positron is opposite of an electron and represented by e+10. Therefore, in the case of positron the atomic number of an element decreases by 1 and there is no change in mass number. The atomic number of cobalt is 27 which decreases by 1 that means resulting element has atomic number 26. The element which has an atomic number 26 is iron.

The resulting equation is shown below.

C2755o→F2655e+e+10

Conclusion

The equation of Co−55 decay is stated above.

Interpretation Introduction

(d)

Interpretation:

The equation for Ti−44 decay by electron capture is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Ti−44 decay by electron capture is shown below.

T2244i+e−10→S2144c

Explanation of Solution

It is given that Ti−44 undergoes decay by electron capture. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by 1, and there is no change in mass number. The atomic number of titanium is 22 which decreases by 1 that means resulting element has atomic number 21. The element which has an atomic number 21 is scandium.

The resulting equation is shown below.

T2244i+e−10→S2144c

Conclusion

The equation of Ti−44 decay is stated above.

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Answer 5

Chapter 18, Problem 10E

Interpretation Introduction

(a)

Interpretation:

The equation for W−160 decay by alpha emission is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for W−160 decay by alpha emission is shown below.

W74160→H72156f+H24e

Explanation of Solution

It is given that W−160 undergoes decay by alpha emission. An alpha particle is H24e. Therefore, in the case of alpha decay the atomic number of an element decreases by 2 and the mass number decreases by 4. The atomic number of tungsten is 74 which decreases by 2 that means resulting element has atomic number 72. The element which has atomic number 70 is Hafnium.

The resulting equation is shown below.

W74160→H72156f+H24e

Conclusion

The equation of W−160 decay is stated above.

Interpretation Introduction

(b)

Interpretation:

The equation for P−32 decay by beta emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for P−32 decay by beta emission is shown below.

P1532→S1632+e−10

Explanation of Solution

It is given that P−32 undergoes decay by beta emission. A beta particle is identical to an electron that is e−10. Therefore, in the case of beta decay the atomic number of an element increases by 1 and there is no change in mass number. The atomic number of phosphorus is 15 which increases by 1 that means resulting element has atomic number 16. The element which has an atomic number 16 is sulfur.

The resulting equation is shown below.

P1532→S1632+e−10

Conclusion

The equation of P−32 decay is stated above.

Interpretation Introduction

(c)

Interpretation:

The equation for Co−55 decay by positron emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Co−55 decay by positron emission is shown below.

C2755o→F2655e+e+10

Explanation of Solution

It is given that Co−55 undergoes decay by positron emission. A positron is opposite of an electron and represented by e+10. Therefore, in the case of positron the atomic number of an element decreases by 1 and there is no change in mass number. The atomic number of cobalt is 27 which decreases by 1 that means resulting element has atomic number 26. The element which has an atomic number 26 is iron.

The resulting equation is shown below.

C2755o→F2655e+e+10

Conclusion

The equation of Co−55 decay is stated above.

Interpretation Introduction

(d)

Interpretation:

The equation for Ti−44 decay by electron capture is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Ti−44 decay by electron capture is shown below.

T2244i+e−10→S2144c

Explanation of Solution

It is given that Ti−44 undergoes decay by electron capture. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by 1, and there is no change in mass number. The atomic number of titanium is 22 which decreases by 1 that means resulting element has atomic number 21. The element which has an atomic number 21 is scandium.

The resulting equation is shown below.

T2244i+e−10→S2144c

Conclusion

The equation of Ti−44 decay is stated above.

Answer 6

Chapter 18, Problem 10E

Interpretation Introduction

(a)

Interpretation:

The equation for W−160 decay by alpha emission is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for W−160 decay by alpha emission is shown below.

W74160→H72156f+H24e

Explanation of Solution

It is given that W−160 undergoes decay by alpha emission. An alpha particle is H24e. Therefore, in the case of alpha decay the atomic number of an element decreases by 2 and the mass number decreases by 4. The atomic number of tungsten is 74 which decreases by 2 that means resulting element has atomic number 72. The element which has atomic number 70 is Hafnium.

The resulting equation is shown below.

W74160→H72156f+H24e

Conclusion

The equation of W−160 decay is stated above.

Answer 7

Chapter 18, Problem 10E

Interpretation Introduction

(a)

Interpretation:

The equation for W−160 decay by alpha emission is to be stated.

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer 8

Expert Solution

Answer to Problem 10E

The equation for W−160 decay by alpha emission is shown below.

W74160→H72156f+H24e

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

It is given that W−160 undergoes decay by alpha emission. An alpha particle is H24e. Therefore, in the case of alpha decay the atomic number of an element decreases by 2 and the mass number decreases by 4. The atomic number of tungsten is 74 which decreases by 2 that means resulting element has atomic number 72. The element which has atomic number 70 is Hafnium.

The resulting equation is shown below.

W74160→H72156f+H24e

Answer 13

Conclusion

The equation of W−160 decay is stated above.

Answer 14

Interpretation Introduction

(b)

Interpretation:

The equation for P−32 decay by beta emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for P−32 decay by beta emission is shown below.

P1532→S1632+e−10

Explanation of Solution

It is given that P−32 undergoes decay by beta emission. A beta particle is identical to an electron that is e−10. Therefore, in the case of beta decay the atomic number of an element increases by 1 and there is no change in mass number. The atomic number of phosphorus is 15 which increases by 1 that means resulting element has atomic number 16. The element which has an atomic number 16 is sulfur.

The resulting equation is shown below.

P1532→S1632+e−10

Conclusion

The equation of P−32 decay is stated above.

Answer 15

Interpretation Introduction

(b)

Interpretation:

The equation for P−32 decay by beta emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer 16

Expert Solution

Answer to Problem 10E

The equation for P−32 decay by beta emission is shown below.

P1532→S1632+e−10

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

It is given that P−32 undergoes decay by beta emission. A beta particle is identical to an electron that is e−10. Therefore, in the case of beta decay the atomic number of an element increases by 1 and there is no change in mass number. The atomic number of phosphorus is 15 which increases by 1 that means resulting element has atomic number 16. The element which has an atomic number 16 is sulfur.

The resulting equation is shown below.

P1532→S1632+e−10

Answer 21

Conclusion

The equation of P−32 decay is stated above.

Answer 22

Interpretation Introduction

(c)

Interpretation:

The equation for Co−55 decay by positron emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Co−55 decay by positron emission is shown below.

C2755o→F2655e+e+10

Explanation of Solution

It is given that Co−55 undergoes decay by positron emission. A positron is opposite of an electron and represented by e+10. Therefore, in the case of positron the atomic number of an element decreases by 1 and there is no change in mass number. The atomic number of cobalt is 27 which decreases by 1 that means resulting element has atomic number 26. The element which has an atomic number 26 is iron.

The resulting equation is shown below.

C2755o→F2655e+e+10

Conclusion

The equation of Co−55 decay is stated above.

Answer 23

Interpretation Introduction

(c)

Interpretation:

The equation for Co−55 decay by positron emission is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer 24

Expert Solution

Answer to Problem 10E

The equation for Co−55 decay by positron emission is shown below.

C2755o→F2655e+e+10

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

It is given that Co−55 undergoes decay by positron emission. A positron is opposite of an electron and represented by e+10. Therefore, in the case of positron the atomic number of an element decreases by 1 and there is no change in mass number. The atomic number of cobalt is 27 which decreases by 1 that means resulting element has atomic number 26. The element which has an atomic number 26 is iron.

The resulting equation is shown below.

C2755o→F2655e+e+10

Answer 29

Conclusion

The equation of Co−55 decay is stated above.

Answer 30

Interpretation Introduction

(d)

Interpretation:

The equation for Ti−44 decay by electron capture is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Expert Solution

Answer to Problem 10E

The equation for Ti−44 decay by electron capture is shown below.

T2244i+e−10→S2144c

Explanation of Solution

It is given that Ti−44 undergoes decay by electron capture. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by 1, and there is no change in mass number. The atomic number of titanium is 22 which decreases by 1 that means resulting element has atomic number 21. The element which has an atomic number 21 is scandium.

The resulting equation is shown below.

T2244i+e−10→S2144c

Conclusion

The equation of Ti−44 decay is stated above.

Answer 31

Interpretation Introduction

(d)

Interpretation:

The equation for Ti−44 decay by electron capture is to be stated

Concept introduction:

Heavy atoms have unstable nuclei. Due to which they are radioactive and disintegrates into smaller nuclei. The decay of radioactive nuclei can be done by alpha particle emission, beta-particle emission, positron decay, gamma decay, and electron capture.

Answer 32

Expert Solution

Answer to Problem 10E

The equation for Ti−44 decay by electron capture is shown below.

T2244i+e−10→S2144c

Answer 33

Expert Solution

Answer 34

Expert Solution

Answer 35

Expert Solution

Answer 36

Explanation of Solution

It is given that Ti−44 undergoes decay by electron capture. Electron capture means that when an electron is used to make any unstable atom stable. Therefore, in the case of electron capture the atomic number of an element decreases by 1, and there is no change in mass number. The atomic number of titanium is 22 which decreases by 1 that means resulting element has atomic number 21. The element which has an atomic number 21 is scandium.

The resulting equation is shown below.

T2244i+e−10→S2144c