The process of transformation of T 230 h to R 226 a

Question

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Answer 1

Question

Chapter 14, Problem 14.15PAE

(a)

Interpretation Introduction

To determine:

The process of transformation of T230h to R226a

(a)

Expert Solution

Explanation of Solution

Atomic numbers of thorium and radium are 90 and 88, respectively.

Consider the transformation:

T90230h→R88226a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 90−88=2 and the atomic mass as 230−226=4. The nucleus with atomic number 2 and mass number 4 is aHelium nucleus. Hence, the process leading to the transformation of T230h to R226a is an alpha emission process.

T90230h→R88226a+H24e

(b)

Interpretation Introduction

To determine:

The process of transformation of C137s to B137a

(b)

Expert Solution

Explanation of Solution

Atomic numbers of Cs and Ba are 55 and 56, respectively.

Consider the transformation:

C55137s→B56137a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 55−56=−1 and the atomic mass as 137−137=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of C137s to B137a is a beta emission process.

C55137s→B56137a+β−10+ν¯

(c)

Interpretation Introduction

To determine:

The process of transformation of K38 to A38r

(c)

Expert Solution

Explanation of Solution

Atomic numbers of K and Ar are 19 and 18, respectively.

Consider the transformation:

K1938→A1838r

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 19−18=1 and the atomic mass as 38−38=0. The particle with atomic number 1 and mass number 0 is a positron. Hence, the process leading to the transformation of K38 to A38r is a positron emission process.

K1938→A1838r+β10+ν

(d)

Interpretation Introduction

To determine:

The process of transformation of Z97r to N97b

(d)

Expert Solution

Explanation of Solution

Atomic numbers of Zr and Nb are 40 and 41, respectively.

Consider the transformation:

Z4097r→N4197b

Thus, balancing the atomic number on both sides, we get that the unknown particle ejected has atomic number of 40−41=−1 and the atomic mass as 97−97=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of Z97r to N97b is a beta emission process.

K1938→A1838r+β10+ν

Conclusion

Therefore, if one of the species from the radioactive decay equation is missing, it can be determined by balancing atomic numbers and atomic mass numbers. Thus, the type of radioactive decay process can be determined if the atomic mass and atomic number of transforming nuclei is given.

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Answer 2

Question

Chapter 14, Problem 14.15PAE

(a)

Interpretation Introduction

To determine:

The process of transformation of T230h to R226a

(a)

Expert Solution

Explanation of Solution

Atomic numbers of thorium and radium are 90 and 88, respectively.

Consider the transformation:

T90230h→R88226a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 90−88=2 and the atomic mass as 230−226=4. The nucleus with atomic number 2 and mass number 4 is aHelium nucleus. Hence, the process leading to the transformation of T230h to R226a is an alpha emission process.

T90230h→R88226a+H24e

(b)

Interpretation Introduction

To determine:

The process of transformation of C137s to B137a

(b)

Expert Solution

Explanation of Solution

Atomic numbers of Cs and Ba are 55 and 56, respectively.

Consider the transformation:

C55137s→B56137a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 55−56=−1 and the atomic mass as 137−137=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of C137s to B137a is a beta emission process.

C55137s→B56137a+β−10+ν¯

(c)

Interpretation Introduction

To determine:

The process of transformation of K38 to A38r

(c)

Expert Solution

Explanation of Solution

Atomic numbers of K and Ar are 19 and 18, respectively.

Consider the transformation:

K1938→A1838r

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 19−18=1 and the atomic mass as 38−38=0. The particle with atomic number 1 and mass number 0 is a positron. Hence, the process leading to the transformation of K38 to A38r is a positron emission process.

K1938→A1838r+β10+ν

(d)

Interpretation Introduction

To determine:

The process of transformation of Z97r to N97b

(d)

Expert Solution

Explanation of Solution

Atomic numbers of Zr and Nb are 40 and 41, respectively.

Consider the transformation:

Z4097r→N4197b

Thus, balancing the atomic number on both sides, we get that the unknown particle ejected has atomic number of 40−41=−1 and the atomic mass as 97−97=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of Z97r to N97b is a beta emission process.

K1938→A1838r+β10+ν

Conclusion

Therefore, if one of the species from the radioactive decay equation is missing, it can be determined by balancing atomic numbers and atomic mass numbers. Thus, the type of radioactive decay process can be determined if the atomic mass and atomic number of transforming nuclei is given.

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Answer 3

Question

Chapter 14, Problem 14.15PAE

(a)

Interpretation Introduction

To determine:

The process of transformation of T230h to R226a

(a)

Expert Solution

Explanation of Solution

Atomic numbers of thorium and radium are 90 and 88, respectively.

Consider the transformation:

T90230h→R88226a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 90−88=2 and the atomic mass as 230−226=4. The nucleus with atomic number 2 and mass number 4 is aHelium nucleus. Hence, the process leading to the transformation of T230h to R226a is an alpha emission process.

T90230h→R88226a+H24e

(b)

Interpretation Introduction

To determine:

The process of transformation of C137s to B137a

(b)

Expert Solution

Explanation of Solution

Atomic numbers of Cs and Ba are 55 and 56, respectively.

Consider the transformation:

C55137s→B56137a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 55−56=−1 and the atomic mass as 137−137=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of C137s to B137a is a beta emission process.

C55137s→B56137a+β−10+ν¯

(c)

Interpretation Introduction

To determine:

The process of transformation of K38 to A38r

(c)

Expert Solution

Explanation of Solution

Atomic numbers of K and Ar are 19 and 18, respectively.

Consider the transformation:

K1938→A1838r

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 19−18=1 and the atomic mass as 38−38=0. The particle with atomic number 1 and mass number 0 is a positron. Hence, the process leading to the transformation of K38 to A38r is a positron emission process.

K1938→A1838r+β10+ν

(d)

Interpretation Introduction

To determine:

The process of transformation of Z97r to N97b

(d)

Expert Solution

Explanation of Solution

Atomic numbers of Zr and Nb are 40 and 41, respectively.

Consider the transformation:

Z4097r→N4197b

Thus, balancing the atomic number on both sides, we get that the unknown particle ejected has atomic number of 40−41=−1 and the atomic mass as 97−97=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of Z97r to N97b is a beta emission process.

K1938→A1838r+β10+ν

Conclusion

Therefore, if one of the species from the radioactive decay equation is missing, it can be determined by balancing atomic numbers and atomic mass numbers. Thus, the type of radioactive decay process can be determined if the atomic mass and atomic number of transforming nuclei is given.

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Answer 4

Question

Chapter 14, Problem 14.15PAE

(a)

Interpretation Introduction

To determine:

The process of transformation of T230h to R226a

(a)

Expert Solution

Explanation of Solution

Atomic numbers of thorium and radium are 90 and 88, respectively.

Consider the transformation:

T90230h→R88226a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 90−88=2 and the atomic mass as 230−226=4. The nucleus with atomic number 2 and mass number 4 is aHelium nucleus. Hence, the process leading to the transformation of T230h to R226a is an alpha emission process.

T90230h→R88226a+H24e

(b)

Interpretation Introduction

To determine:

The process of transformation of C137s to B137a

(b)

Expert Solution

Explanation of Solution

Atomic numbers of Cs and Ba are 55 and 56, respectively.

Consider the transformation:

C55137s→B56137a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 55−56=−1 and the atomic mass as 137−137=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of C137s to B137a is a beta emission process.

C55137s→B56137a+β−10+ν¯

(c)

Interpretation Introduction

To determine:

The process of transformation of K38 to A38r

(c)

Expert Solution

Explanation of Solution

Atomic numbers of K and Ar are 19 and 18, respectively.

Consider the transformation:

K1938→A1838r

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 19−18=1 and the atomic mass as 38−38=0. The particle with atomic number 1 and mass number 0 is a positron. Hence, the process leading to the transformation of K38 to A38r is a positron emission process.

K1938→A1838r+β10+ν

(d)

Interpretation Introduction

To determine:

The process of transformation of Z97r to N97b

(d)

Expert Solution

Explanation of Solution

Atomic numbers of Zr and Nb are 40 and 41, respectively.

Consider the transformation:

Z4097r→N4197b

Thus, balancing the atomic number on both sides, we get that the unknown particle ejected has atomic number of 40−41=−1 and the atomic mass as 97−97=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of Z97r to N97b is a beta emission process.

K1938→A1838r+β10+ν

Conclusion

Therefore, if one of the species from the radioactive decay equation is missing, it can be determined by balancing atomic numbers and atomic mass numbers. Thus, the type of radioactive decay process can be determined if the atomic mass and atomic number of transforming nuclei is given.

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Answer 5

Chapter 14, Problem 14.15PAE

(a)

Interpretation Introduction

To determine:

The process of transformation of T230h to R226a

(a)

Expert Solution

Explanation of Solution

Atomic numbers of thorium and radium are 90 and 88, respectively.

Consider the transformation:

T90230h→R88226a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 90−88=2 and the atomic mass as 230−226=4. The nucleus with atomic number 2 and mass number 4 is aHelium nucleus. Hence, the process leading to the transformation of T230h to R226a is an alpha emission process.

T90230h→R88226a+H24e

(b)

Interpretation Introduction

To determine:

The process of transformation of C137s to B137a

(b)

Expert Solution

Explanation of Solution

Atomic numbers of Cs and Ba are 55 and 56, respectively.

Consider the transformation:

C55137s→B56137a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 55−56=−1 and the atomic mass as 137−137=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of C137s to B137a is a beta emission process.

C55137s→B56137a+β−10+ν¯

(c)

Interpretation Introduction

To determine:

The process of transformation of K38 to A38r

(c)

Expert Solution

Explanation of Solution

Atomic numbers of K and Ar are 19 and 18, respectively.

Consider the transformation:

K1938→A1838r

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 19−18=1 and the atomic mass as 38−38=0. The particle with atomic number 1 and mass number 0 is a positron. Hence, the process leading to the transformation of K38 to A38r is a positron emission process.

K1938→A1838r+β10+ν

(d)

Interpretation Introduction

To determine:

The process of transformation of Z97r to N97b

(d)

Expert Solution

Explanation of Solution

Atomic numbers of Zr and Nb are 40 and 41, respectively.

Consider the transformation:

Z4097r→N4197b

Thus, balancing the atomic number on both sides, we get that the unknown particle ejected has atomic number of 40−41=−1 and the atomic mass as 97−97=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of Z97r to N97b is a beta emission process.

K1938→A1838r+β10+ν

Conclusion

Therefore, if one of the species from the radioactive decay equation is missing, it can be determined by balancing atomic numbers and atomic mass numbers. Thus, the type of radioactive decay process can be determined if the atomic mass and atomic number of transforming nuclei is given.

Answer 6

Chapter 14, Problem 14.15PAE

(a)

Interpretation Introduction

To determine:

The process of transformation of T230h to R226a

(a)

Expert Solution

Explanation of Solution

Atomic numbers of thorium and radium are 90 and 88, respectively.

Consider the transformation:

T90230h→R88226a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 90−88=2 and the atomic mass as 230−226=4. The nucleus with atomic number 2 and mass number 4 is aHelium nucleus. Hence, the process leading to the transformation of T230h to R226a is an alpha emission process.

T90230h→R88226a+H24e

Answer 7

Chapter 14, Problem 14.15PAE

(a)

Interpretation Introduction

To determine:

The process of transformation of T230h to R226a

Answer 8

(a)

Expert Solution

Explanation of Solution

Atomic numbers of thorium and radium are 90 and 88, respectively.

Consider the transformation:

T90230h→R88226a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 90−88=2 and the atomic mass as 230−226=4. The nucleus with atomic number 2 and mass number 4 is aHelium nucleus. Hence, the process leading to the transformation of T230h to R226a is an alpha emission process.

T90230h→R88226a+H24e

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

To determine:

The process of transformation of C137s to B137a

(b)

Expert Solution

Explanation of Solution

Atomic numbers of Cs and Ba are 55 and 56, respectively.

Consider the transformation:

C55137s→B56137a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 55−56=−1 and the atomic mass as 137−137=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of C137s to B137a is a beta emission process.

C55137s→B56137a+β−10+ν¯

Answer 13

(b)

Interpretation Introduction

To determine:

The process of transformation of C137s to B137a

Answer 14

(b)

Expert Solution

Explanation of Solution

Atomic numbers of Cs and Ba are 55 and 56, respectively.

Consider the transformation:

C55137s→B56137a

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 55−56=−1 and the atomic mass as 137−137=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of C137s to B137a is a beta emission process.

C55137s→B56137a+β−10+ν¯

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

To determine:

The process of transformation of K38 to A38r

(c)

Expert Solution

Explanation of Solution

Atomic numbers of K and Ar are 19 and 18, respectively.

Consider the transformation:

K1938→A1838r

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 19−18=1 and the atomic mass as 38−38=0. The particle with atomic number 1 and mass number 0 is a positron. Hence, the process leading to the transformation of K38 to A38r is a positron emission process.

K1938→A1838r+β10+ν

Answer 19

(c)

Interpretation Introduction

To determine:

The process of transformation of K38 to A38r

Answer 20

(c)

Expert Solution

Explanation of Solution

Atomic numbers of K and Ar are 19 and 18, respectively.

Consider the transformation:

K1938→A1838r

Thus, balancing the atomic number on both sides, we get that unknown particle ejected has atomic number of 19−18=1 and the atomic mass as 38−38=0. The particle with atomic number 1 and mass number 0 is a positron. Hence, the process leading to the transformation of K38 to A38r is a positron emission process.

K1938→A1838r+β10+ν

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

(d)

Interpretation Introduction

To determine:

The process of transformation of Z97r to N97b

(d)

Expert Solution

Explanation of Solution

Atomic numbers of Zr and Nb are 40 and 41, respectively.

Consider the transformation:

Z4097r→N4197b

Thus, balancing the atomic number on both sides, we get that the unknown particle ejected has atomic number of 40−41=−1 and the atomic mass as 97−97=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of Z97r to N97b is a beta emission process.

K1938→A1838r+β10+ν

Conclusion

Therefore, if one of the species from the radioactive decay equation is missing, it can be determined by balancing atomic numbers and atomic mass numbers. Thus, the type of radioactive decay process can be determined if the atomic mass and atomic number of transforming nuclei is given.

Answer 25

(d)

Interpretation Introduction

To determine:

The process of transformation of Z97r to N97b

Answer 26

(d)

Expert Solution

Explanation of Solution

Atomic numbers of Zr and Nb are 40 and 41, respectively.

Consider the transformation:

Z4097r→N4197b

Thus, balancing the atomic number on both sides, we get that the unknown particle ejected has atomic number of 40−41=−1 and the atomic mass as 97−97=0. The particle with atomic number -1 and mass number 0 is an electron. Hence, the process leading to the transformation of Z97r to N97b is a beta emission process.

K1938→A1838r+β10+ν

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Conclusion

Therefore, if one of the species from the radioactive decay equation is missing, it can be determined by balancing atomic numbers and atomic mass numbers. Thus, the type of radioactive decay process can be determined if the atomic mass and atomic number of transforming nuclei is given.