(a) in which square in Fig 42.6 of the textbook does the radionuclide 196 I transform when it decays by an electron. (b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Question

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Answer 1

Question

Chapter 42, Problem 1Q

To determine

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) When the radionuclide ¹⁹⁶I decays by an electron, it transforms into ¹⁹⁶Pt.

(b) No further decay occurs after ¹⁹⁶I decays by an electron, and transforms into ¹⁹⁶Pt, because ¹⁹⁶Pt is a stable nucleus.

Explanation of Solution

Given:

i. A radionuclide ¹⁹⁶I decays by emitting an electron.

ii. Refer to Fig 42.6 of the textbook to know how this nuclide is transformed.

Concept

When a nuclide decays via an electron, a neutron is changing or transforming into a proton, hence the new nuclide will have the same mass A, but a higher Z, that is Z will be transformed to Z+1.

XZA YZ+1A because n p + e ^-+ ν-

From Fig 42.6 of the textbook, we get

Fundamentals of Physics, Volume 1, Chapter 1-20, Chapter 42, Problem 1Q , additional homework tip 3

The Z of Ir is 77, and Z of Pt is 78; when ¹⁹⁶I decays by emitting an electron, we get

I77196 Pt78196

From Fig 42.6 of the textbook, we can say that Pt78196 is a stable nucleus; therefore there will be no further decay of this nucleus.

Conclusion

When a nuclide decays by an electron (β^- - decay), its Z transforms to Z+1, but it’s mass A remains unchanged.

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Answer 2

Question

Chapter 42, Problem 1Q

To determine

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) When the radionuclide ¹⁹⁶I decays by an electron, it transforms into ¹⁹⁶Pt.

(b) No further decay occurs after ¹⁹⁶I decays by an electron, and transforms into ¹⁹⁶Pt, because ¹⁹⁶Pt is a stable nucleus.

Explanation of Solution

Given:

i. A radionuclide ¹⁹⁶I decays by emitting an electron.

ii. Refer to Fig 42.6 of the textbook to know how this nuclide is transformed.

Concept

When a nuclide decays via an electron, a neutron is changing or transforming into a proton, hence the new nuclide will have the same mass A, but a higher Z, that is Z will be transformed to Z+1.

XZA YZ+1A because n p + e ^-+ ν-

From Fig 42.6 of the textbook, we get

Fundamentals of Physics, Volume 1, Chapter 1-20, Chapter 42, Problem 1Q , additional homework tip 3

The Z of Ir is 77, and Z of Pt is 78; when ¹⁹⁶I decays by emitting an electron, we get

I77196 Pt78196

From Fig 42.6 of the textbook, we can say that Pt78196 is a stable nucleus; therefore there will be no further decay of this nucleus.

Conclusion

When a nuclide decays by an electron (β^- - decay), its Z transforms to Z+1, but it’s mass A remains unchanged.

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Answer 3

Question

Chapter 42, Problem 1Q

To determine

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) When the radionuclide ¹⁹⁶I decays by an electron, it transforms into ¹⁹⁶Pt.

(b) No further decay occurs after ¹⁹⁶I decays by an electron, and transforms into ¹⁹⁶Pt, because ¹⁹⁶Pt is a stable nucleus.

Explanation of Solution

Given:

i. A radionuclide ¹⁹⁶I decays by emitting an electron.

ii. Refer to Fig 42.6 of the textbook to know how this nuclide is transformed.

Concept

When a nuclide decays via an electron, a neutron is changing or transforming into a proton, hence the new nuclide will have the same mass A, but a higher Z, that is Z will be transformed to Z+1.

XZA YZ+1A because n p + e ^-+ ν-

From Fig 42.6 of the textbook, we get

Fundamentals of Physics, Volume 1, Chapter 1-20, Chapter 42, Problem 1Q , additional homework tip 3

The Z of Ir is 77, and Z of Pt is 78; when ¹⁹⁶I decays by emitting an electron, we get

I77196 Pt78196

From Fig 42.6 of the textbook, we can say that Pt78196 is a stable nucleus; therefore there will be no further decay of this nucleus.

Conclusion

When a nuclide decays by an electron (β^- - decay), its Z transforms to Z+1, but it’s mass A remains unchanged.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 42, Problem 1Q

To determine

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) When the radionuclide ¹⁹⁶I decays by an electron, it transforms into ¹⁹⁶Pt.

(b) No further decay occurs after ¹⁹⁶I decays by an electron, and transforms into ¹⁹⁶Pt, because ¹⁹⁶Pt is a stable nucleus.

Explanation of Solution

Given:

i. A radionuclide ¹⁹⁶I decays by emitting an electron.

ii. Refer to Fig 42.6 of the textbook to know how this nuclide is transformed.

Concept

When a nuclide decays via an electron, a neutron is changing or transforming into a proton, hence the new nuclide will have the same mass A, but a higher Z, that is Z will be transformed to Z+1.

XZA YZ+1A because n p + e ^-+ ν-

From Fig 42.6 of the textbook, we get

Fundamentals of Physics, Volume 1, Chapter 1-20, Chapter 42, Problem 1Q , additional homework tip 3

The Z of Ir is 77, and Z of Pt is 78; when ¹⁹⁶I decays by emitting an electron, we get

I77196 Pt78196

From Fig 42.6 of the textbook, we can say that Pt78196 is a stable nucleus; therefore there will be no further decay of this nucleus.

Conclusion

When a nuclide decays by an electron (β^- - decay), its Z transforms to Z+1, but it’s mass A remains unchanged.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 42, Problem 1Q

To determine

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) When the radionuclide ¹⁹⁶I decays by an electron, it transforms into ¹⁹⁶Pt.

(b) No further decay occurs after ¹⁹⁶I decays by an electron, and transforms into ¹⁹⁶Pt, because ¹⁹⁶Pt is a stable nucleus.

Explanation of Solution

Given:

i. A radionuclide ¹⁹⁶I decays by emitting an electron.

ii. Refer to Fig 42.6 of the textbook to know how this nuclide is transformed.

Concept

When a nuclide decays via an electron, a neutron is changing or transforming into a proton, hence the new nuclide will have the same mass A, but a higher Z, that is Z will be transformed to Z+1.

XZA YZ+1A because n p + e ^-+ ν-

From Fig 42.6 of the textbook, we get

Fundamentals of Physics, Volume 1, Chapter 1-20, Chapter 42, Problem 1Q , additional homework tip 3

The Z of Ir is 77, and Z of Pt is 78; when ¹⁹⁶I decays by emitting an electron, we get

I77196 Pt78196

From Fig 42.6 of the textbook, we can say that Pt78196 is a stable nucleus; therefore there will be no further decay of this nucleus.

Conclusion

When a nuclide decays by an electron (β^- - decay), its Z transforms to Z+1, but it’s mass A remains unchanged.

Answer 6

Chapter 42, Problem 1Q

To determine

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) When the radionuclide ¹⁹⁶I decays by an electron, it transforms into ¹⁹⁶Pt.

(b) No further decay occurs after ¹⁹⁶I decays by an electron, and transforms into ¹⁹⁶Pt, because ¹⁹⁶Pt is a stable nucleus.

Explanation of Solution

Given:

i. A radionuclide ¹⁹⁶I decays by emitting an electron.

ii. Refer to Fig 42.6 of the textbook to know how this nuclide is transformed.

Concept

When a nuclide decays via an electron, a neutron is changing or transforming into a proton, hence the new nuclide will have the same mass A, but a higher Z, that is Z will be transformed to Z+1.

XZA YZ+1A because n p + e ^-+ ν-

From Fig 42.6 of the textbook, we get

Fundamentals of Physics, Volume 1, Chapter 1-20, Chapter 42, Problem 1Q , additional homework tip 3

The Z of Ir is 77, and Z of Pt is 78; when ¹⁹⁶I decays by emitting an electron, we get

I77196 Pt78196

From Fig 42.6 of the textbook, we can say that Pt78196 is a stable nucleus; therefore there will be no further decay of this nucleus.

Conclusion

When a nuclide decays by an electron (β^- - decay), its Z transforms to Z+1, but it’s mass A remains unchanged.

Answer 7

Chapter 42, Problem 1Q

To determine

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Answer 8

Expert Solution & Answer

Answer to Problem 1Q

Solution:

(a) When the radionuclide ¹⁹⁶I decays by an electron, it transforms into ¹⁹⁶Pt.

(b) No further decay occurs after ¹⁹⁶I decays by an electron, and transforms into ¹⁹⁶Pt, because ¹⁹⁶Pt is a stable nucleus.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

i. A radionuclide ¹⁹⁶I decays by emitting an electron.

ii. Refer to Fig 42.6 of the textbook to know how this nuclide is transformed.

Concept

When a nuclide decays via an electron, a neutron is changing or transforming into a proton, hence the new nuclide will have the same mass A, but a higher Z, that is Z will be transformed to Z+1.

XZA YZ+1A because n p + e ^-+ ν-

From Fig 42.6 of the textbook, we get

Fundamentals of Physics, Volume 1, Chapter 1-20, Chapter 42, Problem 1Q , additional homework tip 3

The Z of Ir is 77, and Z of Pt is 78; when ¹⁹⁶I decays by emitting an electron, we get

I77196 Pt78196

From Fig 42.6 of the textbook, we can say that Pt78196 is a stable nucleus; therefore there will be no further decay of this nucleus.

Conclusion

When a nuclide decays by an electron (β^- - decay), its Z transforms to Z+1, but it’s mass A remains unchanged.

Answer 12

Ch. 42 - Prob. 1Q Ch. 42 - Prob. 2Q Ch. 42 - Prob. 3Q Ch. 42 - Prob. 4Q Ch. 42 - Prob. 5Q Ch. 42 - Prob. 6Q Ch. 42 - Prob. 7Q Ch. 42 - Prob. 8Q Ch. 42 - Prob. 9Q Ch. 42 - Prob. 10Q

(a) in which square in Fig 42.6 of the textbook does the radionuclide 196 I transform when it decays by an electron. (b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Concept explainers

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Answer to Problem 1Q

Explanation of Solution

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Chapter 42 Solutions

(a) in which square in Fig 42.6 of the textbook does the radionuclide 196 I transform when it decays by an electron. (b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Concept explainers

To find: (a) in which square in Fig 42.6 of the textbook does the radionuclide 196I transform when it decays by an electron. (b) if any further decays occur, by looking at Fig 42.6 of the textbook.

Answer to Problem 1Q

Explanation of Solution

Want to see more full solutions like this?

Chapter 42 Solutions

To find:

(a) in which square in Fig 42.6 of the textbook does the radionuclide ¹⁹⁶I transform when it decays by an electron.

(b) if any further decays occur, by looking at Fig 42.6 of the textbook.