General, Organic, and Biological Chemistry
General, Organic, and Biological Chemistry
7th Edition
ISBN: 9781285853918
Author: H. Stephen Stoker
Publisher: Cengage Learning
Question
Book Icon
Chapter 11, Problem 11.28EP

(a)

Interpretation Introduction

Interpretation:

Half-life of the radionuclide has to be determined if after 3.2 days, 1/8 fraction of undecayed nuclide is present.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  All the radioactive nuclide do not undergo the decay at a same rate.  Some decay rapidly and others decay very slowly.  The nuclear stability can be quantitatively expressed by using the half-life.

The time required for half quantity of the radioactive substance to undergo decay is known as half-life.  It is represented as t1/2.

The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,

(Amountofradionuclideundecayedafternhalflives) =  (Originalamountofradionuclide)  x  (12n)

(a)

Expert Solution
Check Mark

Answer to Problem 11.28EP

Half-life of the radionuclide is 1.1 days.

Explanation of Solution

Number of half-lives can be determined as shown below,

(12n) = 18(12n) = 1232n = 23

As the bases are equal, the power can be equated.  This gives the number of half-lives that have elapsed as 3 half-lives.

In the problem statement it is given that the time is 3.2 days.  From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,

3.2 days   x  (1 half-lifet1/2) = 3 half-lives t1/2 = 3.2 days3 = 1.1days

Therefore, the half-life of the given sample is determined as 1.1 days.

Conclusion

Half-life of the given sample is determined.

(b)

Interpretation Introduction

Interpretation:

Half-life of the radionuclide has to be determined if after 3.2 days, 1/128 fraction of undecayed nuclide is present.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  All the radioactive nuclide do not undergo the decay at a same rate.  Some decay rapidly and others decay very slowly.  The nuclear stability can be quantitatively expressed by using the half-life.

The time required for half quantity of the radioactive substance to undergo decay is known as half-life.  It is represented as t1/2.

The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,

(Amountofradionuclideundecayedafternhalflives) =  (Originalamountofradionuclide)  x  (12n)

(b)

Expert Solution
Check Mark

Answer to Problem 11.28EP

Half-life of the radionuclide is 0.46 day.

Explanation of Solution

Number of half-lives can be determined as shown below,

(12n) = 1128(12n) = 1272n = 27

As the bases are equal, the power can be equated.  This gives the number of half-lives that have elapsed as 7 half-lives.

In the problem statement it is given that the time is 3.2 days.  From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,

3.2 days   x  (1 half-lifet1/2) = 7 half-lives t1/2 = 3.2 days7 = 0.46day

Therefore, the half-life of the given sample is determined as 0.46 day.

Conclusion

Half-life of the given sample is determined.

(c)

Interpretation Introduction

Interpretation:

Half-life of the radionuclide has to be determined if after 3.2 days, 1/32 fraction of undecayed nuclide is present.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  All the radioactive nuclide do not undergo the decay at a same rate.  Some decay rapidly and others decay very slowly.  The nuclear stability can be quantitatively expressed by using the half-life.

The time required for half quantity of the radioactive substance to undergo decay is known as half-life.  It is represented as t1/2.

The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,

(Amountofradionuclideundecayedafternhalflives) =  (Originalamountofradionuclide)  x  (12n)

(c)

Expert Solution
Check Mark

Answer to Problem 11.28EP

Half-life of the radionuclide is 0.64 day.

Explanation of Solution

Number of half-lives can be determined as shown below,

(12n) = 132(12n) = 1252n = 25

As the bases are equal, the power can be equated.  This gives the number of half-lives that have elapsed as 5 half-lives.

In the problem statement it is given that the time is 3.2 days.  From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,

3.2 days   x  (1 half-lifet1/2) = 5 half-lives t1/2 = 3.2 days5 = 0.64day

Therefore, the half-life of the given sample is determined as 0.64 day.

Conclusion

Half-life of the given sample is determined.

(d)

Interpretation Introduction

Interpretation:

Half-life of the radionuclide has to be determined if after 3.2 days, 1/512 fraction of undecayed nuclide is present.

Concept Introduction:

Radioactive nuclides undergo disintegration by emission of radiation.  All the radioactive nuclide do not undergo the decay at a same rate.  Some decay rapidly and others decay very slowly.  The nuclear stability can be quantitatively expressed by using the half-life.

The time required for half quantity of the radioactive substance to undergo decay is known as half-life.  It is represented as t1/2.

The equation that relates amount of decayed radioactive material, amount of undecayed radioactive material and the time elapsed can be given as,

(Amountofradionuclideundecayedafternhalflives) =  (Originalamountofradionuclide)  x  (12n)

(d)

Expert Solution
Check Mark

Answer to Problem 11.28EP

Half-life of the radionuclide is 0.36 day.

Explanation of Solution

Number of half-lives can be determined as shown below,

(12n) = 1512(12n) = 1292n = 29

As the bases are equal, the power can be equated.  This gives the number of half-lives that have elapsed as 9 half-lives.

In the problem statement it is given that the time is 3.2 days.  From the number of half-lives elapsed and the total time given, the length of one half-life can be calculated as shown below,

3.2 days   x  (1 half-lifet1/2) = 9 half-lives t1/2 = 3.2 days9 = 0.36day

Therefore, the half-life of the given sample is determined as 0.36 day.

Conclusion

Half-life of the given sample is determined.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
reaction scheme for C39H4202 Hydrogenation of Alkyne (Alkyne to Alkene) show reaction (drawing) please
Give detailed mechanism Solution with explanation needed. Don't give Ai generated solution
Show work with explanation needed....don't give Ai generated solution

Chapter 11 Solutions

General, Organic, and Biological Chemistry

Ch. 11.4 - The half-life of cobalt-60 is 5.2 years. This...Ch. 11.4 - Prob. 2QQCh. 11.4 - Prob. 3QQCh. 11.4 - Prob. 4QQCh. 11.4 - Prob. 5QQCh. 11.5 - Prob. 1QQCh. 11.5 - The bombardment reaction involving 1123Na and 12H...Ch. 11.5 - Prob. 3QQCh. 11.5 - Prob. 4QQCh. 11.6 - Prob. 1QQCh. 11.6 - In the 14-step uranium-238 decay series a. all...Ch. 11.7 - Prob. 1QQCh. 11.7 - Prob. 2QQCh. 11.8 - Which of the following is not a form of ionizing...Ch. 11.8 - Prob. 2QQCh. 11.8 - Prob. 3QQCh. 11.8 - Prob. 4QQCh. 11.9 - Prob. 1QQCh. 11.9 - Which of the following correctly orders the three...Ch. 11.10 - Prob. 1QQCh. 11.10 - Prob. 2QQCh. 11.10 - Prob. 3QQCh. 11.11 - Prob. 1QQCh. 11.11 - Prob. 2QQCh. 11.11 - Prob. 3QQCh. 11.12 - Prob. 1QQCh. 11.12 - Prob. 2QQCh. 11.12 - Prob. 3QQCh. 11.12 - Prob. 4QQCh. 11.13 - Prob. 1QQCh. 11.13 - Prob. 2QQCh. 11 - Prob. 11.1EPCh. 11 - Prob. 11.2EPCh. 11 - Prob. 11.3EPCh. 11 - Prob. 11.4EPCh. 11 - Prob. 11.5EPCh. 11 - Prob. 11.6EPCh. 11 - Prob. 11.7EPCh. 11 - Prob. 11.8EPCh. 11 - Prob. 11.9EPCh. 11 - Prob. 11.10EPCh. 11 - Prob. 11.11EPCh. 11 - Prob. 11.12EPCh. 11 - Prob. 11.13EPCh. 11 - Prob. 11.14EPCh. 11 - Prob. 11.15EPCh. 11 - Prob. 11.16EPCh. 11 - Prob. 11.17EPCh. 11 - Prob. 11.18EPCh. 11 - Prob. 11.19EPCh. 11 - Prob. 11.20EPCh. 11 - Prob. 11.21EPCh. 11 - Prob. 11.22EPCh. 11 - Prob. 11.23EPCh. 11 - Prob. 11.24EPCh. 11 - Prob. 11.25EPCh. 11 - Prob. 11.26EPCh. 11 - Prob. 11.27EPCh. 11 - Prob. 11.28EPCh. 11 - Prob. 11.29EPCh. 11 - Fill in the blanks in each line of the following...Ch. 11 - Prob. 11.31EPCh. 11 - Prob. 11.32EPCh. 11 - Prob. 11.33EPCh. 11 - Prob. 11.34EPCh. 11 - Prob. 11.35EPCh. 11 - Prob. 11.36EPCh. 11 - Prob. 11.37EPCh. 11 - Prob. 11.38EPCh. 11 - Prob. 11.39EPCh. 11 - Prob. 11.40EPCh. 11 - Prob. 11.41EPCh. 11 - Prob. 11.42EPCh. 11 - Prob. 11.43EPCh. 11 - Prob. 11.44EPCh. 11 - Prob. 11.45EPCh. 11 - Prob. 11.46EPCh. 11 - Prob. 11.47EPCh. 11 - Prob. 11.48EPCh. 11 - Prob. 11.49EPCh. 11 - Prob. 11.50EPCh. 11 - Prob. 11.51EPCh. 11 - Prob. 11.52EPCh. 11 - Prob. 11.53EPCh. 11 - Prob. 11.54EPCh. 11 - Prob. 11.55EPCh. 11 - Prob. 11.56EPCh. 11 - Prob. 11.57EPCh. 11 - Write a chemical equation that involves water as a...Ch. 11 - Prob. 11.59EPCh. 11 - Prob. 11.60EPCh. 11 - Prob. 11.61EPCh. 11 - Prob. 11.62EPCh. 11 - Prob. 11.63EPCh. 11 - Prob. 11.64EPCh. 11 - Prob. 11.65EPCh. 11 - Prob. 11.66EPCh. 11 - Prob. 11.67EPCh. 11 - Prob. 11.68EPCh. 11 - Prob. 11.69EPCh. 11 - Prob. 11.70EPCh. 11 - Prob. 11.71EPCh. 11 - Prob. 11.72EPCh. 11 - Prob. 11.73EPCh. 11 - Prob. 11.74EPCh. 11 - Prob. 11.75EPCh. 11 - Prob. 11.76EPCh. 11 - Prob. 11.77EPCh. 11 - Prob. 11.78EPCh. 11 - Prob. 11.79EPCh. 11 - Prob. 11.80EPCh. 11 - Prob. 11.81EPCh. 11 - Prob. 11.82EPCh. 11 - Prob. 11.83EPCh. 11 - Prob. 11.84EPCh. 11 - Prob. 11.85EPCh. 11 - Prob. 11.86EPCh. 11 - Prob. 11.87EPCh. 11 - Prob. 11.88EPCh. 11 - Prob. 11.89EPCh. 11 - Prob. 11.90EP
Knowledge Booster
Background pattern image
Recommended textbooks for you
Text book image
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Text book image
Introductory Chemistry For Today
Chemistry
ISBN:9781285644561
Author:Seager
Publisher:Cengage
Text book image
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Text book image
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
Text book image
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Text book image
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning