General, Organic, And Biological Chemistry, Hybrid (with Owlv2 Quick Prep For General Chemistry Printed Access Card)
General, Organic, And Biological Chemistry, Hybrid (with Owlv2 Quick Prep For General Chemistry Printed Access Card)
7th Edition
ISBN: 9781305253070
Author: STOKER, H. Stephen
Publisher: Cengage Learning
Question
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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.

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Chapter 11 Solutions

General, Organic, And Biological Chemistry, Hybrid (with Owlv2 Quick Prep For General Chemistry Printed Access Card)

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
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