Student Solutions Manual for Zumdahl/Zumdahl/DeCoste?s Chemistry, 10th Edition
Student Solutions Manual for Zumdahl/Zumdahl/DeCoste?s Chemistry, 10th Edition
10th Edition
ISBN: 9781305957510
Author: ZUMDAHL, Steven S.; Zumdahl, Susan A.; DeCoste, Donald J.
Publisher: Cengage Learning
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Chapter 19, Problem 34E
Interpretation Introduction

Interpretation: List of isotopes of krypton is given. The most stable and the hottest among them is to be stated. Time of decay of 87.5% of each isotope is to be stated.

Concept introduction: Decay constant is the quantity that expresses the rate of decrease of number of atoms of a radioactive element per second. Half life of radioactive sample is defined as the time required for the number of nuclides to reach half of the original value.

The nuclides having longer half life are more stable while nuclides having shorter half life are less stable.

To determine: The most stable and the hottest isotope among the given isotopes of krypton; the time of decay for 73Kr ; the time of decay for 74Kr ; the time of decay for 76Kr and the time of decay for 81Kr .

Expert Solution & Answer
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Answer to Problem 34E

Answer

The most stable isotope is 81Kr and the hottest one is 73Kr

The time of decay for 87.5%of73Kr is 81.428s_ .

The time of decay for 87.5%of74Kr is. 34.51s_

The time of decay for 87.5%of76Kr is 44.41h_

The time of decay for 87.5%of81Kr is 6.302×105years_

Explanation of Solution

Explanation

The most stable isotope is 81Kr and the hottest one is 73Kr

The nuclides having longer half life are more stable while nuclides having shorter half life are less stable. Thus the most stable isotope is 81Kr and the hottest isotope is 73Kr .

The time of decay for 73Kr is 81.428s_

Explanation

The decay constant can be calculated by the formula given below.

λ=0.693t1/2

Where

  • t1/2 is the half life of nuclide.
  • λ is the decay constant.

Substitute the value of λ in the above expression.

λ=0.69327s1=0.0256s1

The fraction of isotope decayed is 87.5100 .

The fraction remaining =187.5100=12.5100

The time of decay can be calculated by the formula,

t=2.303λlogn0n

Where

  • n0 is the number of atoms initially present.
  • n is the number of atoms after time “t”.

Substitute the values of λ , n0 and n in the above expression.

t=2.303λlogn0nt=2.303λlog10012.5t=2.3030.0256log10012.5t=81.428s_ .

The time of decay for 74Kr is 34.51s_ .

Explanation

The decay constant is calculated by the formula,

λ=0.693t1/2

Where

  • t1/2 is the half life of nuclide.
  • λ is the decay constant.

Substitute the value of t1/2 in the above expression.

λ=0.69311.5min1

The fraction of isotope decayed is 87.5100 .

The fraction remaining =187.5100=12.5100

The time of decay can be calculated by the formula,

t=2.303λlogn0n

Where

  • n0 is the number of atoms initially present.
  • n is the number of atoms after time “t”.

Substitute the values of λ , n0 and n in the above expression.

t=2.303λlogn0nt=2.303λlog10012.5t=2.303×11.50.693log8t=34.51min_

The time of decay for 76Kr is 44.41h_

Explanation

The decay constant can be calculated by the formula given below.

λ=0.693t1/2

Where

  • t1/2 is the half life of nuclide.
  • λ is the decay constant.

Substitute the value of t1/2 in the above expression.

λ=0.69314.8h1

The fraction of isotope decayed is 87.5100 .

The fraction remaining =187.5100=12.5100

The time of decay can be calculated by the formula,

t=2.303λlogn0n

Where

  • n0 is the number of atoms initially present.
  • n is the number of atoms after time “t”.

Substitute the values of λ , n0 and n in the above expression.

t=2.303λlogn0nt=2.303λlog10012.5t=2.303×14.80.693log8t=44.41h_

The time of decay for 81Kr is 6.302×105years_ .

Explanation

The decay constant can be calculated by the formula given below.

λ=0.693t1/2

Where

  • t1/2 is the half life of nuclide.
  • λ is the decay constant.

Substitute the value of t1/2 in the above expression.

λ=0.6932.1×105year

The fraction of isotope decayed is 87.5100 .

The fraction remaining =187.5100=12.5100

The time of decay can be calculated by the formula,

t=2.303λlogn0n

Where

  • n0 is the number of atoms initially present.
  • n is the number of atoms after time “t”.

Substitute the values of λ , n0 and n in the above expression.

t=2.303λlogn0nt=2.303×2.1×1050.693log10012.5t=2.303×2.1×1050.693log8t=6.302×105years_

Conclusion

Conclusion

The most stable isotope is 81Kr and the hottest one is 73Kr

The time of decay for 87.5%of73Kr is 81.428s_ .

The time of decay for 87.5%of74Kr is. 34.51s_

The time of decay for 87.5%of76Kr is 44.41h_

The time of decay for 87.5%of81Kr is 6.302×105years_

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Chapter 19 Solutions

Student Solutions Manual for Zumdahl/Zumdahl/DeCoste?s Chemistry, 10th Edition

Ch. 19 - Prob. 1QCh. 19 - Prob. 3QCh. 19 - Prob. 4QCh. 19 - Prob. 5QCh. 19 - Prob. 6QCh. 19 - Prob. 7QCh. 19 - Prob. 8QCh. 19 - Prob. 9QCh. 19 - Prob. 10QCh. 19 - Prob. 11QCh. 19 - Prob. 12QCh. 19 - Prob. 13QCh. 19 - Prob. 14QCh. 19 - Prob. 15ECh. 19 - Prob. 16ECh. 19 - Prob. 17ECh. 19 - Prob. 18ECh. 19 - Prob. 19ECh. 19 - Prob. 20ECh. 19 - Prob. 21ECh. 19 - Prob. 22ECh. 19 - Prob. 23ECh. 19 - Prob. 24ECh. 19 - Prob. 27ECh. 19 - Prob. 28ECh. 19 - Prob. 29ECh. 19 - Prob. 30ECh. 19 - Prob. 32ECh. 19 - Prob. 34ECh. 19 - Prob. 35ECh. 19 - Prob. 36ECh. 19 - Prob. 37ECh. 19 - Prob. 38ECh. 19 - Prob. 39ECh. 19 - Prob. 40ECh. 19 - Prob. 41ECh. 19 - Prob. 42ECh. 19 - Prob. 43ECh. 19 - Prob. 44ECh. 19 - Prob. 45ECh. 19 - Prob. 46ECh. 19 - Prob. 47ECh. 19 - Prob. 48ECh. 19 - Prob. 49ECh. 19 - Prob. 50ECh. 19 - Prob. 52ECh. 19 - Prob. 53ECh. 19 - Prob. 54ECh. 19 - Prob. 55ECh. 19 - Prob. 56ECh. 19 - Prob. 57ECh. 19 - Prob. 58ECh. 19 - Prob. 59ECh. 19 - Prob. 60ECh. 19 - Prob. 61ECh. 19 - Prob. 62ECh. 19 - Prob. 63ECh. 19 - Prob. 64ECh. 19 - Prob. 65AECh. 19 - Prob. 66AECh. 19 - Prob. 67AECh. 19 - Prob. 68AECh. 19 - Prob. 69AECh. 19 - Prob. 70AECh. 19 - Prob. 71AECh. 19 - Prob. 72AECh. 19 - Prob. 73AECh. 19 - Prob. 74AECh. 19 - Prob. 75AECh. 19 - Prob. 76AECh. 19 - Prob. 77AECh. 19 - Prob. 78AECh. 19 - Prob. 79AECh. 19 - Prob. 80AECh. 19 - Prob. 81CWPCh. 19 - Prob. 82CWPCh. 19 - Prob. 83CWPCh. 19 - Prob. 84CWPCh. 19 - Prob. 85CWPCh. 19 - Prob. 86CWPCh. 19 - Prob. 87CPCh. 19 - Prob. 88CPCh. 19 - Prob. 89CPCh. 19 - Prob. 90CPCh. 19 - Prob. 91CPCh. 19 - Prob. 92CPCh. 19 - Prob. 93CPCh. 19 - Prob. 94CPCh. 19 - Prob. 95IPCh. 19 - Prob. 96IP
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