OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card for Zumdahl/Zumdahl/DeCoste’s Chemistry, 10th Edition
OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card for Zumdahl/Zumdahl/DeCoste’s Chemistry, 10th Edition
10th Edition
ISBN: 9781305957558
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 19, Problem 36E

(a)

Interpretation Introduction

Interpretation: Half-life of copper- 64 is given. Its Decay constant is to be determined. Sample of 28.0mg64Cu is given.

Concept introduction: A process through which, an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.  The cause of instability of a nuclide is its inefficiency in holding the nucleus together.  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 decay constant can be calculated by the formula given below.

λ=0.693t1/2

The time of decay can be calculated by the formula given below,

t=2.303λlogn0n

To determine: The value of decay constant in s-1 .

(b)

Interpretation Introduction

Interpretation: Half-life of copper- 64 is given. Its Decay constant is to be determined. Sample of 28.0mg64Cu is given. Decay events produced in first second is to be determined.

Concept introduction: A process through which, an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.  The cause of instability of a nuclide is its inefficiency in holding the nucleus together.  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 decay constant can be calculated by the formula given below.

λ=0.693t1/2

The time of decay can be calculated by the formula given below,

t=2.303λlogn0n

To determine: The number of decay events in the first second.

(c)

Interpretation Introduction

Interpretation: Time in which one have to do the experiment of measuring radioactivity of copper- 64 so that the radioactivity do not fall below 25% of the initial measured value is to be stated.

Concept introduction: A process through which, an unstable nuclide loses its energy due to excess of protons or neutrons is known as radioactive decay.  The cause of instability of a nuclide is its inefficiency in holding the nucleus together.  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 decay constant can be calculated by the formula given below.

λ=0.693t1/2

The time of decay can be calculated by the formula given below,

t=2.303λlogn0n

To determine: The time for which the given experiment is to be done so that the radioactivity does not fall below 25% of the initial measured value.

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Chapter 19 Solutions

OWLv2 with MindTap Reader, 4 terms (24 months) Printed Access Card 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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