# Radioactive copper-64 decays with a half-life of 12.8 days. a. What is the value of k in s −1 ? b. A sample contains 28.0 mg 64 Cu. How many decay events will be produced in the first second? Assume the atomic mass of 64 Cu is 64.0 u. c. A chemist obtains a fresh sample of 64 Cu and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below 25% of the initial measured value. How long does she have to do the experiment?

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

Chapter
Section

### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 18, Problem 28E
Textbook Problem
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## Radioactive copper-64 decays with a half-life of 12.8 days.a. What is the value of k in s−1?b. A sample contains 28.0 mg 64Cu. How many decay events will be produced in the first second? Assume the atomic mass of 64Cu is 64.0 u.c. A chemist obtains a fresh sample of 64Cu and measures its radioactivity. She then determines that to do an experiment, the radioactivity cannot fall below 25% of the initial measured value. How long does she have to do the experiment?

(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 .

### Explanation of Solution

Explanation

The decay constant is calculated by the formula given below.

λ=0.693t1/2

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

λ=0.69312

(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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