Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)
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Chapter 20, Problem 46E
Interpretation Introduction

Interpretation:

The time required for the decay of technetium-99m from 0.050 μg to 1.0×103 μg, having a half-life of 6.0 hours is to be determined.

Concept introduction:

The half-life of a substance is the numerical value in which the given radioactive substance is assumed to be reduced to half of its initial number. The half-life for a given substance is represented by t1/2.

In case, the decay of a radioactive substance is exponential, it will remain constant for the life time of the substance.

After each half-life period, the amount of the substance is reduced to half of the initial number.

The time required for the decay of the substance to a given amount of substance can be calculated using the formula mentioned below:

ln(NtN0)=kt

In the above equation, ‘Nt’ represents the mass of the radioactive substance after a certain time interval t, ‘N0’ indicates the initial mass of the radioactive material, ‘k’ represents the decay constant and ‘t’ represents the time interval for the half-life (t1/2).

The half-life of the element is determined by the following formula:

t1/2=0.693k

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Chapter 20 Solutions

Mastering Chemistry with Pearson eText -- Standalone Access Card -- for Chemistry: A Molecular Approach (4th Edition)

Ch. 20 - Prob. 1ECh. 20 - Prob. 2ECh. 20 - Prob. 3ECh. 20 - Prob. 4ECh. 20 - Prob. 5ECh. 20 - Prob. 6ECh. 20 - Prob. 7ECh. 20 - Prob. 8ECh. 20 - Prob. 9ECh. 20 - Prob. 10ECh. 20 - Prob. 11ECh. 20 - Prob. 12ECh. 20 - Prob. 13ECh. 20 - Prob. 14ECh. 20 - Prob. 15ECh. 20 - Prob. 16ECh. 20 - Prob. 17ECh. 20 - Prob. 18ECh. 20 - Prob. 19ECh. 20 - Prob. 20ECh. 20 - Prob. 21ECh. 20 - Prob. 22ECh. 20 - Prob. 23ECh. 20 - Prob. 24ECh. 20 - Prob. 25ECh. 20 - Prob. 26ECh. 20 - Prob. 27ECh. 20 - Prob. 28ECh. 20 - Prob. 29ECh. 20 - Prob. 30ECh. 20 - Prob. 31ECh. 20 - Prob. 32ECh. 20 - Prob. 33ECh. 20 - Prob. 34ECh. 20 - Prob. 35ECh. 20 - Prob. 36ECh. 20 - Prob. 37ECh. 20 - Prob. 38ECh. 20 - Prob. 39ECh. 20 - Prob. 40ECh. 20 - Prob. 41ECh. 20 - Prob. 42ECh. 20 - Prob. 43ECh. 20 - Prob. 44ECh. 20 - Prob. 45ECh. 20 - Prob. 46ECh. 20 - Prob. 47ECh. 20 - Prob. 48ECh. 20 - Prob. 49ECh. 20 - Prob. 50ECh. 20 - Prob. 51ECh. 20 - Prob. 52ECh. 20 - Prob. 53ECh. 20 - Prob. 54ECh. 20 - Prob. 55ECh. 20 - Prob. 56ECh. 20 - Prob. 57ECh. 20 - Prob. 58ECh. 20 - Prob. 59ECh. 20 - Prob. 60ECh. 20 - Prob. 61ECh. 20 - Prob. 62ECh. 20 - Prob. 63ECh. 20 - Prob. 64ECh. 20 - Prob. 65ECh. 20 - Prob. 66ECh. 20 - Prob. 67ECh. 20 - Prob. 68ECh. 20 - Prob. 69ECh. 20 - Prob. 70ECh. 20 - Prob. 71ECh. 20 - Prob. 72ECh. 20 - Prob. 73ECh. 20 - Prob. 74ECh. 20 - Prob. 75ECh. 20 - Prob. 76ECh. 20 - Prob. 77ECh. 20 - Prob. 78ECh. 20 - Prob. 79ECh. 20 - Prob. 80ECh. 20 - Prob. 81ECh. 20 - Prob. 82ECh. 20 - Prob. 83ECh. 20 - Prob. 84ECh. 20 - Prob. 85ECh. 20 - Prob. 86ECh. 20 - Prob. 87ECh. 20 - Prob. 88ECh. 20 - Prob. 89ECh. 20 - Prob. 90ECh. 20 - Prob. 91ECh. 20 - Prob. 92ECh. 20 - Prob. 93ECh. 20 - Prob. 94ECh. 20 - Prob. 95ECh. 20 - Prob. 96ECh. 20 - Prob. 97ECh. 20 - Prob. 98ECh. 20 - Prob. 99ECh. 20 - Prob. 100ECh. 20 - Prob. 101ECh. 20 - Prob. 102ECh. 20 - Prob. 103ECh. 20 - Prob. 104ECh. 20 - Prob. 105ECh. 20 - Prob. 106ECh. 20 - Prob. 107ECh. 20 - Prob. 108ECh. 20 - Prob. 109ECh. 20 - Prob. 110QGWCh. 20 - Prob. 111QGWCh. 20 - Prob. 112QGWCh. 20 - Prob. 113QGWCh. 20 - Prob. 114QGWCh. 20 - Prob. 115DIA
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