Loose Leaf for General, Organic and Biological Chemistry with Connect 2 Year Access Card
Loose Leaf for General, Organic and Biological Chemistry with Connect 2 Year Access Card
4th Edition
ISBN: 9781260269284
Author: Janice Gorzynski Smith Dr.
Publisher: McGraw-Hill Education
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Chapter 10, Problem 47P
Interpretation Introduction

(a)

Interpretation:

The number of decays in the first two-minute interval needs to be determined.

Concept Introduction:

All the radioactive reactions follow first order kinetics. According to integrated rate law for the first order reaction, the relation between final concentration, initial concentration and half-life is represented as follows:

  N(t)=No(12)t/t1/2

Here, N(t) is amount at time t, No is initial amount and t1/2 is half-life time.

The average number of decays per unit time or change in the radioactive nuclei present is known as activity. It is represented as follows:

  A(t)=dNdt

Also,

  A(t)=λN(t)

Here, λ is decay constant.

The relation between half-life time and decay constant is as follows:

  t1/2=0.6932λ

Here, t1/2 is half-life time.

Interpretation Introduction

(b)

Interpretation:

The number of decays in the second two-minute interval needs to be determined.

Concept Introduction:

All the radioactive reactions follow first order kinetics. According to integrated rate law for the first order reaction, the relation between final concentration, initial concentration and half-life is represented as follows:

  N(t)=No(12)t/t1/2

Here, N(t) is amount at time t, No is initial amount and t1/2 is half-life time.

The average number of decays per unit time or change in the radioactive nuclei present is known as activity. It is represented as follows:

  A(t)=dNdt

Also,

  A(t)=λN(t)

Here, λ is decay constant.

The relation between half-life time and decay constant is as follows:

  t1/2=0.6932λ

Here, t1/2 is half-life time.

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Chapter 10 Solutions

Loose Leaf for General, Organic and Biological Chemistry with Connect 2 Year Access Card

Ch. 10.2 - Prob. 10.3PPCh. 10.2 - Prob. 10.9PCh. 10.2 - Prob. 10.10PCh. 10.2 - Prob. 10.11PCh. 10.3 - Prob. 10.4PPCh. 10.3 - Prob. 10.12PCh. 10.3 - Prob. 10.13PCh. 10.3 - Prob. 10.14PCh. 10.4 - To treat a thyroid tumor, a patient must be given...Ch. 10.4 - A sample of iodine-131 (t1/2=8.0 days) has an...Ch. 10.4 - Prob. 10.15PCh. 10.5 - Prob. 10.16PCh. 10.5 - Prob. 10.17PCh. 10.5 - Prob. 10.18PCh. 10.6 - Prob. 10.7PPCh. 10.6 - Prob. 10.8PPCh. 10 - Prob. 19PCh. 10 - Prob. 20PCh. 10 - Prob. 21PCh. 10 - Prob. 22PCh. 10 - Prob. 23PCh. 10 - Prob. 24PCh. 10 - Prob. 25PCh. 10 - Prob. 26PCh. 10 - Prob. 27PCh. 10 - Prob. 28PCh. 10 - Prob. 29PCh. 10 - Prob. 30PCh. 10 - Prob. 31PCh. 10 - Prob. 32PCh. 10 - Prob. 33PCh. 10 - Prob. 34PCh. 10 - Prob. 35PCh. 10 - Prob. 36PCh. 10 - Prob. 37PCh. 10 - Prob. 38PCh. 10 - Prob. 39PCh. 10 - Prob. 40PCh. 10 - Prob. 41PCh. 10 - Prob. 42PCh. 10 - Prob. 43PCh. 10 - Prob. 44PCh. 10 - Prob. 45PCh. 10 - Prob. 46PCh. 10 - Prob. 47PCh. 10 - Prob. 48PCh. 10 - Prob. 49PCh. 10 - Prob. 50PCh. 10 - Prob. 51PCh. 10 - Prob. 52PCh. 10 - Prob. 53PCh. 10 - Prob. 54PCh. 10 - Prob. 55PCh. 10 - Prob. 56PCh. 10 - Prob. 57PCh. 10 - Prob. 58PCh. 10 - Prob. 59PCh. 10 - Prob. 60PCh. 10 - Prob. 61PCh. 10 - Prob. 62PCh. 10 - Prob. 63PCh. 10 - Prob. 64PCh. 10 - Prob. 65PCh. 10 - Prob. 66PCh. 10 - Prob. 67PCh. 10 - Prob. 68PCh. 10 - Prob. 69PCh. 10 - Prob. 70PCh. 10 - Prob. 71PCh. 10 - Prob. 72PCh. 10 - Prob. 73PCh. 10 - Prob. 74PCh. 10 - Prob. 75PCh. 10 - Prob. 76PCh. 10 - Prob. 77PCh. 10 - Prob. 78PCh. 10 - Prob. 79PCh. 10 - Prob. 80PCh. 10 - Prob. 81PCh. 10 - Prob. 82PCh. 10 - Prob. 83PCh. 10 - Prob. 84PCh. 10 - Prob. 85PCh. 10 - Prob. 86PCh. 10 - Prob. 87PCh. 10 - Prob. 88PCh. 10 - Prob. 89CPCh. 10 - Prob. 90CP
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