The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days). Concept introduction: According to first order of kinetics: ln ( N N 0 ) = − λ t Here, N and N o are final and initial concentrations, λ is decay constant and t is time. t 1 / 2 ( h a l f − t i m e ) = ln 2 λ Here, t 1 / 2 is half-life time.

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Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213
BuyFind

Principles of Instrumental Analysis

7th Edition
Douglas A. Skoog + 2 others
Publisher: Cengage Learning
ISBN: 9781305577213

Solutions

Chapter 32, Problem 32.3QAP
Interpretation Introduction

Interpretation:

The fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days is to be determined. The samples are iron-59 (44.51 days), titanium-45 (3.078h), calcium-47(4.536days) and phosphorus-33 (25.3 days).

Concept introduction:

According to first order of kinetics:

ln(NN0)=λt

Here, N and No are final and initial concentrations, λ is decay constant and t is time.

t1/2(halftime)=ln2λ

Here, t1/2 is half-life time.

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