Principles of Instrumental Analysis
Principles of Instrumental Analysis
7th Edition
ISBN: 9781305577213
Author: Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher: Cengage Learning
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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.

Expert Solution & Answer
Check Mark

Answer to Problem 32.3QAP

The fraction of radionuclide which will remain after 1 day, 2 days, 3 days and 4 days is −

Radioactive element t hr ln(N/N0) 24 (N/N0) t hr ln(N/N0) 48 (N/N0) t hr ln(N/N0) 72 (N/N0) t hr ln(N/N0) 96 (N/N0)
Iron-59 0.0156 0.98455 -0.0311 0.9693343 -0.0467 0.95436 -0.0622914 0.93961
Titanium-45 -5.4047 0.0045 -10.809 2.021×105 -16.214 9.1×108 -21.618626 4.1×1010
Calcium-47 -0.1528 0.085829 -0.3056 2.021×105 -0.4584 0.63228 -0.6112409 0.54268
Phosphorous-33 -0.0274 0.97297 -0.0548 0.9466799 -0.0822 0.9211 -0.1095885 0.8962

Explanation of Solution

The formula for the half time of sample can be calculated by using the following formula-

t1/2=ln2λ

The formula for the fraction of sample and time can be calculated by using the following formula-

ln(NN0)=λt

Where,

t1/2 = half time

λ = decay constant

N and N0 = number of radioactive nuclei

The half- time of the radioactive elements are given as-

Iron-59 = 44.51 days

Titanium-45 = 3.078h

Calcium-47= 4.536days

Phosphorus-33 = 25.3 days

By using the above formulas following data will be calculated-

Radioactive elements t1/2 t1/2 hr λ hr-1
Iron-59 44.51 days 1068.24 hr 0.00065
Titanium-45 3.078 hr 3.078 hr 0.22519
Calcium-47 4.536 days 108.864 hr 0.00637
Phosphorous-33 25.3 days 607.2 hr 0.00114
For t=24 hr t hr 24 t= 48 hr t hr 48
ln(N/N0) (N/N0) ln(N/N0) (N/N0)
Iron-59 0.0156 0.98455 -0.0311 0.9693343
Titanium-45 -5.4047 0.0045 -10.809 2.021×105
Calcium-47 -0.1528 0.085829 -0.3056 0.7366662
Phosphorous-33 -0.0274 0.97297 -0.0548 0.9466799
t =72 hr t hr 72 t = 96 hr t hr 96
ln(N/N0) (N/N0) ln(N/N0) (N/N0)
-0.0467 0.95436 -0.0622914 0.93961
-16.214 9.1×108 -21.618626 4.1×1010
-0.4584 0.63228 -0.6112409 0.54268
-0.0822 0.9211 -0.1095885 0.8962

Spreadsheet documentation of the above table is −

CellE5 = LN(2)/D5CellB11=E5*C9*1CellC11=EXP(B11)CellE11=E5*F9*1CellF11=EXP(E11)CellC18=E5*D16*1CellD18=EXP(C18)CellF18=E5*G16*1CellG18=EXP(F18)

The fraction remained after 1 day, 2 days, 3 days and 4 days is −

Radioactive element t hr ln(N/N0) 24 (N/N0) t hr ln(N/N0) 48 (N/N0) t hr ln(N/N0) 72 (N/N0) t hr ln(N/N0) 96 (N/N0)
Iron-59 0.0156 0.98455 -0.0311 0.9693343 -0.0467 0.95436 -0.0622914 0.93961
Titanium-45 -5.4047 0.0045 -10.809 2.021×105 -16.214 9.1×108 -21.618626 4.1×1010
Calcium-47 -0.1528 0.085829 -0.3056 2.021×105 -0.4584 0.63228 -0.6112409 0.54268
Phosphorous-33 -0.0274 0.97297 -0.0548 0.9466799 -0.0822 0.9211 -0.1095885 0.8962
Conclusion

The formula for the fraction of sample and half time of sample is used for the calculation of the fraction of radionuclide which will remain after 1 day, 2days, 3 days and 4 days for radioactive elements iron-59, titanium-45, calcium- 47, and phosphorous-33. Therefore,

Radioactive element t hr

ln(N/N0)

24

(N/N0)

t hr

ln(N/N0)

48

(N/N0)

t hr

ln(N/N0)

72

(N/N0)

t hr

ln(N/N0)

96

(N/N0)

Iron-59 0.0156 0.98455 -0.0311 0.9693343 -0.0467 0.95436 -0.0622914 0.93961
Titanium-45 -5.4047 0.0045 -10.809 2.021×105 -16.214 9.1×108 -21.618626 4.1×1010
Calcium-47 -0.1528 0.085829 -0.3056 2.021×105 -0.4584 0.63228 -0.6112409 0.54268
Phosphorous-33 -0.0274 0.97297 -0.0548 0.9466799 -0.0822 0.9211 -0.1095885 0.8962

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