Concept explainers
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:
Here, N and No are final and initial concentrations,
Here,
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 |
24 |
t hr |
48 |
t hr |
72 |
t hr |
96 |
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 | -16.214 | -21.618626 | |||
Calcium-47 | -0.1528 | 0.085829 | -0.3056 | -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-
The formula for the fraction of sample and time can be calculated by using the following formula-
Where,
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 | ||||||
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 | |
Iron-59 | 0.0156 | 0.98455 | -0.0311 | 0.9693343 | ||
Titanium-45 | -5.4047 | 0.0045 | -10.809 | |||
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 | |
-0.0467 | 0.95436 | -0.0622914 | 0.93961 | |||
-16.214 | -21.618626 | |||||
-0.4584 | 0.63228 | -0.6112409 | 0.54268 | |||
-0.0822 | 0.9211 | -0.1095885 | 0.8962 |
Spreadsheet documentation of the above table is −
The fraction remained after 1 day, 2 days, 3 days and 4 days is −
Radioactive element | t hr |
24 |
t hr |
48 |
t hr |
72 |
t hr |
96 |
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 | -16.214 | -21.618626 | |||
Calcium-47 | -0.1528 | 0.085829 | -0.3056 | -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 |
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
|
24
|
t hr
|
48
|
t hr
|
72
|
t hr
|
96
|
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 | -16.214 | -21.618626 | |||
Calcium-47 | -0.1528 | 0.085829 | -0.3056 | -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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Chapter 32 Solutions
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