Half-life of a Radioisotope

Topics: Radioactive decay, Isotope, Radionuclide Pages: 3 (942 words) Published: October 30, 2014

The Half-Life of a Radioisotope:
10-22-14
Abstract: The half-life of a radioisotope is the time required for half the atoms in a given sample to undergo radioactive, or nuclear decay. The amount of radioactive isotope remaining can be calculated using the equation, t1/2 = .693/K. A sample data set was provided due to safety concerns. Using the data set, a half-life of 14.27days -1 was calculated using graphical linear regression analysis.

Introduction:
The purpose of this experiment is to determine the half-life of an unknown radioisotope. Half-life is defined as the time it takes for one half of the atoms in in a radioactive sample to decay. Data will be collected on the activity of a radioactive isotope vs. elapsed time, the half-life will then be determined by two different types of graphical analysis. Methods/Procedure:

Using the provided data in Table 1, the time in days, the normalized activity of the unknown (A’), and the natural log of the normalized activity of the unknown (ln A’) were calculated to determine beta emission half-life, and the half-life of the radioisotope. This data was then plotted on a graph to determine the half-life. The calculation for A’ was A’unk = At (S0/St), where:

S0 = corrected activity of your standard at elapsed time t = 0 St = corrected activity of your standard at elapsed time t. At = corrected activity of your unknown at elapsed time t.
A’unk = normalized activity of your unknown at elapsed time t. Example of calculations:
A’unk = At (S0/St)
Data taken from table for Half-life of a Radioisotope
Time = 0 days:
S0 = 324515 counts – 418 counts (background) / 5 min = 64819.4 counts/min At = (105993 counts – 418 counts (background) / 5 min = 21115 counts/min A’unk = 21115 counts/min (64819.4 counts/min / 64819.4 counts/min) = 21115 counts/min. ln A’unk = ln (21115) = 9.96

Time = 3 days:
St = 324434 counts – 396 counts / 5 min = 64807.6 counts/min At = 91656 counts – 396 counts / 5 min = 18252...
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