# Half-Life of Aluminum-28

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• Published : September 26, 2008

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Abstract:

Finding the half-life of Aluminum-28 is the purpose of this lab. The class met at the Radiation Center and exposed small amounts of aluminum to energetic neutrons and watched the samples decay. After creating a table with background count and gross counts of the aluminum remaining at fifteen different one-minute intervals, this allowed the net counts per minute at each interval to be found as well. Next a graph of the net cpm verses time was plotted. After each interval was plotted the K-value (slope) was found along with the half-life. The actual half-life of Aluminum is 2.25 minutes. The half-life found in our group was 2.23 minutes.

Procedure:

The procedure is as follows on pages 63-67 of the Chemistry Laboratory Manual. No changes were made to this procedure. Data:
Background in 30 minutes = 553
Counts per minute (cpm) = 18.45
TimeGross CountsTimeGross cpmBk cpmNet cpm
12203.2 min1101518.4510996.5
21662.2 min831018.458291.5
31288.2 min644018.456421.5
4946.2 min473018.454711.5
5734.2 min367018.453651.5
6540.2 min270018.452681.5
7425.2 min212518.452106.5
8325.2 min162518.451606.5
9238.2 min119018.451171.5
10183.2 min91518.45896.5
11132.2 min66018.45641.5
1278.2 min39018.45371.5
1375.2 min37518.45356.5
1464.2 min32018.45301.5
1543.2 min21518.45196.5

Calculations:

Calculating Gross cpm example (at time 1)

Gross cpm = gross count (y) x 5

Gross cpm = 2203 x 5

Gross cpm = 11015

Calculating Net cpm example (at time 2)

Net cpm = gross cpm – background cpm

Net cpm = Gross cpm – 18.45

Net cpm = 11015 – 18.45

Net cpm = 10996.57

Find the slope from paper graph:

K = slope

K= [ln(net cpm12)- ln(net cpm2)] / [t12-t2]

K = [ln(371.55) – ln(8291.57)] / (12-2)

K = .3105 min

Calculating Half-life from estimated K-value

T1/2= ln2
K

T1/2=2.23

Graph of ln(cpm) versus time:

Calculating slope from Excel Graph:

K= slope

Slope of best-fit line:
Y = mx+b

Y = -0.2881x + 9.6201

M = slope = K = -0.2881

Calculating Half-life from estimated K-value

T1/2= ln2
K
T1/2 = [.693]/ |-0.28|

T1/2 = 2.47 min

The class values are as follows:

GroupHalf-life
12.34
22.31
32.39
42.39
52.48
62.4
72.23

Standard Deviation = .0792

Final Data

K (minutes) = -0.25

T1/2 (minutes) = 2.47

Reliability of the T1/2 values( + minutes) = 0.098

Calculating time for cpm to equal background value:

Ln(N) = -kt + ln(N0)

T = [ln(N/ N0)] / -k

T = [ln(11015/18.45)] / 0.25

T = 25.5 minutes

Balanced equations for activation and decay:

11449 In + 10n →11549In + Ɣ

11449 In→ 11448 + 0-1e +Ɣ

Error Analysis:
Since mainly computers did this experiment, the only error that could have happened was calculator error. The counts per minutes came strictly from the computers. Our group’s final answer was .02 off from the correct number for Aluminum-28 half-life, concluding that there was a little error in our experiment. The mistake that most likely happened and caused the percent errors to occur was not exactly calculating the slope for the k-value. Having an imprecise k-value could throw off the half-life value by quite a bit. Conclusion:

Finding the half-life of the aluminum isotope 28Al was the objective of this experiment. The class average for half-life of this experiment was 2.38 minutes. The specific lab group was 2.57 minutes. These results show that the class did a good job with the lab and the calculations, and the specific groups did an okay job. References:

CH221 Laboratory Manual Winter 2008. Haak, Margie. Myles, Daniel. Pages 63-67.

Van Schandevijl, R.. "Half Life of Al-28." Journal of Radioanalytical and Nuclear Chemistry 9(1971): 55-60.