Solve Half-Life Problem: Iodine-131 Activity Calculation

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In summary, we calculated the activity of a sample of iodine-131 containing 8.5 x 10^16 nuclei to be 3.307 x 10^-22 Ci. We also determined that if the half-life of iodine-131 were one-fourth of its actual value, the activity of the sample would increase. Finally, we found that the activity of the sample would increase by a factor of 4 if the half-life of iodine-131 were reduced to 2.01 days.
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Homework Statement



The half-life of iodine-131, an isotope used in the treatment of thyroid disorders, is 8.04 d.
(a) If a sample of iodine-131 contains 8.5 1016 nuclei, what is the activity of the sample? Express your answer in curies.
(b) If the half-life of iodine-131 were only one-fourth of its actual value, would the activity of this sample be increased or decreased? Explain.
(c) Calculate the factor by which the activity of this sample would change under the assumptions stated in part (b).

Homework Equations



R = l deltaN/deltat l = lambdaN
1 Ci = 3.7 x 10^10 decays/s

The Attempt at a Solution


For part (a)
8.04 d x 24hr/1d x 60min/1hr x 60s/1min = 694,656 s
R = l deltaN/deltat l = lambdaN = (8.5 x 10^16 nuclei)/(694,656 s) = 1.2236 x 10^11
1.2236 x 10^11 x 1s/3.6 x to^10 decays = 3.307 x 10^-22 Ci
Can you please check if I did this correctly?

For Part (b), if the half-life of iodine-131 were only one-fourth of its actual value, the activity of this sample would be increased because there is an inverse relationship between the two.

I do not know how to approach part (c), so I would like to request help for that section. Thanks.
 
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  • #2
Remember that the decay constant, lambda, is equal to ln2 divided by the half life, and that activity is the number of nuclei times the decay constant. this means your answer for a) is off by a factor of ln2.
Your reasoning on part b) is correct.
Part c seems to be asking that is the half life was actually 1/4 of the 8.04 days (i.e., 2.01 days), by how much would the activity be multiplid. Since activity and half life are inversely proportional, if half-life is reduce by a factor, what is the factor by which the activity increases.
 
  • #3
daveb said:
Remember that the decay constant, lambda, is equal to ln2 divided by the half life, and that activity is the number of nuclei times the decay constant. this means your answer for a) is off by a factor of ln2.
Your reasoning on part b) is correct.
Part c seems to be asking that is the half life was actually 1/4 of the 8.04 days (i.e., 2.01 days), by how much would the activity be multiplid. Since activity and half life are inversely proportional, if half-life is reduce by a factor, what is the factor by which the activity increases.

Thanks for the help!
 
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Related to Solve Half-Life Problem: Iodine-131 Activity Calculation

1. What is the half-life of Iodine-131?

The half-life of Iodine-131 is 8.02 days. This means that it takes 8.02 days for half of the original amount of Iodine-131 to decay into other elements.

2. How is the activity of Iodine-131 calculated?

The activity of Iodine-131 is calculated using the formula A = A0e-kt, where A is the current activity, A0 is the initial activity, k is the decay constant, and t is the time elapsed.

3. What is the decay constant for Iodine-131?

The decay constant for Iodine-131 is 0.0866 days-1. This means that 8.66% of the original amount of Iodine-131 will decay every day.

4. How can I use the half-life of Iodine-131 to determine its activity after a certain amount of time?

To determine the activity of Iodine-131 after a certain amount of time, you can use the formula A = A0(1/2)t/h, where A is the current activity, A0 is the initial activity, t is the time elapsed, and h is the half-life.

5. Why is it important to calculate the activity of Iodine-131?

Calculating the activity of Iodine-131 is important because it allows us to understand its radioactive properties and potential impact on the environment and human health. It also helps us to monitor and regulate the use of Iodine-131 in various industries, such as nuclear medicine and nuclear power production.

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