Solve Half-Life Problem: Iodine-131 Activity Calculation

[FlipStyle1308]

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.

 

[daveb]

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.

 

[FlipStyle1308]

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!