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  1. Problems
  2. /Iodine-131 is used to treat the thyroid…
MediumChemistryHigh School (11–12)Chemical Kinetics › Integrated Rate Equations and Half-LifeNumericalAPA-LevelIBAEEE (Amrita)AMUEEEAP EAPCET
+17 more exams
BCECECOMEDK UGETCUET UGGUJCETJCECEJKCETKCETKEAMKIITEELPUNESTMET (Manipal)MHT-CETOJEESRMJEEETS EAPCET

Iodine-131 is used to treat the thyroid…

StemPy LabContributed by StemPy Lab

Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Iodine-131 is used to treat the thyroid. Its half-life is 8.02 d8.02\ \text{d}8.02 d, and its decay follows a first order law. A hospital receives a sample with an activity of 480 MBq480\ \text{MBq}480 MBq. What is the activity after 20 d20\ \text{d}20 d? How long must the sample be kept for the activity to fall to 30 MBq30\ \text{MBq}30 MBq?

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MBq
d
Need a reminder? The key ideaShowHide

For a first order reaction, t1/2=0.693kt_{1/2} = \dfrac{0.693}{k}t1/2​=k0.693​, and the fraction left after nnn half-lives is (12)n\left(\tfrac{1}{2}\right)^n(21​)n. For a zero order reaction, t1/2=[R]02kt_{1/2} = \dfrac{[R]_0}{2k}t1/2​=2k[R]0​​. In a pseudo first order reaction, k′=k[B]0k' = k[\text{B}]_0k′=k[B]0​.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

Step 1. After 20 d20\ \text{d}20 d. The number of half-lives is 20 d8.02 d=2.494\dfrac{20\ \text{d}}{8.02\ \text{d}} = 2.4948.02 d20 d​=2.494. The activity is 480 MBq×(12)2.494=480 MBq×0.1775=85.2 MBq480\ \text{MBq} \times \left(\tfrac{1}{2}\right)^{2.494} = 480\ \text{MBq} \times 0.1775 = 85.2\ \text{MBq}480 MBq×(21​)2.494=480 MBq×0.1775=85.2 MBq.

Step 2. To 30 MBq30\ \text{MBq}30 MBq. The factor is 48030=16=24\dfrac{480}{30} = 16 = 2^430480​=16=24. That is 4 half-lives, so 4×8.02 d=32.08 d4 \times 8.02\ \text{d} = 32.08\ \text{d}4×8.02 d=32.08 d.

Check. At 20 d20\ \text{d}20 d the sample is between 2 and 3 half-lives old. The activity is between 4804=120 MBq\dfrac{480}{4} = 120\ \text{MBq}4480​=120 MBq and 4808=60 MBq\dfrac{480}{8} = 60\ \text{MBq}8480​=60 MBq. We found 85.2 MBq85.2\ \text{MBq}85.2 MBq, which is in that range.

Answer: 85.2 MBq85.2\ \text{MBq}85.2 MBq after 20 d20\ \text{d}20 d, and 32.08 d32.08\ \text{d}32.08 d to reach 30 MBq30\ \text{MBq}30 MBq.

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  1. Half-life and pseudo first order reactionsIntegrated Rate Equations and Half-Life

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