12. if 10 mg of iodine 131 is given to a patient, how much is left after 24 days? the half - life of iodine…

12. if 10 mg of iodine 131 is given to a patient, how much is left after 24 days? the half - life of iodine - 131 is 8 days.\na) 10mg\nb) 10g\nc) 1.25g\nd) 1.25mg\n13. \n|half - life|original carbon - 14 remaining (%)|number of years|\n|----|----|----|\n|0|100|0|\n|1|50|5,700|\n|2|25|11,400|\n|3||17,100|\n|4||\n|5||\nafter 22,800 years, approximately what percentage of the original carbon - 14 remains?\na) 3.125%\nb) 6.25%\nc) 15%\nd) 12.5%\n14. what is the half - life of iodine - 131?\na) 8 days\nb) 24 days\nc) 32 days\nd) 16 days

12. if 10 mg of iodine 131 is given to a patient, how much is left after 24 days? the half - life of iodine - 131 is 8 days.\na) 10mg\nb) 10g\nc) 1.25g\nd) 1.25mg\n13. \n|half - life|original carbon - 14 remaining (%)|number of years|\n|----|----|----|\n|0|100|0|\n|1|50|5,700|\n|2|25|11,400|\n|3||17,100|\n|4||\n|5||\nafter 22,800 years, approximately what percentage of the original carbon - 14 remains?\na) 3.125%\nb) 6.25%\nc) 15%\nd) 12.5%\n14. what is the half - life of iodine - 131?\na) 8 days\nb) 24 days\nc) 32 days\nd) 16 days

Answer

12.

Answer:

d) 1.25mg

Explanation:

Step1: Calculate number of half - lives

$n=\frac{24}{8}=3$

Step2: Use decay formula

$N = N_0\times(\frac{1}{2})^n$, where $N_0 = 10$mg and $n = 3$. So $N=10\times(\frac{1}{2})^3=10\times\frac{1}{8}=1.25$mg

13.

Answer:

b) 6.25%

Explanation:

Step1: Calculate number of half - lives

$n=\frac{22800}{5700}=4$

Step2: Calculate remaining percentage

The remaining percentage $P = 100\times(\frac{1}{2})^n$. Substituting $n = 4$, we get $P=100\times(\frac{1}{2})^4=100\times\frac{1}{16}=6.25%$

14.

Answer:

a) 8 days

Explanation:

The half - life is the time it takes for the amount of a substance to reduce to half of its initial amount. From the graph or the information given in question 12, the half - life of iodine - 131 is 8 days.