Question

1. 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.

a) 10mg
b) 10g
c) 1.25g
d) 1.25mg
13.

half - life	original carbon - 14 remaining (%)	number of years
1	50	5,700
2	25	11,400
3		17,100
4
5

after 22,800 years, approximately what percentage of the original carbon - 14 remains?
a) 3.125%
b) 6.25%
c) 15%
d) 12.5%

1. what is the half - life of iodine - 131?

a) 8 days
b) 24 days
c) 32 days
d) 16 days

Explanation:

12.

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.

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.

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.

Answer:

d) 1.25mg