7. iodine - 131 is used in the treatment of thyroid disease and has a half - life of 8.0 days. how many…
7. iodine - 131 is used in the treatment of thyroid disease and has a half - life of 8.0 days. how many grams of an original 160 mg sample will remain after 40 days?\n8. the half - life of zn - 71 is 2.4 minutes. if one had 100.0g at the beginning, how many grams would be left after 7.2 minutes has elapsed?\n9. carbon - 14 has a half - life of 6595 years. if you have 500mg of carbon - 14, how much will remain after 4 half - lives?\n10. the half - life of i - 123 is 13 hr. how much of a 64mg sample of i - 123 is left after 39 hours?
Answer
7.
Explanation:
Step1: Calculate number of half - lives
The number of half - lives $n=\frac{40}{8}=5$.
Step2: Use decay formula
The amount remaining $A = A_0\times(\frac{1}{2})^n$, where $A_0 = 160\ mg$. So $A=160\times(\frac{1}{2})^5$. $A = 160\times\frac{1}{32}=5\ mg = 0.005\ g$.
Answer:
$0.005\ g$
8.
Explanation:
Step1: Calculate number of half - lives
The number of half - lives $n=\frac{7.2}{2.4}=3$.
Step2: Use decay formula
The amount remaining $A = A_0\times(\frac{1}{2})^n$, where $A_0 = 100.0\ g$. So $A = 100\times(\frac{1}{2})^3$. $A=100\times\frac{1}{8}=12.5\ g$.
Answer:
$12.5\ g$
9.
Explanation:
Step1: Use decay formula
The amount remaining $A = A_0\times(\frac{1}{2})^n$, where $A_0 = 500\ mg$ and $n = 4$. $A=500\times(\frac{1}{2})^4$. $A = 500\times\frac{1}{16}=31.25\ mg$.
Answer:
$31.25\ mg$
10.
Explanation:
Step1: Calculate number of half - lives
The number of half - lives $n=\frac{39}{13}=3$.
Step2: Use decay formula
The amount remaining $A = A_0\times(\frac{1}{2})^n$, where $A_0 = 64\ mg$. So $A=64\times(\frac{1}{2})^3$. $A = 64\times\frac{1}{8}=8\ mg$.
Answer:
$8\ mg$