Respuesta :
Answer:
23 days.
Step-by-step explanation:
Let the original amount of radioactive material be 100.
We have been given that the half life of a radioactive material is 8 days. It is safe to feed the hay to cows when 14% of the radioactive isotope remains. We are asked to find the number of days, the farmers need to wait to use the radioactive contaminated hay.
We will use half-life formula to solve our given problem.
[tex]A=a\cdot (\frac{1}{2})^{\frac{t}{h}}[/tex], where,
A = Amount left after t time,
a = Initial amount,
h = Half-life.
14% of 100 would be 14.
[tex]14=100\cdot (\frac{1}{2})^{\frac{t}{8}}[/tex]
[tex]\frac{14}{100}=\frac{100\cdot (0.5)^{\frac{t}{8}}}{100}[/tex]
[tex]0.14=(0.5)^{\frac{t}{8}}[/tex]
Now, we will take natural log of both sides.
[tex]\text{ln}(0.14)=\text{ln}((0.5)^{\frac{t}{8}})[/tex]
Using natural log property [tex]\text{ln}(a^b)=b\cdot \text{ln}(a)[/tex], we will get:
[tex]\text{ln}(0.14)=\frac{t}{8}\times \text{ln}(0.5)[/tex]
[tex]\text{ln}(0.14)=\frac{t\times \text{ln}(0.5)}{8}[/tex]
[tex]\frac{\text{ln}(0.14)}{\text{ln}(0.5)}=\frac{t\times \text{ln}(0.5)}{8\times \text{ln}(0.5)}[/tex]
[tex]\frac{-1.966112856}{-0.69314718055}=\frac{t}{8}[/tex]
[tex]2.836501267=\frac{t}{8}[/tex]
[tex]\frac{t}{8}=2.836501267[/tex]
[tex]\frac{t}{8}*8=2.836501267*8[/tex]
[tex]t=22.6920101377[/tex]
[tex]t\approx 23[/tex]
Therefore, the farmers need to wait for 23 days.
The farmer needs to wait for about 38 days to use the hay.
Half life
The half life is the amount of time that it takes for a substance to decay to about half of its initial value. It is given by:
N(t) = N(1/2)^(t/T)
Where N is the initial value, N(t) is the amount of value of t years and T = half life.
Given:
N(t) = 4% of N = 0.04N, t = 8 days, hence:
0.04N = N(1/2)^(t/8)
t = 37.15 days
The farmer needs to wait for about 38 days to use the hay.
Find out more on Half life at: https://brainly.com/question/2320811
