Question

After the release of radioactive material into the atmosphere from a nuclear power plant in a country in 1984​, the hay in that country was contaminated by a radioactive isotope​ (half-life 8 ​days). If it is safe to feed the hay to cows when 14​% of the radioactive isotope​ remains, how long did the farmers need to wait to use this​ hay?

Answer 1

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.

$A=a\cdot (\frac{1}{2})^{\frac{t}{h}}$, where,

A = Amount left after t time,

a = Initial amount,

h = Half-life.

14% of 100 would be 14.

$14=100\cdot (\frac{1}{2})^{\frac{t}{8}}$

$\frac{14}{100}=\frac{100\cdot (0.5)^{\frac{t}{8}}}{100}$

$0.14=(0.5)^{\frac{t}{8}}$

Now, we will take natural log of both sides.

$\text{ln}(0.14)=\text{ln}((0.5)^{\frac{t}{8}})$

Using natural log property $\text{ln}(a^b)=b\cdot \text{ln}(a)$, we will get:

$\text{ln}(0.14)=\frac{t}{8}\times \text{ln}(0.5)$

$\text{ln}(0.14)=\frac{t\times \text{ln}(0.5)}{8}$

$\frac{\text{ln}(0.14)}{\text{ln}(0.5)}=\frac{t\times \text{ln}(0.5)}{8\times \text{ln}(0.5)}$

$\frac{-1.966112856}{-0.69314718055}=\frac{t}{8}$

$2.836501267=\frac{t}{8}$

$\frac{t}{8}=2.836501267$

$\frac{t}{8}*8=2.836501267*8$

$t=22.6920101377$

$t\approx 23$

Therefore, the farmers need to wait for 23 days.