Question

1. the image on the right shows a decay of 20mg of oxygen - 15.

a. what will remain after 3 half - lives?
b. after 5 half - lives?

1. looking at the image on the right answer the following questions:

a. when the original magnesium - 28 sample has decreased by half, how many hours have passed?
b. after 3 half - lives, what fraction of the sample remains/

Explanation:

Step1: Formula for radioactive - decay

The amount of a radioactive substance remaining $A$ after $n$ half - lives, given an initial amount $A_0$ is $A = A_0\times(\frac{1}{2})^n$.

Step2: Solve 11a

Given $A_0 = 20$mg and $n = 3$. Then $A=20\times(\frac{1}{2})^3=20\times\frac{1}{8}=2.5$mg.

Step3: Solve 11b

Given $A_0 = 20$mg and $n = 5$. Then $A = 20\times(\frac{1}{2})^5=20\times\frac{1}{32}=0.625$mg.

Step4: Solve 12a

The half - life is the time it takes for the amount of a radioactive substance to decrease by half. For Magnesium - 28, from the table or graph, when the fraction remaining is 0.5, the time $t = 21$ hours.

Step5: Solve 12b

After $n = 3$ half - lives, using the formula for the fraction remaining $f=(\frac{1}{2})^n$. Substituting $n = 3$, we get $f=\frac{1}{8}$.

Answer:

a. 2.5mg
b. 0.625mg

1. a. 21 hours

b. $\frac{1}{8}$