Question

the half-life of the radioactive element unobtanium-53 is 5 seconds. if 80 grams of unobtanium-53 are initially present, how many grams are present after 5 seconds? 10 seconds? 15 seconds? 20 seconds? 25 seconds?

the amount left after 5 seconds is 40 grams.

the amount left after 10 seconds is \boxed{} grams.

Explanation:

Step1: Understand half - life concept

The half - life of a radioactive substance is the time it takes for half of the substance to decay. For unobtanium - 53, the half - life \(t_{1/2}=5\) seconds. This means that every 5 seconds, the amount of the substance is reduced by half.

Step2: Calculate the amount after 10 seconds

We know that the initial amount \(A_0 = 80\) grams. After 5 seconds (one half - life), the amount \(A_1=\frac{A_0}{2}=\frac{80}{2} = 40\) grams. Now, 10 seconds is two half - lives (since \(10\div5 = 2\)). So after the second half - life (another 5 seconds after the first 5 seconds), we need to take half of the amount we had after 5 seconds.
The formula for the amount of a radioactive substance after \(n\) half - lives is \(A = A_0\times(\frac{1}{2})^n\), where \(n\) is the number of half - lives. For \(t = 10\) seconds, \(n=\frac{10}{5}=2\).
So \(A=80\times(\frac{1}{2})^2=80\times\frac{1}{4} = 20\) grams. We can also think of it as taking half of the amount at 5 seconds: \(A=\frac{40}{2}=20\) grams.

Answer:

20