Question

The half-life of the radioactive element unobtainum-31 is 5 seconds. If 80 grams of unobtianum-31 are initally present, how many grams are present after 5 seconds? A. The amount left after 5 seconds is ___ grams.

Answer 1

For this case we know the initial value 80 and the half life if 5 seconds

Then we can create the following equation:

A = 80 (1/2)^(t/5)

Then we can replace for each case like this:

A

`A(5)=80\cdot((1)/(2))^1=40`

B

`A(10)=80\cdot((1)/(2))^2=20`

C

`A(15)=80\cdot((1)/(2))^3=10`

D

`A(20)=80\cdot((1)/(2))^4=5`

E

`A(25)=80\cdot((1)/(2))^5=(5)/(2)=2.5`