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.

Respuesta :

Solution

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

[tex]A(5)=80\cdot(\frac{1}{2})^1=40[/tex]

B

[tex]A(10)=80\cdot(\frac{1}{2})^2=20[/tex]

C

[tex]A(15)=80\cdot(\frac{1}{2})^3=10[/tex]

D

[tex]A(20)=80\cdot(\frac{1}{2})^4=5[/tex]

E

[tex]A(25)=80\cdot(\frac{1}{2})^5=\frac{5}{2}=2.5[/tex]