Respuesta :
Answer:
2.5 days.
Step-by-step explanation:
The number of half lives is n where
12800(0.5)^n = 800
12800(0.5)^4 = 800
so n = 4
So the half life is 10/4
= 2.5 days.
Using an exponential function, it is found that the half-life of the substance is of 2.5 days.
The amount of radioactive substance is proportional to the number of counts per minute, thus the situation is modeled by the following differential equation:
[tex]\frac{dQ}{dt} = -kQ[/tex]
By separation of variables, we have that:
[tex]\frac{dQ}{Q} = -kdt[/tex]
[tex]\int \frac{dQ}{Q} = \int -kdt[/tex]
[tex]\ln{Q} = -kt + K[/tex]
[tex]Q(t) = Q(0)e^{-kt}[/tex]
- Initially, 12800 counts per minute, thus, Q(0) = 12800.
- After 10 days, 800 counts per minute, thus, Q(10) = 800, and this is used to find k.
[tex]Q(t) = Q(0)e^{-kt}[/tex]
[tex]800 = 12800e^{-10k}[/tex]
[tex]e^{-10k} = \frac{8}{128}[/tex]
[tex]\ln{e^{-10k}} = \ln{\frac{8}{128}}[/tex]
[tex]-10k = \ln{\frac{8}{128}}[/tex]
[tex]k = -\frac{\ln{\frac{8}{128}}}{10}[/tex]
[tex]k = 0.277259[/tex]
Thus
[tex]Q(t) = Q(0)e^{-0.277259t}[/tex]
The half-life is the value of t for which:
[tex]Q(t) = 0.5Q(0)[/tex]
Then
[tex]0.5Q(0) = Q(0)e^{-0.277259t}[/tex]
[tex]e^{-0.277259t} = 0.5[/tex]
[tex]\ln{e^{-0.277259t}} = \ln{0.5}[/tex]
[tex]-0.277259t = \ln{0.5}[/tex]
[tex]t = -\frac{\ln{0.5}}{0.277259}[/tex]
[tex]t = 2.5[/tex]
The half-life of the substance is of 2.5 days.
A similar problem is given at https://brainly.com/question/21046174
