Answer:
r=-2.8%, The equation is [tex]A=A_0e^{-0.028t}[/tex]
Step-by-step explanation:
Chemical decay can be represented by the equation [tex]A=A_0e^{rt}[/tex]. We can find the rate at which it decays by using t=24.3 days and A=0.5 or half the amount at that time. This means [tex]A_0=1[/tex] in this context.
[tex]A=A_0e^{rt}\\0.5=1e^{r(24.3)}[/tex]
[tex]0.5=e^{(24.5)r}[/tex]
To solve for r, we will take the natural log of both sides and use log rules to isolate r.
[tex]ln 0.5=ln e^{(24.5)r}\\ln 0.5=24.5r (ln e)\\\frac{ln0.5}{24.5} =r[/tex]
We know [tex]lne=1[/tex] so we were able to cancel it out and divide both sides by 24.5.
We solve with a calculator [tex]\frac{ln0.5}{24.5} =r\\-0.028=r[/tex]
We change -0.028 into a percent by multiplying by 100 to get -2.8% as the rate.
The equation is [tex]A=A_0e^{-0.028t}[/tex]
Answer:
The value of a is a=-0.028524.
Step-by-step explanation:
The function describe the half-life of phosphorus-32 is
[tex]A(t)=A_0e^{at}[/tex]
Where, [tex]A_0[/tex] is the initial amount.
According to question,
[tex]A(t)=\frac{A_0}{2}[/tex] and t=24.3
Substitute in the formula,
[tex]\frac{A_0}{2}=A_0e^{24.3a}[/tex]
[tex]0.5=e^{24.3a}[/tex]
Taking ln both side,
[tex]\ln(0.5)=\log(e^{24.3a})[/tex]
[tex]\ln(0.5)=24.3a[/tex]
[tex]a=\frac{\ln(0.5)}{24.3}[/tex]
[tex]a=-0.028524[/tex]
Therefore, The value of a is a=-0.028524.
