Answer :
Answer:
Given that,
The number of grams A of a certain radioactive substance present at time, in years
from the present, t is given by the formula
[tex]A=45e^{-0.0045(t)}[/tex]a) To find the initial amount of this substance
At t=0, we get
[tex]A=45e^{-0.0045(0)}[/tex][tex]A=45e^0[/tex]We know that e^0=1 ( anything to the power zero is 1)
we get,
[tex]A=45[/tex]The initial amount of the substance is 45 grams
b)To find thehalf-life of this substance
To find t when the substance becames half the amount.
A=45/2
Substitute this we get,
[tex]\frac{45}{2}=45e^{-0.0045(t)}[/tex][tex]\frac{1}{2}=e^{-0.0045(t)}[/tex]Taking natural logarithm on both sides we get,
[tex]\ln (\frac{1}{2})=-0.0045(t)^{}[/tex][tex](-1)\ln (\frac{1}{2})=0.0045(t)[/tex][tex]\ln (\frac{1}{2})^{-1}=0.0045(t)[/tex][tex]\ln (2)=0.0045(t)[/tex][tex]0.6931=0.0045(t)[/tex][tex]t=\frac{0.6931}{0.0045}[/tex][tex]t=154.02[/tex]Half-life of this substance is 154.02
c) To find the amount of substance will be present around in 2500 years
Put t=2500
we get,
[tex]A=45e^{-0.0045(2500)}[/tex][tex]A=45e^{-11.25}[/tex][tex]A=45\times0.000013=0.000585[/tex][tex]A=0.000585[/tex]The amount of substance will be present around in 2500 years is 0.000585 grams
