Respuesta :
Answer:
a) A(t) = 800 * e^-0.1386t
b) A = 3.1287 mg
c) t = 48.23 days
Explanation:
First, this is a problem of exponential decay, so the general expression to use is the following:
A(t) = Ao*e^-kt (1)
Where:
Ao: Innitial mass, concentration, population.....
k: relative decay rate
t: time
A(t): mass, concentration or population left after t has passed.
Knowing this, the problem states that after 5 days, the bismuth decays to it's half life. So we can calculate the value of k, if we have a sample of 800 mg, after 5 days, we can say that this sample was reduced to 400 mg (Half of 800), therefore, we can replace the data on the above expression and find the formula for a mass remaining after t days:
A(t) = 400
Ao = 800
t = 5
k = ?
Replacing in equation (1) we have:
400 = 800 e^-5k ---> solving for k:
1/2 = e^-5k
ln(0.5) = -5k
k = -ln(0.5)/5
k = 0.1386
This is the value of the relative decay after t days. So if we replace in the original equation, we can find the formula for t days:
A(t) = 800 * e^-0.1386t
b) With the above formula, we only replace time and we'll get the remaining mass:
A = 800 * e^-0.1386(40)
A = 800 * 3.911x10^-3
A = 3.1287 mg
c) To know when the mass is reduced to 1 mg, we just solve for t from equation 1:
1 = 800 * e^-0.1386t
1/800 = e^-0.1386t
ln(1/800) = -0.1386t
t = -ln(1/800) / 0.1386
t = 48.23 days
Answer:
a) [tex]y(t)=800e^{-0.1386t}[/tex]
b)[tex]3.1 mg[/tex]
c[tex]48.2 days[/tex]
Explanation:
Let us first find the value of k, the rate constant;
k = (-ln0.5)/half life
= (-ln0.5)/5
= 0.1386
a) [tex]y(t) = y_{0}e^{-kt}\\y(t) = 800e^{-0.1386t}[/tex]
b)[tex]y(t) =800e^{-0.1386t} = 800e^{-0.1386*40} = 3.1 mg[/tex]
c[tex]y(t)=800e^{-0.1386t}\\1=800e^{-0.1386t}\\e^{-0.1386t}=1/800\\-0.1386tlne=ln(1/800)\\t=ln(1/800)/(-0.1386) = 48.2 days[/tex]
