The halflife of a certain isotope is 73 years a Given an ini

.The half-life of a certain isotope is 73 years.

a. Given an initial amount of A grams of this isotope, at t=0, find an exponential decay model, A(t)=Ae^Kt that gives the amount of the isotope at time t , t > 0 Round your kk value to 3 decimal places.   

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b. Use the function you found in part a to calculate the time required for the isotope to decay to numerator is 1 and the denominator is 5 than Round to 3 decimal places

Solution

a)

A(t)=Ae^Kt

given The half-life of a certain isotope is 73 years.

(A_/2)=Ae^k*73

e^k*73=(1_/2)

k*73=ln(1_/2)

k=-0.009495

k=-0.009 per year

A(t)=Ae^-0.009495t

b)isotope to decay to (1_/5) of initial amount

=>(A_/5)=Ae^-0.009495t

=>e^-0.009495t =(1_/5)

=>-0.009495t=ln(1_/5)

=>t=169.501 years