the half - life of a certain isotope is 70 years.\n\na. given an initial amount of ( a ) grams of this…

the half - life of a certain isotope is 70 years.\n\na. given an initial amount of ( a ) grams of this isotope, at ( t = 0 ), find an exponential decay model,\n( a(t)=ae^{kt} ) that gives the amount of the isotope at time ( t,t > 0 ). round your ( k ) value to 3\n\ndecimal places.\n\n( a(t)= )\n\n\nb. use the function you found in part ( a ) to calculate the time required (in years) for the isotope to\n\ndecay to ( \frac{1}{5}a ). round to 3 decimal places.\n\n

the half - life of a certain isotope is 70 years.\n\na. given an initial amount of ( a ) grams of this isotope, at ( t = 0 ), find an exponential decay model,\n( a(t)=ae^{kt} ) that gives the amount of the isotope at time ( t,t > 0 ). round your ( k ) value to 3\n\ndecimal places.\n\n( a(t)= )\n\n\nb. use the function you found in part ( a ) to calculate the time required (in years) for the isotope to\n\ndecay to ( \frac{1}{5}a ). round to 3 decimal places.\n\n

Answer

Explanation:

Step1: Find the value of (k)

The half - life formula for (A(t)=Ae^{kt}) is when (t = T) (half - life), (A(T)=\frac{A}{2}). Substitute into the formula: (\frac{A}{2}=Ae^{kT}). Since (T = 70), we have (\frac{1}{2}=e^{70k}). Take the natural logarithm of both sides: (\ln(\frac{1}{2})=\ln(e^{70k})). Using the property (\ln(e^{x})=x), we get (\ln(\frac{1}{2}) = 70k). Since (\ln(\frac{1}{2})=-\ln(2)\approx - 0.693), then (k=\frac{-\ln(2)}{70}\approx\frac{- 0.693}{70}\approx - 0.010). So the model is (A(t)=Ae^{-0.010t}).

Step2: Calculate the time (t) when (A(t)=\frac{1}{5}A)

Substitute (A(t)=\frac{1}{5}A) into (A(t)=Ae^{-0.010t}). We get (\frac{1}{5}A=Ae^{-0.010t}). Cancel out (A) (since (A\neq0)), so (\frac{1}{5}=e^{-0.010t}). Take the natural logarithm of both sides: (\ln(\frac{1}{5})=\ln(e^{-0.010t})). Using (\ln(e^{x}) = x), we have (\ln(\frac{1}{5})=-0.010t). Since (\ln(\frac{1}{5})=-\ln(5)\approx - 1.609), then (t=\frac{-\ln(5)}{-0.010}=\frac{1.609}{0.010}=160.944).

Answer:

a. (A(t)=Ae^{-0.010t}) b. (160.944)