Question

The model for radioactive decay Is y = yor". A radioactive substance has a half-life of 60 years. Ir 100 grams are present today, in how many years will 68 es artis bepresent? While solving this problem, round the value of k to seven decimal places. Round your answer to two decimal places.KeypadAnswerHow to enter your answer opens in new windowadhat

Answer 1

We are given that the radioactive decay of a substance is given by the following equation:

$y=y_0e^{kt}$

We need to determine the value of "k". To do that we will use the fact that the half-life of the quantity is 60 years. The half-life is the time for the quantity to be half the initial value, therefore, we have:

$\frac{y_0}{2}=y_0e^{k(60)}$

We can cancel out the initial quantity:

$\frac{1}{2}=e^{k(60)}$

Now, we take the natural logarithm to both sides:

$ln(\frac{1}{2})=ln(e^{k(60)})$

Now, we use the following property of logarithms:

$ln(x^y)=ylnx$

Applying the property we get:

$ln(\frac{1}{2})=k(60)lne$

We also have:

$lne=1$

Substituting we get:

$ln(\frac{1}{2})=k(60)$

Now, we divide both sides by 60:

$\frac{1}{60}ln(\frac{1}{2})=k$

Now, we solve the operations:

$-0.011552=k$

Now, we substitute the value of "k":

$y=y_0e^{-0.011552t}$

We are given that 100 grams is present today. If today is the value when time "t" is zero then 100 grams is the initial quantity, therefore, we substitute:

$y=100e^{-0.011552t}$

Now, we are asked to determine the time when "y = 68g":

$68=100e^{-0.011552t}$

Now, we solve for "t". First, we divide both sides by 100:

$\frac{68}{100}=e^{-0.011552t}$

Now, we take the natural logarithm:

$ln(\frac{68}{100})=lne^{-0.011552t}$

Now, we apply the property of logarithms:

$ln(\frac{68}{100})=-0.011552tln(e)$

Applying "ln (e) = 1";

$ln(\frac{68}{100})=-0.011552t$

Now, we divide both sides by -0.011552:

$-\frac{1}{0.011552}ln(\frac{68}{100})=t$

Solving the operation:

$33.38=t$

Therefore, the time required is 33.38 years.