An Initial amount of 100 grams of Thorium-234 decays according to the model: dP/dt=-0.02828P (where t is measured in years).

a) Find the equation for the amount of Thorium left after t years.
b) Determine the half-life of this radioactive isotope
c) How long will it take for the sample to reduce to 25 grams?

To answer these questions, we need to solve the differential equation for radioactive decay. The given equation is dP/dt = -0.02828P, where P represents the amount of Thorium-234 at time t.

a) To find the equation for the amount of Thorium left after t years, we need to solve the differential equation. We can do this by separating variables and integrating. Rearranging the equation, we have dP/P = -0.02828dt. Integrating both sides, we get ∫(1/P)dP = ∫(-0.02828)dt.

This simplifies to ln|P| = -0.02828t + C, where C is the constant of integration. Taking the exponential of both sides, we have |P| = e^(-0.02828t+C). Since e^C is just another constant, let's combine it into a single constant, say A. So, we have |P| = Ae^(-0.02828t).

However, since P represents the amount of Thorium-234, it cannot be negative. Therefore, we take the absolute value of the equation. That gives us P = Ae^(-0.02828t), where A is a constant determined by the initial amount of Thorium-234.

b) The half-life of a radioactive isotope is the time it takes for the amount to decrease by half. In this case, the initial amount is 100 grams, so we want to find the time it takes for P to reach 50 grams.

Using the equation we found in part (a), we set P = 50 and solve for t:
50 = Ae^(-0.02828t)
Dividing both sides by A, we have:
e^(-0.02828t) = 1/2
Taking the natural logarithm of both sides to eliminate the exponential, we get:
-0.02828t = ln(1/2)
Solving for t gives us:
t = ln(1/2) / -0.02828

c) To find how long it will take for the sample to reduce to 25 grams, we use the equation from part (a) and set P = 25. Solving for t:
25 = Ae^(-0.02828t)
Dividing both sides by A, we have:
e^(-0.02828t) = 1/4
Taking the natural logarithm of both sides:
-0.02828t = ln(1/4)
Solving for t, we get:
t = ln(1/4) / -0.02828

Now you have the equations and the values to solve parts (a), (b), and (c) of the problem. Plug in the values and calculate to get the final answers.