Question

The quantity R, in grams, of a certain radioactive substance decreases according to the exponential decay model dRdt=−0.05R, where t is measured in seconds. During an experiment, a scientist determines that the rate of decay of a second substance with the quantity S, in grams, can be represented by a linear model dSdt=−4, where t is measured in seconds. If at time t=0, R(0)=100 and S(0)=125, at what time t, in seconds, will there be equal quantities of both substances?

Answer 1

Answer:

• 24 seconds

Explanation:

1. Quantiy R

Decay model:

       $\dfrac{dR}{dt}=-0.05R$

That is a first order reaction, whose solution is:

        $R=R_0e^{-0.05t}$

The value at t = 0, R(0) = 100 is R₀.

Thus, the law is:

        $R(t)=100e^{-0.05t}$

2. Quantity S

Decay model:

       $\dfrac{dS}{dt}=-4$

That is a zero order kinetic, whose solution is:

       $S=S_0-4t$

The value at t = 0, S(0) = 125 is S₀

Thus, the law is:

      $S(t)=125-4t$

3. At what time will there ve equal quantities of both substances?

You must solve the system doing R(t) = S(t)

      $100e^{-0.05t}=125-4t$

That equation must be solved by graphing or by consecutive iterations

Note that when t = 0 R(t) = 100 and S(t) = 125. From, that the quantities decreases.

Subsitute with t = 10.

• The left-hand side yields: 61 and the right-hand side yields 85.

Then, the time is greater than 10.

Substitute with t = 15

• Left-hand side = 47

• Right-hand side = 65

t = 20

• Left-hand side = 37

• 45

t = 25

• Left-hand side = 29

• Right-hand side =25

t = 24

• Left-hand side =3 0

• Right hand side = 29

Hence, 24 seconds is a satisfactory solution.

The graph attached shows that the intersection point is at t ≈ 23.548 seconds, confirming that 24 seconds is the best solution rounded to an integer number.