Need help on radioactive decay calc problem. urgent.?

Question

A laboratory sample consists of a mixture of two radioactive substances, both of which decay into inert gases which evaporate into the air. The initial composition of the sample by weight is 20% of substance A and 80% of substance B and the initial weight of the sample is 264 grams. After 2 days the sample weighs 217.612240257459 grams and after 4 days it weighs 179.666588758521 grams.
a) What is the half-life of substance A?
b) What is the half-life of substance B?
(c) What will the sample weigh after 25 days?

I'm confused about how to deal with the 2 substances in the sample, so I don't even really know where to start with this one.

Answer 1

A model for the mass (in grams) at time t (in days) is

m(t) = 52.8 * 2^(-t / t_A) + 211.2 * 2^(-t / t_B)

where t_A is the half-life (in days) of substance A and t_B is the half-life of substance B.

We have two data points:

m(2) = 52.8 * 2^(-2 / t_A) + 211.2 * 2^(-2 / t_B)
m(4) = 52.8 * 2^(-4 / t_A) + 211.2 * 2^(-4 / t_B)

where m(2) = 217.612240257459 and m(4) = 179.666588758521

Let x = 2^(-2 / t_A) and y = 2^(-2 / t_B). Then
2^(-4 / t_A) = [2^(-2 / t_A)]² = x², and similarly for 2^(-4 / t_B).

So the above equations may be written

m(2) = 52.8 x + 211.2 y
m(4) = 52.8 x² + 211.2 y²

or even simpler-looking,

x + 4y = p
x² + 4y² = q

where p = m(2)/52.8 and q = m(4)/52.8

If you solve the first of these for x and substitute that into the second, you'll get a quadratic equation for y. It has two real solutions. This is not surprising; graphically, the two equations above represent a line and an ellipse, and those two objects may intersect in 0, 1, or 2 points. If they intersect in one point, the line is tangent to the ellipse at the point of intersection. (That is nearly the case for this problem.)

Then, using a y-solution, you can solve for t_B:

y = 2^(-2 / t_B)
log(y) = log[2^(-2 / t_B)] = (-2 / t_B) log(2)

So

t_B = -2 log(2)/log(y)

One of the y-solutions you'll get gives t_B very close to 8 (and the corresponding value for t_A is very close to 5). Since this is a made-up problem, I suspect that those are the half-life values that were used to generate this problem.

Answer 2

A(t) = A0e^-kt .ninety 3 = a million e^- 7k // Ao = a million , A(t) = .ninety 3 Ao , t =7 ln .ninety 3 = -7 ok ok = - ln .ninety 3 / 7 so A(t) = Ao e^ ln .ninety 3 t/7 while t =24 A(24) = 500 e^ 24ln .ninety 3 / 7 = you calculate