Half life of two substances A and B, The initial compostiton is 40% A and 60% B?

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 40% of substance A and 60% of substance B and the initial weight of the sample is 297 grams. After 7 days the sample weighs 124.419451315581 grams and after 14 days it weighs 55.0505357006202 grams.
What is the half-life of substance A?
What is the half-life of substance B?
What will the sample weigh after 25 days?

Answer 1

The radioactive decay process is the classic exponential decay. The equation is:

N(t) = N0 e^(- r t)
where :
N0 is the initial number of atoms
r is the rate of decay
t is time
e is the base of natural logs (2.718...)

http://en.wikipedia.org/wiki/Exponential_decay

In this case, you have two substances decaying simultaneously. Each has a different weight and a different decay rate. So the equations become:

Wa(t) = (Da)(Na(t)) and Wb(t) = Db(Nb(t))

You are given that:
Wa(0) = (2/3)Wb(0)
Wa(0) + Wb(0) = 297

so you can easily compute Wa(0) and Wb(0).

You are also given:

Wa(7) + Wb(7) = 124.419...
Wa(14) + Wb(14) = 55.050 ...

It appears that you have 6 unknowns: the rate, the density, and N0 for each of A and B, but only 4 equations.

That isn't the case. You never need to find N0 and the density separately, all you care about is their product. So rewrite the decay equations as:

Wa(t) = Wa(0)e^(- Ra t) and Wb(t) = Wb(0)e^(- Rb t)

which leaves you with only two unknowns, Ra and Rb (since you already have Wa(0) and Wb(0)) and two equations, the one for t = 7 and the one for t = 14

We know that x^(a+b) = (x^a)(x^b) so

e^(-Ra 14) = (e^(-Ra 7))(e^(-Ra 7))

if we let x = (e^(-Ra 7)) and y = (e^(-Rb 7)) then

W(7) = Wa(0)x + Wb(0)y
W(14) = Wa(0)x^2 + Wb(0)y^2

These two equations you can solve for x and y. (Get x in terms of y from the first and substitute for x^2 in the second. This gives you a quadratic in y which you can solve, etc.)

Once you have x and y, you can compute ln(x) and ln(y). Since ln x = -7 Ra, you can get Ra and similarly Rb.

The half life for A is just (ln 2)/Ra
http://en.wikipedia.org/wiki/Half-life

Messy in terms of numerical calculation, but still straightforward. A word of caution, exponential problems such as this are frequently numerically ill conditioned. The problem statement has all those decimal places for a reason. Use them. That means a good calculator.

Answer 2

A. The decay consistent is .5 initiate with 2 grams Day__1____2 grams to start Day__4____1 grams ultimate Day__8____0.5 grams ultimate Day_10____0.375_grams ultimate (½ of ½)