Question

Nuclide X has decay constant 1.08 h-1. It can decay in 2 branches, to nuclide Y or to nuclide Z:
X --> Y, with decay constant 0.43 h-1, and
X --> Z, with unknown decay constant.
If there are 10 moles of nuclide X at time t = 0, how many moles of each daughter will there be at (a) t = 45 min, and at (b) t = 72 min? At each time verify the conservation of the total number of moles.

Answer 1

ktotal = k1+k2
1.08 = 0.43 + k2
k2 = 1.08-0.43 = 0.65
(You can verify that if you wish this way. Since k = 0.693/t_1/2 plug in each k and solve for t_1/2. Then Ttotal = t1*t2/(t1+t2), solve for t2, then plug that back into the k = 0.693/t<sub>1/2</SUB> and solve for k.

Here is the 45 min.
ln(No/N) = k(0.75 hr)
ln(10/N) = 0.65(0.75)
Solve for N and I obtained 6.14 X left. That means 10-6.14 = 3.86 X has decayed to form Z.

At the same time X has been decaying to form Y. Note that you start with only 6.14 of X (the rest of it has decayed and you can't count that).

ln(6.14/N) = 0.43(0.75)
N = 4.45 mols X left
6.144-4.45 = 1.69 X has decayed to form Y.
Total Y + Z = 3.86 + 1.69 = 5.55 mols total Y + Z.

Then use the total k of 1.08.
ln(10/N) = 1.08(0.75)
N = 4.45 X left
10 - 4.45 = 5.55 Y + Z mols and that agrees with the above.

The 1.2 hours (72 min) is worked the same way.
You should confirm all of this. Check my reasoning.