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Several students considered this problem
with NRICH as a group. Whilst most thought initially to use
algebra, they soon realised that there was no unique solution. So,
the problem reduced to finding ranges of possible solutions.This
was approached with a mixture of algebra and numerical
Before attempting any calculations I tried to make sense of the
numbers, as suggested in the problem.
The first measurement point was $10$ days. Samples of cobalt $60,
57, 56, 58$ would not decay very much in this time.
Since the sample loses about one third of its mass in this time, a
large component must be cobalt $58$. However, since there is still
about half the sample remaining after $360$ days there must be at
most about half of cobalt-58.
I worked out after 10, 90 and 360 days what fraction of a lump of
sample of each pure isotope would remain. So , after 10 days about
99.6% of cobalt 60 and 0.63% of cobalt 55 would remain; after 360
days 87.8% of a mass of cobalt-60 would remain but 6.98E-80 of
cobalt-55 would remain. Since this felt experimental, I made a
spreadsheet where I could work out the total mass remaining for
given starting sample
Using common sense (that Co-60 decays very slowly and Co-55 very
quickly) I quickly found this solution, which matched the target
masses very closely.
There are obviously other solutions, but experimentation indicated
that there had to be 30g of Co-55 to match the data.
Since there are only 4 conditions given (initial mass followed by
three observations) and five possible isotope masses the system is
under-determined, so there will be no unique solution. Looking at
the data, the half lives of Co-56 and Co-58 are similar, so I will
make the simplifying assumption that the masses of these two
isotopes are equal.
If I choose the initial masses to be $A$, $B$, $C$, $C$, $D$ then
my four equations will be
A+B+C+C+D &=& 100\cr
I chose to work with the numbers to 4 decimal places, as this
seemed about right given the accuracies in the problem (since the
masses remaining were given to 3 dp and the half lives also only to
a couple of places)
This gave me equations
A&+&B&+&2C&+&D &=& 100\cr
These looked quite intimidating to solve, but then I just got down
to it and it only took a few minutes using a spreadsheet to do the
I found that $A=41.91, B=15.57,C= 6.18, D=30.14$
I then wondered about the other solutions. I found these
From this table it seems reasonable to assert that
1) There are $30$ grams of Co-55.
2) The combined mass of Co-56 and Co-58 is between 12 and 18
3) There is at most 16 grams of Co-57
4) There is at least 41 g of Co-60.