6560
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2011-02-01T00:00:01
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</ul>
<div> </div>
<p>The element cobalt has one stable isotope: $^{59}$Co.</p>
<p><br></br>
Chemists can construct several different isotopes of this element with varying numbers of neutrons. All are unstable; the five most stable are given in the following table (y=years, d=days, m=minutes)<br></br>
</p>
<table style="" border="1">
<tbody>
<tr>
<td>Isotope</td>
<td>Half life</td>
</tr>
<tr>
<td>$^{60}$Co</td>
<td>5.2714 y</td>
</tr>
<tr>
<td>$^{57}$Co</td>
<td>271.79 d</td>
</tr>
<tr>
<td>$^{56}$Co</td>
<td>77.27 d</td>
</tr>
<tr>
<td>$^{58}$Co</td>
<td>70.86 d</td>
</tr>
<tr>
<td>$^{55}$Co</td>
<td>17.53 h</td>
</tr>
</tbody>
</table>
<br></br>
<div>I have a lump of pure cobalt, called sample $X$. I know that sample X is uniformly composed of differing quantities of the various isotopes of cobalt (including, possibly, some of the more unstable isotopes not listed in the table above) in nice ratios.</div>
<br></br>
<div>I take a piece of sample $X$ of weight $100$ g and leave it to decay.</div>
<br></br>
<div>After exactly $10$ days $68.402$g of cobalt remains in the sample. What information does this give you about the possible initial composition of the sample? For example, are there any isotopes which you can work out must be, or cannot be, present in the sample? Use your common sense and simple calculations before attempting a numerically detailed analysis!</div>
<br></br>
<div>After $90$ days in total $58.283$g of cobalt remains. After $360$ days $48.359$g of cobalt remains.</div>
<br></br>
<div>How accurately can you determine the composition of the original sample?</div>
<br></br>
<br></br>
<div class="framework"><br></br>
NOTES AND BACKGROUND<br></br>
<br></br>
Data sourced from <a href="http://ie.lbl.gov/education/isotopes.htm">http://ie.lbl.gov/education/isotopes.htm</a><br></br>
<br></br>
The following Wikipedia page might also be of interest: <a href="http://en.wikipedia.org/wiki/Cobalt-59">http://en.wikipedia.org/wiki/Cobalt-59</a></div>
<br></br>
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<span class="editorial">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
methods.</span><br></br>
<br></br>
<span class="editorial">Steve thought</span>:<br></br>
<br></br>
Before attempting any calculations I tried to make sense of the
numbers, as suggested in the problem.<br></br>
<br></br>
The first measurement point was $10$ days. Samples of cobalt $60,
57, 56, 58$ would not decay very much in this time.<br></br>
<br></br>
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.<br></br>
<br></br>
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<br></br>
<br></br>
<mdo:image width="594" height="159" alt="" src="steveSoln.png"></mdo:image><br></br>
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.<br></br>
<br></br>
There are obviously other solutions, but experimentation indicated
that there had to be 30g of Co-55 to match the data.<br></br>
<br></br>
<span class="editorial">Tom thought</span>:<br></br>
<br></br>
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.<br></br>
<br></br>
If I choose the initial masses to be $A$, $B$, $C$, $C$, $D$ then
my four equations will be<br></br>
<br></br>
$$<br></br>
\begin{eqnarray}<br></br>
A+B+C+C+D &=& 100\cr<br></br>
A2^{-10/1924.06}+B2^{-10/271.79}+C2^{-10/77.27}+C2^{-10/70.86}+D2^{-10/1.37}&=&68.402\cr<br></br>
A2^{-90/1924.06}+B2^{-90/271.79}+C2^{-90/77.27}+C2^{-90/70.86}+D2^{-90/1.37}&=&68.402\cr<br></br>
A2^{-360/1924.06}+B2^{-360/271.79}+C2^{-360/77.27}+C2^{-360/70.86}+D2^{-360/1.37}&=&68.402<br></br>
\end{eqnarray}<br></br>
$$<br></br>
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)<br></br>
<br></br>
This gave me equations<br></br>
<br></br>
$$<br></br>
\begin{eqnarray}<br></br>
A&+&B&+&2C&+&D &=& 100\cr<br></br>
0.9964A&+&0.9748B&+&1.8210C&+&0.0063D&=&68.402\cr<br></br>
0.9681A&+&0.7949B&+&0.8607C&&&=&58.283\cr<br></br>
0.8784A&+&0.3993B&+&0.0691C&&&=&48.359<br></br>
\end{eqnarray}<br></br>
$$<br></br>
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
algebra.<br></br>
<br></br>
I found that $A=41.91, B=15.57,C= 6.18, D=30.14$<br></br>
<br></br>
I then wondered about the other solutions. I found these
solutions:<br></br>
<br></br>
<table border="1">
<tbody>
<tr>
<td>Isotope</td>
<td>Mass</td>
<td>Mass</td>
<td>Mass</td>
<td>Mass</td>
</tr>
<tr>
<td>Co-60</td>
<td>52.16</td>
<td>42.31</td>
<td>41.49</td>
<td>41.91</td>
</tr>
<tr>
<td>Co-57</td>
<td>0</td>
<td>15.49</td>
<td>15.67</td>
<td>15.58</td>
</tr>
<tr>
<td>Co-56</td>
<td>12.90</td>
<td>0</td>
<td>12.68</td>
<td>6.18</td>
</tr>
<tr>
<td>Co-58</td>
<td>4.90</td>
<td>12.077</td>
<td>0</td>
<td>6.18</td>
</tr>
<tr>
<td>Co-55</td>
<td>30.03</td>
<td>30.121</td>
<td>30.15</td>
<td>30.13</td>
</tr>
</tbody>
</table>
<br></br>
<br></br>
From this table it seems reasonable to assert that<br></br>
<br></br>
1) There are $30$ grams of Co-55.<br></br>
2) The combined mass of Co-56 and Co-58 is between 12 and 18
grams.<br></br>
3) There is at most 16 grams of Co-57<br></br>
4) There is at least 41 g of Co-60.<br></br>
<br></br>
<br></br>
<br></br>
<br></br>
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<h3>Why do this problem?</h3>
This <a href="http://nrich.maths.org/6560&part=">problem</a> gives an interesting problem solving challenge for students aged 16/17 with an interest in both chemistry and mathematics. It gives a good exercise in mathematical modelling: a physical process needs to be represented by equations which may then be analysed mathematically. It is not simply routine, nor is it especially difficult if
considered in the right ways. However, it is a good example of a situation in which simply trying to 'do the algebra' will not work.
<h3>Possible approach</h3>
<div>This problem works well for individual use. It could be solved numerically, using trial and error, or graphically using many ideas from linear algebra.</div>
<br></br>
<div>If done as a group or set as a homework, it would be interesting to compare solutions, as there are several different possible routes to a conclusion. What do students think of each others answers? Which seem most powerful? Neatest?</div>
<h3>Key questions</h3>
<div>Could the sample be a pure sample of a single isotope? Why?</div>
<div>How many pieces of information are given to us in the question?</div>
<div>Is there any way of distinguishing between the different isotopes for the timescales given?</div>
<div>Chemically, what is happening in the decay? How does decay affect the mass of the sample? Is this a significant effect?</div>
<h3>Possible extension</h3>
<div>Chemical extension:</div>
<ul>
<li>Consider carefully the effects of the decay on the mass of the sample, taking into account that some matter will be lost.</li>
<li>Energy is radiated from the system during decay. Energy is related to mass via Einstein's equation $E=mc^2$. Is this effect relevant or significant?</li>
</ul>
<div>Mathematical Extension:</div>
<ul>
<li>Explore how many more data points would be needed to give a unique solution to the problem. What is the minimum number of extra data points needed to give a unique solution?</li>
</ul>
<h3>Possible support</h3>
<div>Encourage trial and error, starting with equal masses of each sample. Then raise or lower the proportions of two of the masses, using a spreadsheet to do the numerical calculations.</div>
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<mdoxml version="1.0">Do you think it will be easy to distinguish between each of the isotopes of cobalt?<br></br>
<br></br>
Imagine to start that that the sample consists of a single isotope of the cobalt. Would such a sample have too much or too little cobalt at each of the times indicated? How could you adjust the sample to bring it in line with the numbers indicated?<br></br>
<br></br>
<span style="font-style: italic;">This problem makes use of chemistry concepts found in A-level. The main fact of importance is that in each half-life half of the cobalt will transform into other matter, the vast majority of which will remain in the sample. Mathematically, you will need to use linear algebra, simultaneous equations such as</span> <span style="font-style: italic;">typically
encountered at post-16 level, although a simpler numerical analysis will allow younger students to engage with the problem</span>.<br></br>
<br></br>
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Cobalt decay
Investigate the effects of the half-lifes of the isotopes of cobalt
on the mass of a mystery lump of the element.
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