## Secular and Transient Equilibrium Formula

I will give the full derivation since although it is quite long and looks complicated it uses nothing more than A-level mathematics. Despite this the solution of the differential equations is usually given as a standard solution rather than derived.

Consider a parent isotope (P) decaying into a daughter product (D).

The rate of decay of P (which is also the rate of formation of D) is given by

The solution of this is the standard formula for decay

The rate of change of D is the rate of formation minus the rate of decay

Substituting the formula for N_{p} from above we get

Â

We may be tempted just to integrate both sides. However, we cannot do this sine we do not know how N_{D} changes with time.

The way to solve such equations is to remember the formula for the derivative of a product –

So now rearrange our equation

OK the left hand side does not look like the derivative of a product at the moment (probably because it isn’t) but if we multiply everything by e^{Î»Dt} we get

So the left is now a derivative of a product so we can now integrate both sides

Multiplying both sides by e^{-Î»Dt}

We now have to work out what our integration constant is. If we take time t=0 all the exponent terms become 1 since e^{0} is 1 and N_{D} becomes the initial value of N_{D}

We can now substitute this into our original equation for N_{D} to give

Rearranging we get

Note that the last term is the decay of the initial concentration of the daughter product.

## Decay Curves

From now on we are going to assume that there initially not daughter product. The plots are arbitrary units. If you want to play about with them yourself then I have loaded the spreadsheets inÂ xls andÂ ods formats.

You can see the typical decay curve for the parent when the two decay constant are very similar. The daughter builds up and then slowly decreases as it and its source (the parent) decays.

NOTE: From now on the y-axis is logarithmic.

This is similar to the graph above it but using a logarithmic scale.

If the parent is longer lived than the daughter after a certain period of time the amount of the daughter depends mainly on the rate of formation.

If we take our formula and ignore the last term – i.e. assume there is no daughter product initially then we get

If the parent decay constant is significantly smaller than that of the daughter then after a reasonable length of time.

and

The formal now becomes

and since

we have

i.e. the amounts become proportional to each other as seen on the plot above. This is known as transient equilibrium.

Now we consider the case where the decay constant for the parent is negligible compared to that of the daughter.

Since the amount of the daughter product depends mainly on its rate of formation and due to the small decay constant of the parent this becomes constant.

Now we consider the activity – i.e. the number of decays per second from the daughter and parent. The activity is just the amount times the decay constant:

First of all multiply through by Î»_{D}

Subsitituting this into our formula we have:

Now if the decay constant for the parent is a lot smaller than that of the daughter.

and therefore

This is secular equilibrium.

## Grand Daughter and Greate Gand Daughter

You can apply the same method to grand daughter and great grand daughter products. The general formula is:

I shall put this into a spread sheet for you to use at some point. Note that it is you may think that you can just apply the first formula several times. However, this does not work well since you have to set your time scales to be very small to avoid losing some intermediate isotopes. Since the halflife of some isotopes vary by many orders of magnitude.

1-A radioactive needle contains 222

86Rn (T1/2 = 3.83 days) in secular equilibrium

with 222

88Ra (T1/2 = 1600 years). How long is required for the 222

86Rn to decay

to half of its original activity?

2-In nature, 222

88Ra (T1/2 = 1600 years) exists in secular equilibrium with 238

92U

+(T1/2 = 4.5 Ã— 109 years). What fraction of the world’s supply of radium will be

left after 1600 years?

3-How many atoms and grams of 90Y are in secular equilibrium with 50 mCi of

90Sr?

4- How many MBq of 132I (T1/2 = 2.3 hours) should be ordered so that the sample

activity will be 500 MBq when it arrives 24 hours later?

Just a quick note to mention that this problem is best approached using Laplace Transforms – particularly if you are looking at grand-daughter products and beyond. It can be done with the method above but you have to use definite integrals and it gets very messy.

I did have a look for the Laplace method on the Internet but could not find anything – if you know of anything then please let me know. I did intend to do it myself but did not get round to it.

There was at one point some interest in secular equilibrium amongst the people I ran the ‘Basic Nuclear Physics’ course for and I was being asked questions about it. Again I was going to write more on the reasons for this interest but never got round to it – so much to do and so little time.

Hi,

Just wanted to know if you ever created the spreadsheet for grand daughter products. I am trying to observe whether an equilibrium is set up between Radon-222 (Parent) and Lead-214 (Grand daughter).

Thanks

You have to be careful if you try to do it numerically on a spreadsheet because some of the daughter products have very different half lives. If you set your timescale too long then the amounts will be wrong or even zero.

You can use the Bateman formula to calculate them or have a look at some online calculators such as http://www.wise-uranium.org/calc.html or I think that there is an app that you can download to do this for you.

Hopefully it is obvious that if you are looking at a sample of natural uranium then the answer is quite easy. The decay of the Uranium 238 would be the rate defining. Therefore the amount of Radon-222 would be constant since the rate of decay would equal the rate of production – i.e. the activity of the Radon-222 would be equal to the activity of U-238. It is therefore easy to calculate the amount of Radon-222 in your sample. The same would be true of the Lead-214.

To get secular equilibrium then enough time must have past – you must wait for the red line to flatten out in the diagram above. Also none of the isotopes must escape your sample area – your Ra-228 has not leached out or your Rn-222 bubbled away.

However, even if you relax these two conditions then calculating the amounts assuming secular equilibrium would give you a maximum upper bound for the daughter products. With the possibility of leaching it will still give you the total amount of daughter product but it will not be in your sample.

If you have isolated the Rn-222 then the story is quite different since it will no longer be created at a ‘constant’ rate and will decay with time and you will get transient equilibrium.