Nuclear Physics Tutorial 4 - Inverse Square Law of Gamma Radiation



To measure how the intensity of gamma radiation varies with distance, we can set up the following apparatus:



We carry the experiment out like this:


If we plot distance 2 (y-axis) against 1/Count Rate, we will get a straight-line graph:



Notice that the line makes an intercept with the y-axis below the origin.  This is because the gamma source is deep within its container.  It would clearly be most undesirable to have it exposed immediately to the room.  The intercept gives us the count rate right at the source.


We find that the intensity (the number of counts per second) decreases with the square of the distance.  This means that if we double the distance, the intensity goes down four times.  The relationship is:


[I intensity; I0 intensity at the source; k constant; x the distance from the source]


We can explain this by reference to this diagram



Let us look at the square, which is one unit away from the source.  It has a side of one unit.  Suppose we move away from the source by two units.  We find that the square has side of length two units.  Whereas the area of the little square is 1 square unit, the area of the big square is four square units.  So the same amount of energy is spread out over four times as much area.


This is true for any kind of wave radiation, not just gamma.


In three dimensions, we find that waves propagate from a point source spherically, but the same argument applies.



A sample of radioisotope emits 1.2 1012 g photons per second in all directions.  Assuming that the area of a person is about 1 m2, estimate the number of photons per second received by someone standing (a) 2 m; (b) 4 m; (c) 5.5 m from the source if they were accidentally exposed.

We need to work out the area of a sphere 2 m radius:

A = 4pr2 = 4 p 22 = 16 p =  50.27 m2

Now we can work out the fraction of the radiation received.

Fraction = 1 m2 50.27 m2 =  0.0199

Now we can calculate the dose received:

          Dose = 0.0199 1.2 1012 = 2.39 1010 s-1

We use the inverse square law to work this out.  Double the distance, the intensity goes down by 4 times.  So the dose is:

Dose = 2.39 1010 s-1 4 = 6.00 109 s-1

The problem here is that 5.5 is not an easy multiple of 2.  What we need to do is to use

I = k(I0)


                        2.39 1010 = k(I0)



                        I = k(I0)


We combine these two to give:

2.39 1010 4.84 = 30.25I

I = 2.39 1010 4.84 = 3.82 109 s-1



This calculation tells us that the further away from a gamma source, the lower the rate of radiation exposure.  This has important implications for work with radioactive materials.  The rule is quite simple; the further away, the better.  Hence we handle radioactive sources with tongs.  School radioactive sources are very weak, but it is still essential to handle them carefully.


Background Radiation

Radiation detectors will always detect radiation, even if there is no source in the room.  There is always background radiation coming from all sorts of sources:


The amount should be constant, but actually it does vary, so when you do experiments, you should always do a background radiation check as a matter of good practice.  You should always subtract the background radiation from your results to get a corrected result.


In some parts of the country, especially where there are granite rocks, the background radiation is quite high.  Some houses have fans to reduce the build up of radon gas.


When doing intensity calculations, it is more common that we do not know what Io is.  Instead we have a count I1 at point 1 and a count I2 at point 2 at distances x1 and x2 respectively.  So we can write:




we can combine these in a rearranged form to give us:


I1(x1)2 = kIo = I2(x2)2


If we rearrange further we can get a ratio:




This relationship is easier to use and you will have a go at it in the next question.




Question 1


A detector placed 0.20 m from a sealed gamma ray source receives a mean count rate of 2550 counts per minute.  The mean background radiation count is 50 counts per minute.  The experiment is set up as shown.



Calculate the least distance between the source and the detector if the count rate is not to exceed 6000 counts per minute.





Intensity is the number of photons per unit area.

The intensity and distance from the source are related by an inverse square law.

The basic relationship is I1 = kIo  


A more useful relationship is: