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:
Place the
GM tube near the source.
Count for
1 minute, and divide by 60 to get the count rate per second.
Move away
in steps of 5 cm until very low count rates are obtained.
Use
increased time samples for low rates to get an accurate count rate per second.
Find out
the background count rate.
Correct
the count rate using the background count rate.
If we plot distance ^{2} (yaxis) against 1/Count Rate, we will get a straightline graph:
Notice that the line makes an intercept with the yaxis 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.
This is a required practical (Number 12).
Gamma rays obey the inverse square law, as they are an electromagnetic radiation.
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 (W m^{2});
I_{0} – intensity at the source
(W m^{2});
k
– constant;
x – the
distance from the source (m)]
The unit for intensity is watt per square metre (W m^{2}).
We can explain the inverse square law 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.
Worked Example A sample of radioisotope emits 1.2 ´ 10^{12} g photons per second in all directions. Assuming that the area of a person is about 1 m^{2}, estimate the number of photons per second received by someone standing (a) 2.0 m; (b) 4.0 m; (c) 5.5 m from the source if they were accidentally exposed. 
Answer
We
need to work out the area of a sphere 2.0 m radius: A = 4pr^{2} = 4 ´ p ´ (2.0 m)^{2} = 16 m^{2} ´ p = 50.27 m^{2}
Now
we can work out the fraction of the radiation received.
Fraction = 1 m^{2} ¸ 50.27 m^{2} = 0.0199
Now
we can calculate the dose (number of photons per second) received:
Dose = 0.0199 ´ 1.2 ´ 10^{12} s^{1} = 2.39 ´ 10^{10} 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 ´ 10^{10} s^{1} ¸ 4 = 6.00 ´ 10^{9} 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(I_{0})
x^{2}
We need to get rid of the constant and the original intensity, so we use the 2.0 m result as a reference:
Þ 2.39
´ 10^{10} =
k(I_{0}) (2.0 m)^{2}
And we then write:
I = k(I_{0}) (5.5m)^{2}
We
combine these two to give:
2.39 ´ 10^{10} s^{1} ´ 4.0 m^{2} = 30.25 m^{2} × I
Þ I = 2.39 × 10^{10} s^{1} × 4.00 m^{2} = 3.16 × 10^{9 }s^{1}^{ } 30.25 m^{2} 
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.
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:
Cosmic
rays
Radioactive
material in the bricks of the building.
Small
amounts from medical and industrial uses.
Yourself
(you are alive with Carbon14 amongst other radioisotopes)
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 I_{0} is. Nor do we know the constant k. Neither of these matter. Instead we have a count I_{1} at point 1 and a count I_{2} at point 2 at distances x_{1} and x_{2} respectively. So we can write:
We can combine these in a
rearranged form to give us:
I_{1}(x_{1})^{2}
= kI_{0} = I_{2}(x_{2})^{2}
So we can write:
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.

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.
