reply to post by Wrabbit2000
The problem with a Becquerel, is that it only tells you the amount of atoms decaying per second.
It does not tell you the Type of radiation, or the energy level of the radiation.
Potassium-40 (Found in bananas and the human body) gives off beta radiation (Electron Ejection) at an energy level of 1.33 MegaelectronVolts
Meanwhile, Tritium gives off beta radiation (Again, electron Ejection) at an energy level of only 0.018590 MegaelectronVolts
So, when calculating your radiation dose, you need various equations to help estimate how much of a threat the radiation is to you.
These calculations are:
1. Becquerel (The amount of decays per second from the source, this measurement includes the Half-Life (how quickly it decays) and the amount of
2. Quality (The energy level of the radiation, Higher energy levels pose more of a threat than lower levels; and certain types of radiation are
intrinsically more dangerous) For example, Beta radiation is typically less hazardous than Gamma Radiation.
3. Exposure (How many of these decays are actually impacting upon your tissues, which is a function of your distance from the source of radiation, the
size of the silhouette you present to the radiation (area), and how much attenuation the materials in between you and the source represent (how much
radiation is soaked up by the air, water, concrete, lead, etcetera)
The international unit for Radiation Dose Received is the Seivert, in which One sievert is equal to one joule of radiation energy impinging upon one
kilogram of tissue.
So, let's say that you have a chunk of Tritium that is undergoing 3,000,000 Becquerel. (decays per second)
And this tritium is at the bottom of a swimming pool (the deep end) approximately 4 meters underwater.
You are standing on the side of the pool, over the deep end, and you want to see what your Effective Dose is (Sieverts)
So, we have a point source of Tritium emitting 3,000,000 Becquerel in all directions.
The decay energy of tritium is 18.590 KeV (18,590 electron volts), and an electron volt is equivalent to 1.60217657 × 10^-19 joules
So, this means that the total energy released by the Point source of tritium is:
(18,590 electron volts) * (3,000,000 Becquerel) * (0.000000000000000000160217657 joules per electron volts)
Equals: 8.93533873 × 10^-9 (0.00000000893533873 Joules) emitted per second from our point source of radiation.
Now, this is what is EMITTED... to see how much of that HITS you, we need to use the Inverse Square Law:
Now, if we are to assume that our point source is a sphere with a radius of one centimetre, and a surface area of 12.5663706 square centimetres (and
the total energy output of the radiation source is evenly distributed around the surface of the sphere) then all we have to do is increase the radius
of our hypothetical sphere to the distance that you are standing away from the source of radiation.
In this case, we would go from a radius of 1 centimetre, to a radius of 400 centimetres, which increases our surface area from 12.5663706 centimetres
to 2,010,619.3 square centimetres.
Now, the ratio of 12.5663706 to 2,010,619.3 is 160,000:1... which means that the energy output passing through a single square centimetre AT THE
SOURCE, is going to be reduced 160,000 times by the time it reaches YOU, standing on the edge of the pool.
Next, we need to calculate the attenuation of the radiation from the water in-between you and the source.
Water has a Half Value Thickness of 5.823 centimetres (I am rounding the radiation energy WAY up) which means that it takes 5.823 centimetres of water
to reduce the total radiation passing through it by HALF.
And since there are 400 centimetres of water between you and the source, this equals approximately 68 total "Halvings" of the radiation energy level
(Joules) which equates to 1/4624 the original energy level by the time it reaches you at the surface of the pool.
So, now we need to calculate how many Square Centimetres our silhouette takes up. Since you are facing the Source of radiation feet first, this means
that you are presenting a silhouette of approximately one third of a meter by one third of a meter.... or... about 1,089 square centimetres.
So, to finish the calculation:
0.00000000893533873 joules at source (reduced by 160,000 (angular)) gives us 5.58458671 × 10^-14 joules per square centimetre at your distance.
5.58458671 × 10^-14 joules attenuated through 400 centimetres of water gives us 1/4624th the energy which is 1.20773934 × 10^-17 joules
And since your silhouette is about 1,089 square centimetres, you receive 1.31522814 × 10^-14 joules
Given an average body weight of 60kg, that makes your effective dose:
0.0000000000000131522814 joules divided by 65 kilograms