Dosimetry & Safety

Activity The term 'Activity' of a source describes the (in)stability of the atoms within a substance. One atom decaying per second is called 1 Becquerel (Bq). Uranium is a very unstable substance with an activity of 12,000 Bq. Sea water is a stable substance with an activity of 11 Bq. Each atom that decays will release potentially harmful radiation, in the form of Alpha and Beta particles, Gamma rays (and sometimes Protons, Neutrons and X-rays).AlphaBetaGamma rays

Activity & Half-Life You should already be aware that the Activity of some substances will quickly decrease. The time taken for the Activity to reduce by 50% is called the Half-Life. Uranium 238 has a very long half life of years while the half life of Protactinium 238 is only 2 minutes. (More half life information can be found here, or a quick revision of half-life calculations can be found here, otherwise continue to the next slide)here next slide

Biological Effect The radiation is harmful to living things, only if it is absorbed by the cells of that living thing. Absorption will cause ionisation in the cells. The risk of biological harm is affected by the following main factors:  The quantity of radiation energy absorbed by the body  The type of radiation energy absorbed by the body.  The type of tissue or organ etc that is receiving the energy.  The amount (mass) of tissue that is receiving the energy.  The time (exposure) period over which the energy was absorbed.

Absorbed Dose Absorbed Dose (D) accounts for the quantity of energy absorbed per unit mass. (Absorbed Dose Rate, is the Absorbed Dose over a period of time) Ü Units are Gys -1, Gyh -1, etc i.e. How many Joules absorbed per Kilogram Ü 1 JKg -1 absorbed is called 1 Gray.

Equivalent Dose Equivalent Dose (H) accounts for the quantity and type of energy absorbed. A high  R value is more dangerous where  R is the radiation weighting factor. Ü Units are Sieverts (Sv)

Equivalent Dose Rate The Equivalent Dose Rate accounts for the quantity and type of energy absorbed over a period of time. (This can also be expressed in two other ways) Ü Units are Svh -1, Svyr -1

Effective Equivalent Dose The Effective Equivalent Dose takes account of all of these, and is used to indicate the risk to health from ionising radiation. The average Annual Effective Equivalent Dose for people in the UK is about 2 milli-Sieverts (2 mSv) Ü This is due to background radiation In any one year, the public should not be exposed to more than 5 mSv. On average over a longer term, 1 mSvyr -1 should not be exceeded.

Background Radiation Radioactivity from rocks and soil on the Earth’s surface. Radioactivity which is naturally present in the human body. Radioactive gases such as radon and thoron. Cosmic radiation from the sun and outer space.

Example 1 By law, a radiation worker should not be exposed to more than 50 mSv in any one year. A worker at Hunterston Power Station receives 200  Gy of alpha radiation during an average working month. Will this worker exceed the annual effective dose equivalent limit? D = 200x10 -6 Gray  R = 20 H = ? H = D  R H = (200x10 -6 ) x 20 H = 4x10 -3 Sv i.e. Equivalent Dose Rate is 4mSv per month, and hence 48mSv over one year. The annual effective equivalent dose limit is not exceeded.

Example 2 A hospital radiographer receives 15  Gy due to alpha radiation and 400  Gy due to X-Rays, in a 35-hour working week. Calculate her average equivalent dose per working hour. Due to alpha H = D  R H = (15x10 -6 ) x 20 H = 3.0x10 -4 Sv Due to X-Rays H = D  R H = (400x10 -6 ) x 1 H = 4.0x10 -4 Sv Equivalent Dose Rate= 7.0x = 2x10 -5 Svhr -1 Ü Total equivalent dose, H = 7x10 -4 Sv over 35 hours

Reducing Biological Risk The 3 main ways in which we can reduce risk are By reducing our exposure time wherever possible, By increasing our distance from the source, By using shielding between source and ourselves.shielding

Half-Value Thickness …of shielding used to absorb gamma rays. Shielding cannot reduce the Activity of a source, nor can it affect the half-life time for a particular source…. ÙThe half-life of Cobalt 60 is 5.3 years ÙThe half-life of Uranium 238 is 4.5 billion years The purpose of shielding is to reduce the number of ‘counts per second’, as received by the Geiger-Muller tube (ie the radiation received by the user) ÞCommon shielding materials include lead, aluminium, concrete and water.

The count rate of a source is measured every minute for 10 minutes using a Geiger - Müller tube and counter. (The count rate is corrected for background) The results are shown below: The results show that as a greater thickness of lead shielding is used, the count rate detected behind the lead shield decreases. What thickness of lead shielding is required for the count rate to fall by one half? We usually need to plot a graph of results ……….. Experiment to find Half-Value Thickness

The graph can be used to find the half-value thickness

Select a point on the line which crosses grid lines on both axes. t1t1 At t 1 the count rate = 480 counts/second 480

t1t1 Now find the point on the graph where the count rate is half the previous value, (one half of 480 is 240). Call this t 2 t2t

t1t1 t2t2 The half-value thickness is the thickness of lead shielding required to reduce the count rate by half. hvt 1 = t2 t2 - t1 t1 = = mm

It is sometimes possible to calculate the half-value thickness of an absorber without using a graph. Example A source has a measured count rate of cps when no shielding is used. Using 24 cm of concrete the activity has dropped to only 750 cps. What is one half-value thickness? 11 22 33  24 cm of concrete have “halved” the count rate 4 times. ie 24 cm is equal to 4 hvts. One hvt = = 6 cm Reducing Biological Risk

Revision of Half-Life Calculations To say “the half life of Protactinium 238 is 2 minutes” means that in 2 minutes, 50% of the nuclei within that Protactinium will have decayed. (By emitting a Beta particle each nucleus decays to become a Uranium nucleus) By a further 2 minutes later, half of the remaining 50% of the Protactinium nuclei will also have decayed, and so on. For example…

The count rate of a source is measured every minute for 10 minutes using a Geiger - Müller tube and counter. (The count rate has been corrected for background) The results are shown below: The results show that as time passes the count rate of the source is decreasing. How long does it take for the count rate to fall by one half? We usually need to plot a graph of results ………..

The graph can be used to find the half-life of the source.

Select a point on the line which crosses grid lines on both axes. t1t1 At t 1 the count rate = 480 counts/second 480

t1t1 Now find the point on the graph where the count rate is half the previous value, (one half of 480 is 240). Call this t 2 t2t

t1t1 t2t2 The half-life is the time taken for count rate to drop by half. t 1/2 = t2 t2 - t1 t1 = = s = 2 minutes

It is sometimes possible to calculate the half-life of a source without using a graph. Example A source has an original activity of Bq. After 24 days the activity has dropped to only 750 Bq. What is the half life of the source? 11 22 33  4 half lives have passed in 24 days. One half life = = 6 days Return to Activity