Assistance with Calculation problems
Calculations
0.1. Decay Nt = Noe-0.693t/T where t is the time elapsed and T is the half life. Or Nt = No
(0.5)t/T
0.2. 1 mrad = 62,400 MeV per gram. Rule of thumb: 100 beta particles per cm per
second = 10 mrad/hr exposure to the dermal layer of skin with high energy betas.
Problem calculate the exposure if a thin window gm tube detector gives 30,000 counts per
minute the counter has 5 cm2 surface area and 100% counting efficiency.
We can say that 30,000 counts are picked up in a 5 square cm area or we have 6,000
counts per minute through each square centimeter. This is equal to 100 counts per cm per
second or about 10 mrad/hr. If you had been given the precise energy of the beta particle
then it is a simple matter to convert electron volts to rad and come up with a precise
answer. For example 100 beta's per second from 32P with an average energy of 0.7
Mev/beta and a range of 0.8 cm will deposit all of their energy in the first cm of tissue.
Thus 100 X 0.7 = 70MeV of energy is deposited in one gram (1cm3 soft tissue is about
one gram). The rest is a simple conversion problem. Please note the exposure to the
dermal layer close to the skin surface will be much higher than the dose to deeper tissue
due to the fact that beta particles have a high LET.
For a dilution problem the following explanation is provided:
6. 0.5 liters of tritiated water containing 0.5 Ci 3H is mixed with the water in
a tank. A 1 ml sample of water is removed from the tank mixed with
appropriate scintillation counting fluid and counted on a scintillation
counter. You may disregard quenching effects. The efficiency of the
counter is determined to be 55% and the observed count is 5000 counts per
minute. What was the volume of water in the tank prior to the addition of
the tritiated water?
a.
b.
c.
d.
e.
300,000 liters
244,200 liters
122,100 liters
3,765,000 liters
9,000 liters
Basic equations are Df = Vi/ Vi +Vp (the dilution factor is equal to the volume
added, divided by the volume added plus the volume present
Ad = Ao * Df The final diluted activity is equal to the initial activity times the
dilution factor
Substitute Vi/Vi+Vp into the equation and rearrange for the following
(Ao *Vi) – Vi
------------------ = Vp
Ad
The initial activity is given as 0.5 Ci/0.5 liters or 1 Ci/liter which is 2.22 *10^12
dpm/liter The final activity is 5000 cpm per ml or at a .55% efficiency 9091
dpm/ml we can convert this to dpm/liter since 1000 ml = 1 liter and the final
activity is 9,091,000 dpm/l Vi is 0.5 liters
Plugging the numbers into the equation above the answer is
(2.22*10^12 * 0.5) -0.5
---------------------------------- =122,098 (due to rounding error))
9,091,000
0.3. Dilution Problems. Activity X dilution factor = Activity after dilution
Be prepared to complete dilution problems
Dilution factor = volume added/volume added + volume present
For serial dilutions the total dilution is derived by multiplying the individual
dilution factors.
For a simple serial dilution: you calculate the dilution factors for each tube and multiply
them to get the total dilution: example:
If you add 0.1 ml of radioactive liquid to 0.9 the dilution factor is 0.1/0.1+0.9 = 1/10 if I
then take 0.1 ml from the first dilution tube and place it in 0.9 ml of diluent the second
tube also has a dilution factor of 1/10 but the total dilution is df1 X df2 = 1/100
Dose rate calculation given known skin surface contamination. First we need to know
the approximate dose rate of the nuclide in question. In the case of 32P that is 90
mGray/hr for one microcurie covering an area of one square centimeter. Given a count
rate I need to determine how many microcuries/cm2 are contaminating the surface (and
multiply by 90mGray/microcurie/cm2). So if I measure 200 counts per second using a
meter with 80% 2pi efficiency, I know that the actual number of betas emitted in one half
sphere (the counting geometry of all flat survey meters, is 200 cps/.8 = 250 dps and
adjusting for the geometry (0.5) yields a total number of disintegrations per second of
500. Now I need to know how concentrated the material is over a surface area. If
contamination is evenly distributed, I can simply divide by the surface area of the meter
(let’s say the meter has a surface area of 5 cm2 then there are 100 dps/cm2. Now we
need to change dps/cm2 to microcuries/cm2. Since one microcurie = 3.7*104 dps we
simply divide 100 dps by that number and get .0027 microcuries/cm2. Multiplying by
our 90mGray/hr factor per micurcurie/cm2 we get 0.24 mGray/hr as the final answer.
Decay problems are pretty straight forward. If you have 5 mCi of a material with a 20
day half life how much is left after 50 days? At= Ao (.5) t/T where At is the activity at
time t, Ao is the initial activity and t is the elapsed time and T is the half life. So for this
problem A(50) = 5 mCi (0.5) 50/20 = 0.88 mCi
For calculating surface contamination given a specific survey meter count the answer is a
variation on the calculation above for dose rate due to skin contamination. Example we
measure a hot spot with 200 counts per second and the spot covers an area of 10 cm2.
We know that the 2 pi efficiency for the counter for the nuclide in question is 0.33. Find
the total surface contamination in microcuries. First we correct the count for efficiency
(200 cps/.33 = 606 2pi dps ) then we correct for the geometry go from 2pi to 4pi (since
we can only count 50% of the total emissions due to the fact that we have a flat detector)
which yields 1212 dps total. Next we convert dps to microcuries (in the case of
becquerels one dps is one Becquerel so if the problem asked for becquerels you are
finished) 3.7*104 dps = 1 microcurie and we have a total of 1212/3.7*104 dps =0.032
microcuries.
For calculating beta flux: What is the beta flux 1 cm away from a 0.2 mCi 32P point
source. We know that 0.2 mCi 32P source emits 0.2* 3.7*107 beta particles per second in
a uniform spherical pattern. So at one centimeter from the source that number of beta
particles go through the total spherical surface each second. The total area of a sphere is
4 pi r2 hence the total surface area of the sphere at one centimeter is 12.57 centimeters
and the flux is 588,843 betas/cm2/s
I believe that covers most of the problems that you see on the test.