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- since 1996 -
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John Porter
United Kingdom, Joined Oct 2014, 5

John Porter

United Kingdom,
Joined Oct 2014
5
00:08 Oct-16-2014

Could anyone please help me with trying to understand how to work out the following questions. I am not asking for the answers, just to understand the formulas to work these questions out.

1) Selenium-75 has a half life of approximately 120 days. If the initial activity A0 is 100 GBq, which of the following is closest to the value of the activity after 60 days?

2) If Polonium-210 has an initial activity of 50 GBq, and is found to have an activity of 8 GBq after 1 year (365 days), which of the following is nearest to its half-life?

3) A Se-75 source has an activity of 400 GBq and a dose rate constant of 5.5 x 10-2 mSv m2/hGBq. What will the effective dose rate be at a distance of 100 cm?

Any assistance appreciated.
Thanks
Johnny

johnny
United Kingdom, Joined Oct 2014, 5

johnny

United Kingdom,
Joined Oct 2014
5
13:34 Oct-16-2014
In Reply to John Porter at 00:08 Oct-16-2014 (Opening).

OK, I think I have solved the first problem:

A = Ao/2^t/T

100/2^0.5=70.71 GBq

I am still trying to work out questions 2 and 3, if anyone can help that would be great.

Thanks
Johnny

johnny
United Kingdom, Joined Oct 2014, 5

johnny

United Kingdom,
Joined Oct 2014
5
14:10 Oct-16-2014
In Reply to johnny at 13:34 Oct-16-2014 .

Also think I have solved question 3, if anyone would like to verify?

3) A Se-75 source has an activity of 400 GBq and a dose rate constant of 5.5 x 10-2 mSv m2/hGBq. What will the effective dose rate be at a distance of 100 cm?

0.5/70=0.0071

400/2^0.0071=398 GBq

398 x 0.055 / 1 = 21.89 mSv/h

RENGARAJU. B. N.
NDT Inspector, NDT Level-III/Radiation Safety Officer(RSO)
Marine and Heavy Engineering Company, MALAYSIA., India, Joined Sep 2014, 106

RENGARAJU. B. N.

NDT Inspector, NDT Level-III/Radiation Safety Officer(RSO)
Marine and Heavy Engineering Company, MALAYSIA.,
India,
Joined Sep 2014
106
16:05 Oct-16-2014
In Reply to John Porter at 00:08 Oct-16-2014 (Opening).

I checked and found Johnny's answer for 1 and 3 to be correct.

For problem no: 2, using the formula A = Ao x e^-0.693/HVL X Number of days elapsed(365), HVL for Polonium-210 is nearest to 138 days.

In Problem No: 3, dose rate constant is given as 5.5 x 10-2 mSv m2/hGB. There is an error in this as dose rate constant is related to distance and not area. So, i feel it should be m/hGB.

Regards
Rengaraju. B. N.

RENGARAJU. B. N.
NDT Inspector, NDT Level-III/Radiation Safety Officer(RSO)
Marine and Heavy Engineering Company, MALAYSIA., India, Joined Sep 2014, 106

RENGARAJU. B. N.

NDT Inspector, NDT Level-III/Radiation Safety Officer(RSO)
Marine and Heavy Engineering Company, MALAYSIA.,
India,
Joined Sep 2014
106
16:19 Oct-16-2014
In Reply to John Porter at 00:08 Oct-16-2014 (Opening).

Dear Sir,

The third problem can also be explained simply like this:

As i said earlier, the dose rate constant is 5.5 x 10-2 mSv m/hGBq. Now distance of 100cm is nothing but 1 meter. So, if for 1 GBq the dose is 5.5 x 10-2 mSv/hr, then for 400 GBq the dose rate shall be 400 x 5.5 = 22 mSv/hr.

Regards
Rengaraju. B. N.

Uli Mletzko
R & D, Retired
Germany, Joined Nov 1998, 89

Uli Mletzko

R & D, Retired
Germany,
Joined Nov 1998
89
21:45 Oct-16-2014
In Reply to RENGARAJU. B. N. at 16:05 Oct-16-2014 .

Rengaraju,

in your mail you are writing:

... In Problem No: 3, dose rate constant is given as 5.5 x 10-2 mSv m2/hGB. There is an error in this as dose rate constant is related to distance and not area. So, i feel it should be m/hGB. ...

In my opinion this comment might contain an error.

Imagine, that you have a small source, which is radiating homogenous into all directions. Then, if you have a sphere of a certain radius = distance around the source, then at all points of the surface of that sphere you will measure the same radiation dose.

If you are taking the double distance, the surface of the new spere will be four times of the surface of the first spere. Therefore you will not measure half of the first dose rate, but only one quarter of the first dose rate. If you are using half of the original distance, you will not have two times of the first dose rate, but four times !!! etc.

This basic law of correlation between radiation dose rates and distances is known as 'inverse distance square law'.

Therefore in all those calculations you have to use the square of the distance, and the 'm2' in the mail of John is O.K.

Regards
Uli

RENGARAJU. B. N.
NDT Inspector, NDT Level-III/Radiation Safety Officer(RSO)
Marine and Heavy Engineering Company, MALAYSIA., India, Joined Sep 2014, 106

RENGARAJU. B. N.

NDT Inspector, NDT Level-III/Radiation Safety Officer(RSO)
Marine and Heavy Engineering Company, MALAYSIA.,
India,
Joined Sep 2014
106
06:26 Oct-18-2014
In Reply to Uli Mletzko at 21:45 Oct-16-2014 .

Dear Sir,

The dose rate constant, by definition, is amount of dose (mSv/hr) emitted by a source per unit of Strength or Activity (GBq) at a unit distance or length(meter or inches or foot etc)

So the unit for dose rate constant is mSV/hr/GBq at a unit distance or length which can be in meters or inches or any other unit of length. It cannot be m2.

As respect with explanation regarding homogeneous source and sphere, please note, radius is measured from the centre which is again length or distance where units can be m and not m2.

Inverse square law shall be applicable only when you want to know the dose at a distance different from the distance at which the dose rate constant is specified.

Hence, I still stand by observation regarding the error in the 3rd question of John.

Rengaraju. B. N.

Frank Lund
R & D,
United Kingdom, Joined Apr 2005, 222

Frank Lund

R & D,
United Kingdom,
Joined Apr 2005
222
13:47 Oct-18-2014
In Reply to RENGARAJU. B. N. at 06:26 Oct-18-2014 .

At any given distance from any point source, the intensity of undirected radiation will have a constant value at any point on the surface of the imaginary sphere having its centre coincident with that point source and its radius equal to that given distance.

For any given solid angle of that sphere, the number of photons will be constant for that solid angle regardless of the distance at which it intersects the surface of the imaginary sphere.

Given that this number will of photons is constant within this solid angle but passing though an area which is a function of the square of radius at which this area on the surface of the sphere is calculated, the flux density will be an inverse function of the square of the radius.

Simples!

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