In summary, the conversation discusses the calculation of radiation exposure for a 2 Ci 137Cs source, taking into account factors such as lead shielding and distance from the source. The equations used include the inverse square law, linear attenuation coefficient, and specific gamma ray constant. The conversation also mentions the use of a .85 factor when detecting 660 keV photons for calculating activity. However, it is noted that this factor may not be appropriate for calculating exposure rate.

## Homework Statement

I have attached a picture of the question.

A 2 Ci 137Cs source is equal to 2 x (3.7x1010 Bq) = 7.4x1010 s-1

8 hours a day, 5 days a week and for 50 weeks is a total duration in seconds of 7.2x106 s

An 8 cm lead shield = 0.08 m

IRR99 regulations set an effective dose limit as 20 millisieverts per year. Therefore, 10% of of their effective dose limit is 2 millisieverts = 0.02 m2s-2

None

## The Attempt at a Solution

[7.4x1010 s-1 x 0.08 m] / 7.2x106 s = 0.02 m2s-2 - d

d = 2.4x10-5 m

I jiggle the equation about in different ways but keep getting small values for d.

#### Attachments

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How does your calculation take into account that there is a lead door, as opposed to some other kind of door, between the source and the worker?
How does your calculation take into account that the farther away from the source, the less radiation the worker receives?

Hi,
There are a lot of confusion in your calculation :
Do you know the inverse square rule ? The exponential attenuation ? The intensity of emission (85% for Cs-137) ...
PSR

A 2 Ci 137Cs source is equal to 2 x (3.7x1010 Bq) = 7.4x1010 s-1

You need to find the the specific gamma rate constant/dose rate constant which relates the dose rate (Sv/hr) at a specified distance to the activity(Bq) of the source which a published for radionuclides and sources made from them.

I haven't a clue to be honest guys as this isn't in my lecture notes.

I've got a value of 3.7x107 Bq/hr (source strength?)

The inverse square law is I = S/4 pie x r2

I don't know how to work out a value for I (intensity at surface of sphere).

Last edited:
I've got a value of 3.7x107 Bq/hr (source strength?)

You can have any source with a specific decay rate but the radiation dose rate at a specified distance depends on the isotope. Each isotope has its own specific gamma ray constant and each particular source made from this isotope may have that significantly modified by the encapsulation material due to attenuation of that material.

What are the equations I need?

I have a dose constant of 0.38148 Rem/hr for 137Cs at a distance of 1m from a 1 point curie source.

Hi,
A small, help,
For Cs137 you have the specific dose rate 84 microSv/h/GBq at one meter
For 2 Ci it gives 6216 microSv/h
Your worker works 2000 h and your objective is 2mSv per year so your objective dose rate is 1 microSv/h
The attenuation coefficient is for lead and 660 keV 1.4 /cm
With that you must conclude
I have 30 cm
PSR

PSRB191921 said:
The intensity of emission (85% for Cs-137) ..

gleem said:
It is a data you can get here:
http://www.nucleide.org/Laraweb/index.php
It means that for 1000 Bq of Cs137 850 photons per seconds are emitted with an energy of 662 keV

How do you get a value for the linear attenuation coefficient for lead, anyway?

I'm close to giving up because I don't know what equations I need / what order to do this question in. Do I use I = I0e-ux then the inverse square law? I have no idea how to approach this question.

Yes, first you use the exponential to find how much radiation makes it through the door, then you use the inverse square law to find how much distance you need to put between the worker and the door to get the desired overall attenuation factor. Look at post #8 for the linear attenuation coefficient for lead.

I = I0e(-ux)

I = (7.4x10-10)e(-1.4x0.08)
I = 6.61x10-10 s-1
I = 1.83x10-13 /hr

I = S/4 pie r2
r = square root [(10-6)/(4pie)(1.83x10-13)]
r = 659

lmao I don't know what I am doing. I'm going to quit the course.

Thanks for the help.

PSRB191921 said:
It is a data you can get here:
http://www.nucleide.org/Laraweb/index.php
It means that for 1000 Bq of Cs137 850 photons per seconds are emitted with an energy of 662 keV

The use of the .85 factor is appropriate when you are detecting the 660 keV photons and are trying to determine the activity of the source. It is not appropriate when we are interested in the exposure rate of a given source activity.

gleem said:
The use of the .85 factor is appropriate when you are detecting the 660 keV photons and are trying to determine the activity of the source. It is not appropriate when we are interested in the exposure rate of a given source activity.
Of course it is !
The dose for Cs137 is due to the 662 keV
You know that
H=1.5E-10×A×i×E (in mSv/h)
Ain Bq , E in MeV and i=.85
See, for example chapter 4 of http://www.springer.com/us/book/9783319486581

PSRB191921 said:
Of course it is !
Only if you are calculating the exposure rate constant from the spectrum of photons emitted. The exposure rate constants quoted in the literature have taken that fact into account or they are measured in which case it is irrelevant.

Of course, but it is interessting to know how to calculate this constant dose rate from a source flux, no ?
It is useful to know with a activity to calculate flux, fluence at a distance d, and dose equivalent with the conversion factor of the ICRP. But it is only my opinion

= (7.4x10-10)e(-1.4x0.08)
The attenuation coefficient is in 1.4 cm-1 so the door thickness should be expressed in centimeters not meters. This should change your answer significantly.

Using the thickness in cm does change the answer but it's still wrong by a lot. Thanks for the help.

I need someone to type out the solution as it's the only way I can understand this question. I don't know what I'm doing and have wasted too much time on this one question.

Using the following data

dose at occupied area = 2 mSv
attenuation coefficient for lead and Cs-137 gamma rays = 1.24 cm ( gives a Half Value Layer of 0.56 cm the published value)
shielding 8 cm lead.giving an attenuation of 0.492 x10-4
Dose rate constant = 0.003815 Sv/Ci/hr at 1 meter
time of occupation = 2000 hrs
Source activity = 2 Ci

I get 61.3 cm

Thank you.

Can anyone else please type out their worked solution? This type of question might come up in my exam and I still don't understand how to do it. I need to know how to do it step by step.

Sorry, forum rules expressly prohibit what you ask. You will find them here if you have not already seen them.
Under item 8:
"Complete solutions can be provided to a questioner after the questioner has arrived at a correct solution. If the questioner has not produced a correct solution, complete solutions are not permitted, whether or not an attempt has been made."

Read post #20 by @gleem and see if you can reproduce the numbers show.

Hi PSR, why is 84 divided by GBq in the first line? Is it a given value in a textbook?
PSRB191921 said:
Hi,
A small, help,
For Cs137 you have the specific dose rate 84 microSv/h/GBq at one meter
For 2 Ci it gives 6216 microSv/h
Your worker works 2000 h and your objective is 2mSv per year so your objective dose rate is 1 microSv/h
The attenuation coefficient is for lead and 660 keV 1.4 /cm
With that you must conclude
I have 30 cm
PSR

Could you tell me the website / book which gives the specific dose rate (84sv/h/GBq). Thank you

See my post 15
"The dose for Cs137 is due to the 662 keV
You know that H=1.5E-10×A×i×E (in mSv/h) with A in Bq , E in MeV and i=.85 See, for example chapter 4 of
http://www.springer.com/us/book/9783319486581 "
84 microSv/h/GBq means 84 microSv/h for 1 GBq at one meter

How do I take a reading for a value of u (in cm) for 662 kev from this data?
First you need to convert 662 keV to MeV in the format 6.62×10y MeV. (What should y be?). Then you find the closest value in the Table and interpolate.
Be sure to read how to use the value that you found here.
https://physics.nist.gov/PhysRefData/XrayMassCoef/chap2.html

For 6.62x10-1 the closest value is 6.00000E-01, which gives a reading of 1.248E-01 for u/p. Not sure what to do now.

Right so 6.00000E-01 cm2g-1 x 11.34 gcm-3 gives ~ 1.4 cm-1 for u. When I put all of this into my formula I get 28.98cm which is apporoximately 30cm. I'm satisfied that this is the correct answer unless anyone thinks my value of u is incorrect?

## 1. What is radiation dose calculation?

Radiation dose calculation is the process of determining the amount of radiation exposure a person or object has received. It involves measuring the radiation dose in a specific unit, such as gray (Gy) or sievert (Sv), and taking into account factors such as the type of radiation, duration of exposure, and the sensitivity of the body or material being exposed.

## 2. Why is radiation dose calculation important?

Radiation dose calculation is important because it helps us understand the potential health risks associated with exposure to radiation. By accurately calculating the dose, we can determine if a person or object has received a safe or unsafe level of radiation and take appropriate measures to protect their health and safety.

## 3. How is radiation dose calculated?

Radiation dose can be calculated using various methods, depending on the type of radiation and the purpose of the calculation. Some common methods include using dosimeters, which are devices that measure radiation exposure, or using mathematical models to estimate dose based on factors such as distance from the source and shielding.

## 4. What factors can affect radiation dose calculation?

There are several factors that can affect radiation dose calculation, including the type of radiation, the energy of the radiation, the duration of exposure, and the sensitivity of the body or material being exposed. Other factors such as shielding, distance from the source, and the presence of other materials can also impact the calculated dose.

## 5. How is radiation dose calculation used in different fields?

Radiation dose calculation is used in a variety of fields, including medicine, nuclear power, and industrial settings. In medicine, it is used to determine the appropriate dosage for radiation therapy and to monitor the exposure of patients and healthcare workers. In nuclear power, it is used to ensure the safety of workers and the public. In industrial settings, it is used to monitor and control radiation exposure for workers and the environment.

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