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Technical Discussions
07:47 Mar-03-2012

John M. Champion
Other, UT/AUT Scanner Technician,
International Union of Operating Engineers; (NDT) Local #112 ,
Joined Apr 2009
state card exam
Dear Colleagues,

I am preparing to take my state card exam in Radiation safety and I am have a bit of difficulty with several of the practice exam questions dealing with the amount of radiation you will receive at a given distance with a given Ci amount. I have attempted to utilize the inverse square law to answer the questions, but I can not seem to match the answer key answers. I have attempted different ways to match the answer key answers with no success. can anyone detail the proper procedure for getting the correct answers please. Thank you for your time and assistance.

the first questions is thus:
1) How much radation will you recieve if you are 20 ft. from a 100 Ci IR192 source for 30 seconds?
Using the information from my textbook for the constant for Iridium 192 source @ .3m (1ft.) = 5.2R/h. 100Ci x 5.2 R/h (IR192 constant) = 520 R/h @ 1ft. ( I 1) ; 20ft (D 2)2 =400 ft.2 (squared); (D 1)2 (squared) = 1ft. so, 520 (I1)x 1[(D1)2] / 400 [(D2)2] = 1.3 (R/h) (I2) / 60 minutes in an hour = 0.021666 or 22mR/minute = 11mR/ 1/2 minute (30 seconds) correct? The answer in the answer key is 12.3mR what am I doing incorrectly? I am aware that the IR192 constant is sometimes 5.9R/h, but here it is 5.2.

2) Similar question:
What is the total amount of radiation that you will receive if you are 50ft. from a 50Ci (curie) IR192 source for 5 minutes and 40ft. from a 1Ci CO60 (Cobalt-60) source for 30 seconds? (answer key answers are 5.9mR and 0.075mR)

I used the same I1/I2 = (D2)2/(D1)2 inverse square law to do the same calculation as above, but it did not match answer key. How do I properly calculate these "problems"

and the last question:
3) How long could you be at 10ft. from a 50Ci IR192 source and receive a maximum of 200mR? (answer key answer is 0.067hours)
These have me stumped...I've tried to look at it a few times, but cannot seem to come close to the answers given...Can anybody help me?
Thank you...I appreciate your help in advance...
11:07 Mar-04-2012


Engineering, QA/QC/NDT,
NOV (National Oilwell Varco),
Joined Mar 2006
Re: state card exam
In Reply to John M. Champion on 07:47 Mar-03-2012 (Opening).

Hi John,

Ans - 1: 10.9 mR/30 sec
Ans - 2: Ir 192 8.33 mR/5min, Co60 0.073 mR/30 sec
Ans - 3: 0.07 hrs.

I used this formula: Radiation Level (RL) = Curie * Rhm/ Distance ²
Constant I have used Ir 192 1 Curie @ 1ft 5.2 R/h and Co 60 1 Curie @ 1 ft 14 R/hr


12:27 Mar-04-2012

John M. Champion
Other, UT/AUT Scanner Technician,
International Union of Operating Engineers; (NDT) Local #112 ,
Joined Apr 2009
Re: state card exam
In Reply to S.Senthilkumar on 11:07 Mar-04-2012 .

Thank you for your reply Mr. Senthilkumar,
I appreciate you taking the time to try to assist me on this, however, the answers in the answer key are as follows:
Ans - 1: 12.3 mR/hr
Ans - 2: Ir192= 5.9 mR; Co60= 0.075 mR
Ans - 3: 0.067 hours ( so yes it is appears you rounded up to 0.07 hours; a difference of .003)
I am a bit confused on your formula. I understand: Curies (multiply by) Rhm (is that roentgen per hour per minute?) divided by (Distance2) squared? The inverse square law correct? Intensity1 (divided by) Intensity2 = Distance2 squared (divided by) Distance1 squared ie. I1 / I2 = (D2)squared / (D1)squared ?

Thank you for your time
Best regards

20:38 Mar-08-2012

Csaba Hollo

RSO, QA Rep,
Acuren Group Inc.,
Joined Feb 2010
Re: state card exam
In Reply to John M. Champion on 12:27 Mar-04-2012 .

First of all, .3m is not a 1ft. Its 0.3048. So your calculations will be a little off.
Are you using 480 mr/hr/m/ci for your IR constant? Try that for your calculations

10:54 Mar-09-2012

John M. Champion
Other, UT/AUT Scanner Technician,
International Union of Operating Engineers; (NDT) Local #112 ,
Joined Apr 2009
Re: state card exam
In Reply to Csaba Hollo on 20:38 Mar-08-2012 .

Hello Csaba Hollo...Thank you for your reply and your time at responding to my request for help. Yes, .3m is not precisly 1ft, however it is the "legend" given for the constants. As to the constanst I am using, again they are the constants "given" in the ASNT Radiation Safety text book...they are as follows: IR192 - 5.2R/hr and .052 Sv/hr for SI unit constant @ .3m (1ft). for Co60 the constants are 14.0R/hr and for SI unit constant 0.14 Sv/hr @ .3m (1ft). It was also explained to me that the "constants" are sometimes referenced as 5.9 R/hr or .059 Sv/hr (IR192) and 14.4 R/hr or 0.144 Sv/hr (Co60). I was instructed that my method was correct; using the Inv.Sq. law, however the questions were presented without reference to the constant being used to calculate the solution. As such, using the constant 5.2 R/hr was the incorrect constant, and after calculating with the constant 5.9 R/hr my calculations were "spot on" for the IR192 problem and 14.4 R/hr for Co60 also gave the correct answer. I was also instructed that I may use the equasion of: Dose (divided by) Dose Rate X Time, when I have the DOSE given, and the TIME, at a certain distance of a specified Activity Level (ie. 50 Ci of IR192).

So (question 3)
How long could you be at 10 ft. from a 50 Ci (IR192) source and receive a maxium dose of 200mR? Using the constant 5.9 R/hr, the constant needs to be changed to mR/hr (5.9 R/hr X 1000 = 5900 mR/hr). 5900 mR/hr X 50 Ci (IR192) = 295,000 mR/hr @ .3m (divided by) 10 ft. Squared = 100ft (2) = 2950 mR/hr . Using the D/ DR X T: [DOSE] is 200 mR (divided by) [DOSE RATE] 2950 mR/hr = 0.0677966101 etc. = 0.067 hours. Now I need to multiply by 60 minutes/hr = 4.067 minutes @ 10ft to receive a dose of 200 mR. I am able to check this by reversing my calculations... 0.067 X 2950 = 199.715 mR...this is not precisly 200 mR , however if it is rounded up it will equaly 200 mR.

Thank you for your time and I hope this helps others.
Best ragards

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