
#1
Nov810, 03:58 PM

P: 15

[b]1. If the radiationlevel is 400mGy/hr at 2m from a point radiation source, what will be the radiation level at 6m
[b]2. I figured that if the radiation was 400mGy/hr at 2m it have been 1600mGy/hr when emmitted from the sorce. [b]3. If the inverse square law works like at 2m from the source it covers an area of 4m2 and at 3m it covers an area of 9m2 i figured that at 6m it covers anarea of 36m2 so the radiation would be less by the same proportion? 



#2
Nov910, 09:52 AM

P: 596

Do you want to determine the radiation level 6 m from the source ? If yes, then you no need to find the source strength (just to avoid confusion with units!). May I know how did you get the source strength as 1600 mGy/hr? Any why dont you take pi (=3.14) value in your computation for finding the source strength? 



#3
Nov1010, 07:20 PM

P: 15

Hi,
Thanks for the reply Yes i want to know the radiation level at 6m The only thing i know is the level is 400mGy/hr at 2m from a point radiation source. I do not know how to calculatate the rest. I thought 400mGy/hr at 2m might be a quarter of the original strength?? 



#4
Nov1110, 06:25 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

xradiation inverse square law
hi nonphysical!
and anyway you don't need to find the source strength as Rajini points out, the surface area of a sphere is 4πr^{2}, not r^{2} … but the important point is the ratio … all you need say is, if you double the radius, you quadruple the area, so if you triple the radius (3 = 6/2), then … ? 



#5
Nov1110, 09:38 AM

P: 596

Apply inverse square law first to 2 m distance. From this you get the source strength. You no need to compute and get numerical value..just substitute in the inverse square formula with their units. Then for 6 m distance again using inverse square law substitute for source strength that you obtained for 2 m and get your answer in mG/hr. 



#6
Nov1110, 11:40 PM

P: 15

Thanks TinyTim,
i think i'm getting closer 400mGy/hr at 2 meters should still be 400mGy/hr at 6m? then 400mGy/hr divided by the square of the distance ie 6m is that how it works? 



#7
Nov1210, 01:47 AM

Sci Advisor
HW Helper
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P: 26,167

hi nonphysical!
(just got up …) it isn't, it's only a measure of the intensity at the receiver you have to divide by the ratio of the squares of the distances. ie not by 6^{2}, but by (6/2)^{2} (btw, the "original strength" would be the intensity at distance r times 4πr^{2} … 



#8
Nov1210, 02:16 AM

P: 596





#9
Nov1410, 07:25 PM

P: 15

Thanks again TinyTim/Rajani
Firstly Rajani, no i'm lost i cannot write the inverse square law formula. Secondly, TinyTim again i'm lost in the meaning of (6/2)2 if you could put it more simply maybe i could grasp it and go from there. Thanks again 



#10
Nov1510, 04:01 AM

P: 596

see this link http://en.wikipedia.org/wiki/Inversesquare_law In that link look into the first example, which starts as: 'Let the total power radiated from a point source...' read it and use that formula to obtain source strength using 2 m, and then substitute the source strength into the same formula to derive for 6 m. 



#11
Nov1610, 04:13 PM

P: 15

rajini,
I've googled the inverse square law prior to asking these questions, and if i fully understood it i would not be asking the question. as i understand it so far the radiation strength at D1 is squared. ie 400mGy/hr @ 2m = 1600mGy/hr Then the distance at D2 is squared ie 6m = 36 The result of the square of the strength at D1 (1600 mGy/hr) is divided by the square of D2 Example 400mGy/hr @ 2m = 1600mGy/hr divided by 36 = 44.44mGy/hr I cannot write all the physics symbols, so i have to write in long hand. Thanks again 



#12
Nov1710, 04:30 AM

P: 596

Hi,
I guess your answer is correct. How i usually derive: For 2 m: [tex] 400 mGy/hr=\frac{S}{4\pi(2m^2)}\rightarrow S=(400 mGy/hr) 4\pi(2 m)^2 [/tex] For 6 m: [tex]?mG/hr=\frac{S}{4\pi(6m^2)}[/tex] Substitute 'S' from 1st equation into 2nd equation to obtain '?'. If you want to find the source strength take care of units! 



#13
Dec2110, 01:09 PM

P: 15

Hi Rajani
Thanks for all of your help May you have a very merry christmas and a healthy and prosperous 2011 cheers. 


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