x-radiation inverse square law


by nonphysical
Tags: inverse, square, xradiation
nonphysical
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#1
Nov8-10, 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?
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Rajini
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#2
Nov9-10, 09:52 AM
P: 597
Quote Quote by nonphysical View Post
[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?
Hi,

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?
nonphysical
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#3
Nov10-10, 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??

tiny-tim
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#4
Nov11-10, 06:25 AM
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x-radiation inverse square law


hi nonphysical!
Quote Quote by nonphysical View Post
2. I figured that if the radiation was 400mGy/hr at 2m it have been 1600mGy/hr when emmitted from the sorce.
i'm sorry, this is nonsense

and anyway you don't need to find the source strength
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?
(try using the X2 icon just above the Reply box )

as Rajini points out, the surface area of a sphere is 4πr2, not r2

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 ?
Rajini
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#5
Nov11-10, 09:38 AM
P: 597
Quote Quote by nonphysical View Post
Hi,
Thanks for the reply
Yes i want to know the radiation level at 6m
Okay 6 m away from source
The only thing i know is the level is 400mGy/hr at 2m from a point radiation source.
this is already given !
I do not know how to calculatate the rest.
I thought 400mGy/hr at 2m might be a quarter of the original strength??
No!
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.
nonphysical
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#6
Nov11-10, 11:40 PM
P: 15
Thanks Tiny-Tim,
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?
tiny-tim
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#7
Nov12-10, 01:47 AM
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hi nonphysical!

(just got up )
Quote Quote by nonphysical View Post
400mGy/hr at 2 meters should still be 400mGy/hr at 6m?
you're still thinking of mGy/hr as a measure of the source (the "original strength")

it isn't, it's only a measure of the intensity at the receiver
then 400mGy/hr divided by the square of the distance ie 6m
is that how it works?
not exactly

you have to divide by the ratio of the squares of the distances.

ie not by 62, but by (6/2)2
(btw, the "original strength" would be the intensity at distance r times 4πr2

that is a constant for any value of r )
Rajini
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#8
Nov12-10, 02:16 AM
P: 597
Quote Quote by nonphysical View Post
Thanks Tiny-Tim,
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?
can you write the inverse square law formula?
nonphysical
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#9
Nov14-10, 07:25 PM
P: 15
Thanks again Tiny-Tim/Rajani
Firstly Rajani, no i'm lost i cannot write the inverse square law formula.

Secondly, Tiny-Tim 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
Rajini
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#10
Nov15-10, 04:01 AM
P: 597
Quote Quote by nonphysical View Post
Thanks again Tiny-Tim/Rajani
Firstly Rajani, no i'm lost i cannot write the inverse square law formula.
Why you are not showing any interest in finding the formula? If you have access to internet you can easily google with 'inverse square law'..In fact tim already gave good ints for you and please understand what a inverse square law means.
see this link
http://en.wikipedia.org/wiki/Inverse-square_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.
nonphysical
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#11
Nov16-10, 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
Rajini
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#12
Nov17-10, 04:30 AM
P: 597
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!
nonphysical
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#13
Dec21-10, 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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