Radiation Levels: Inverse Square Law Outcome at 6m

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Homework Help Overview

The discussion revolves around the application of the inverse square law in the context of radiation levels from a point source. The original poster presents a scenario where the radiation level is measured at a specific distance and seeks to determine the level at a greater distance.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the implications of the inverse square law, questioning how radiation levels change with distance. There are attempts to clarify the relationship between distance and radiation intensity, with some participants suggesting that the radiation level decreases as distance increases.

Discussion Status

The discussion is active, with participants providing insights and questioning the assumptions made about radiation levels at different distances. Some guidance is offered regarding the interpretation of radiation intensity and area coverage, but no consensus has been reached on the specific calculations or implications.

Contextual Notes

There are discussions about the limitations of applying the inverse square law directly at the point of radiation and the need for clarification on how radiation dose rates relate to area coverage at varying distances. Participants also raise questions about the necessity of calculating shielding for x-radiation based on dose rates.

nonphysical
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Using the inverse square law could anyone tell me what the outcome of this problem is?
If the radiation level is 400mGy/hr at 2m from the point radiation source, what will be the radiation level at 6m?
I figured it would be 1600mGy/hr at the point, and therefore 44.4mGy/hr at 6m
being distance squared of 6m

Thanks
 
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Not quite. You can't use the inverse square law to find the radiation level AT the point, because the distance would be zero, and therefore the radiation infinite. Notice however that 6m is 3 times farther away than 2m... so by what factor should the radiation decrease?
 
Zhermes
Q. If the radiation level is 400mGy/hr at 2m from a point source, what will be the radiation level be at 6m?

Are you saying the radiation level will decrease by a factor of three?

If at the piont of radiation the level is 1600mGy/hr, then at 2m it should represent a factor of four , ie 400mGy/hr at 2m. therefore at 3m should represent a factor of nine, etc.

Does this mean the radiation dose level drops or just the area increases, or both?
 
nonphysical said:
If at the piont of radiation the level is 1600mGy/hr, then at 2m it should represent a factor of four , ie 400mGy/hr at 2m.
Again, you can't talk about the amount of radiation AT the point.
nonphysical said:
therefore at 3m should represent a factor of nine, etc.
There you go.

nonphysical said:
Does this mean the radiation dose level drops or just the area increases, or both?
Both. The total radiation is constant no matter how far away you are; the area over which that radiation is distributed increases as the distance squared. Therefore the radiation per unit area (which is proportional to the radiation dose) drops as the distance squared.
 
Thanks for the reply

OK so the radiation level at 6m is still 400mGy/hr but covering an area of 36m2

as per 400mGy/hr at 2m covering an area of 4m2

How come the x-radiation shielding HVL system has to be calculated?
Why not just shield for the given dose rate regardless of distance, if the dose remains constant?

If the dose remains the same but proportional given the area, does this therefore mean at 2m the dose rate is 100mGy/hr for each square metre
And the same would apply at a distance of 6m being 36m2 each sqare metre having radiation penetration equivanlent to the proportion of the original dose spread over that area ie 11.11each?
 

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