Finding Sphere Diameter for Radiation Equilibrium?

kmoh111
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Homework Statement


This problem comes from Frank Attix's book Intro to Radiological Physics and Rad Dosimetry.
From ch. 4 - problem #1:

Approximately what diameter for a sphere of water would be required to approach radiation equilibrium wihin 1% at its center, assuming it contains a uniform dilute solution of 60Cobalt (1.25 MeV gamma rays). Use u(en) and u as approximations to the effective gamma ray attenuation coefficient; this will over- and under-estimate the size respectively.



Homework Equations



u - is the attenuation coefficient for primary photons only
u(en) - is the effective attenuation coefficient for ideal broad beam attenuation


Attix gives the solutions as 312 cm and 144 cm


The Attempt at a Solution



I come close to one of the answers but I'm not sure how u and u(en) come into play here.

(4/3) pi r^3 = .01(1.25MeV).

This gives r = .144
 
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This thread can be disregarded. I was able to solve the problem.

The solution to this problem is to use exponential attenuation:

N = N(0) exp(-uL). The approach is to solve for L for given u and u(en) values when 99% of the photons are allowed to pass thru the sphere.
 
kmoh111 said:
This thread can be disregarded. I was able to solve the problem.

The solution to this problem is to use exponential attenuation:

N = N(0) exp(-uL). The approach is to solve for L for given u and u(en) values when 99% of the photons are allowed to pass thru the sphere.





hi..one of my classmates has asked about this problem too, i was thinking of this solution too, but i can't find the real values for u(en) and u...

i would like to ask, where can i find those values, i keep searching the net but gave me none... hope you would help us.

thanks :)

andy from philippines

God bless
 
thank you so much kmoh111

God bless :)
 
hi me again :(

ive tried to solve the problem using exponential attenuation but i always got a different answer...however i tried to use equation 3.14 (build up factor) in attix book... where B is approximately equal to 1.06... here's how it went..

0.99 = (1.06) exp(-uL) and 0.99 = (1.06) exp(u(en)L)

i used u = 0.46903 and u(en)= 0.216626 (this is base from the link that you've given me before)

but i got, L = 0.145 cm and L = 0.313 cm :( Attix gives the solutions as 312 cm and 144 cm



did i lack something in my equation? :(

pls help us...any comments will be very much appreciated

million thanks..


andy :(
 
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To solve this, I first used the units to work out that a= m* a/m, i.e. t=z/λ. This would allow you to determine the time duration within an interval section by section and then add this to the previous ones to obtain the age of the respective layer. However, this would require a constant thickness per year for each interval. However, since this is most likely not the case, my next consideration was that the age must be the integral of a 1/λ(z) function, which I cannot model.

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