Radiation Safety UML: Neutron Capture Exposure Rate at 1 Meter

[MattC1606]

Homework Statement

A 1.0g Co(59) target is exposed to a constant fluence rate of thermal neutrons equal to 1E+12 n cm-2 s. Cross section (n,gamma) is 37.18barn and density is 8.9g cm3. What is the initial exposure rate at 1 meter due to neutron capture?

Homework Equations

1R=2.58e-4C/kg
gamma = 7.49MeV

The Attempt at a Solution

I calculated on the order of 10^6 R/hr. Just curious to see if that value is reasonable for 1 gram of Co59 being exposed.

 

[dalexander]

Hey Matt,

I actually got an answer on the order of milli-Roentgen... Where did you get your gamma energy from? I used our old Q-value technique to calculate a gamma energy and got 6.98MeV (French recommended this in class, saying that the Q-value is essentially equal to the gamma energy as the recoil of the Co atom is so small)
n + Co-59 --> Co-60 + Tgamma + Q

 

[dalexander]

While I'm here, I'm having trouble on the third problem part a. The prompt is as follows: Use the mean value theorem to derive the equation for the mean free path of a photon in terms of its linear attenuation coefficient.
We are given the fact that mfp x=$$\frac{1}{\mu}$$ and the mean value theorem: $$\frac{\int f(x)w(x)dx}{\int w(x)dx}$$ with the range a=0, b=$$\infty$$
The professor also said that w(x) is a distribution function.

My problem is figuring out what f(x) and w(x) are. Could w(x) be the n(x) = n(0)e$$^{-\mu x}$$ attenuation equation he gave us in class?

Thanks, guys!

 

[Henry Zalman]

I tried the problem several different ways, mostly with variations on the fluence the e/w-bar.

I finally ended up with an answer on the order of 10^1... don't know if this is correct, but consider this:
200 mR per YEAR is considered acceptable background (per Wikipedia, anyway), and 500 R/hr. for 5 hours is lethal. And Dr. French mentioned that a 1 R exposure is pretty high. So an answer on the order of 10^6 seems high to me, as does the mR answer. Just my two cents worth...

 

[Chad Smith]

Hi Everyone,

As far as problem 1 is concerned - I calculated a photon energy of 7.491 Mev using the change in Binding energy guy for the Q value. I ended up with ~23 R/h...which seems like an awfully large dose rate.

For number 3 - I used V2 thumb 55 as a launch point. The function to integrate in the numerator would be x = 1/u(ln(n0/n))
denomenator would be just dn
Where n is the uncolliding photons in the medium and n0 is the total at start.
Since you are trying to integrate from total number of photons to no photons you integrate from n0 to 0.
Page 55 gets you the answer - from there it is just integration.

 

[Henry Zalman]

Anyone have any guidance on no. 5? Ignoring the geometry for the moment, I'm having issues doing the calculations...
a) do the mass attenuation coefficients get used as is, or do they need to be multiplied by the density? (No units are given)

Does the problem just get treated as an attenuation problem?

 

[dalexander]

Thanks Chad.
I also have an update for my problem 1 result. I forgot to factor in the conversion from C/kg to R, and my gamma energy was off (7.5ish is correct). My new exposure rate is around 20R/hr.

 

[Chad Smith]

Hey guys,
for number 4 I calculated (tediously) 4.85%. Anyone else in that ballpark? I'm going to start number 5 soon and hopefully have something to post.

 

[dalexander]

Hi Chad,
I got the 'opposite' answer (95.11%). I believe the equation for the fraction that passes through the thickness is 1-e^-$$\mu*x$$
Thoughts?

Dan

 

[Chad Smith]

Hi Dan,
I think you might have it backwards...page 56 V2. Your answer is the fraction of uncollided projectiles interfacing with the thickness. Whereas the same equation without the (1-) is the fraction of uncollided projectiles remaining in the beam.

 

[Chad Smith]

Henry,
For number 5 I have completed parts a and b. I am stuck on part c for setting up the mean value theorem. As for part a I ended up multiplying the mu of air by the density of the concrete. Then I multiplied that value by the fluence rate (units of cm^-3/s)
For part b I just used the decay/attentuation function for a distance of 10 cm through the concrete (units of cm^-3/s).
If anyone has any idea for part c, please post.

 

[Henry Zalman]

Chad,

I finally did the same, since it is the MASS attenuation, and Dr. French confirms that (I asked him today).

Part B, did the same as you did.

For part C on 5, see http://www.mathwords.com/m/mean_value_theorem_integrals.htm for some guidance on the MVT.

Good luck

 

[dalexander]

Chad,
You were absolutely right, haha. I had it backwards.
For 5c, you want to integrate Rv over the integral of dV. As the prof said in class, dV= the surface area of the target times dx, the "depth-slice." So the MVT equation becomes:
The integral of [Rv times the area of the face of the slab times dx] divided by the integral of [the area of the face times dx], integrated from 0cm to 10cm (the thickness of which dx is a slice)

For number 6... I believe part a is the same as 5a, but then on part b does the new equation become the (S/4*pi*d^2)*e^(-mu*w) we discussed in class? and if so, how does one get from phi to S (what area or distance squared is factored into the fluence to get the photon production rate)?