How Do You Calculate Radiation Count Rate for Different Energies?

[v_pino]

Homework Statement

I've attached the problem sheet and my attempt at the problem. I've completed all the parts except for the last part, where I have to calculate the count rate of the radioactive source. The answers should be 6000 counts for 185MeV and 4650 counts for 1MeV.

I'm not sure if it's simply 1000x6 for the 185MeV one. And I couldn't get the right answer for the 1MeV one.

Thank you.

Homework Equations

The Attempt at a Solution

 

[mfb]

Those 185keV-photons are absorbed quickly - the depth of the material (as seen by the detector) does not matter, only the surface area is relevant (this is different from your result). You can just multiply the number with 6 to get a good approximation.
If all photons would penetrate the material, the orientation of the bar would not matter.

For the 1MeV-photons, you have to take the depth into account: For an emission x cm away from the air (towards the detector), how many photons reach the detector? Can you integrate this over the whole material?

 

[v_pino]

Doesn't the mass attenuation coefficient take into account of a 3D source? Should I use it in the equation I=I_0 exp(-mu * x) ? Can I say that the final intensity I for case (A) is 1000? Then use this to find the initial intensity in case (A). Then assuming the initial intensity is same in (A) and (B), use it to find the final intensity in (B)? Using this method, I get 21600 counts which isn't correct.

 

[mfb]

Doesn't the mass attenuation coefficient take into account of a 3D source?

What do mean with that? I don't find an interpretation where the answer is "yes".

Should I use it in the equation I=I_0 exp(-mu * x) ?

That is the point of this coefficient.

Can I say that the final intensity I for case (A) is 1000?

Sure, that is given. That is not the I in your formula, however, as you don't have a constant x for all regions of the source.