What is the Half Value Thickness for Lead at 0.4 MeV and 1.59 cm^-1?

[foxandthehen]

1.
Determine the half value thickness, X(1/2) for lead where;
I = 0.4 MeV
u = 1.59cm^-1

2.
using;
I = Io e^-ux3.
ln(I/Io) = ln(e) -ux
ln(2) / -1.59x10^-2 = X(1/2)

=> X(1/2) = -1.7523m
Is that right?

 

[mgb_phys]

Firstly I=0,4MeV is an energy not an intensity.
Your 'u' seems to already be a half thickness.
Are you sure of the units of \mu, it's normally a bulk absorption ie cm²/g

 

[foxandthehen]

Firstly I=0,4MeV is an energy not an intensity.
Your 'u' seems to already be a half thickness.
Are you sure of the units of \mu, it's normally a bulk absorption ie cm²/g

Hi, thanks for your answers!

The mU I have stated is the linear absorption coefficient and therefore rated in the units of "per cm" and your right that that's energy, not intensity, but as you would need to do the same things to both the first (I) and the second value (Io) to convert them to intensity (and its one divided by the other), it will always come out as 2 for (I/Io) in order to get the half value thickness, therefore you can make 'I' any value and any units and the answer will be the same, its just the linear absorption coefficient that's important... right? Or perhaps I am missing the point?

 

[mgb_phys]

You can't say I=0.4mev - just checking you weren't confused with something else.
Your method is almost correct

You want the intensity to reduce to 0.5, so I=0.5 if Io=1
so so ln(I/Io) = ln(0.5) = -ux

There's no need to convert into m.
Then you can check your answer by putting it back into the equation and checking you get
0.5 = e ^ -1.59 x

 

[foxandthehen]

ahhh, I see! I was on the whole, the symbol with the 'o' after it is normally lower, where as for this one its higher as it decreases over distance! Must have been a long day as I totally missed that one! Plus, as its 'per cm' I was going the wrong way on converting it to m!

So we have;
X(1/2) = ln(0.5) / (-1.59) = 0.436cm

Thank you!