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

In summary, the conversation discusses determining the half value thickness, X(1/2), for lead using the linear absorption coefficient u and the equation I = Io e^-ux. It is clarified that I is an energy, not an intensity, and u is normally rated in the units of "per cm". The correct method is to set I to 0.5 and Io to 1, and then solve for X(1/2) using the equation ln(I/Io) = -ux. The final answer is 0.436cm.
  • #1
foxandthehen
13
0
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?
 
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  • #2
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 [tex]\mu[/tex], it's normally a bulk absorption ie cm2/g
 
  • #3
mgb_phys said:
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 [tex]\mu[/tex], 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?
 
  • #4
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
 
  • #5
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!
 

What is the half value thickness for lead?

The half value thickness for lead is the amount of lead that is needed to reduce the intensity of a particular type of radiation by half.

Why is the half value thickness for lead important?

The half value thickness for lead is important because it is used to measure the effectiveness of lead as a shielding material against radiation. It helps determine the amount of lead needed to protect individuals from harmful radiation exposure.

How is the half value thickness for lead calculated?

The half value thickness for lead is calculated using the logarithmic equation: HVT = 0.693/μ, where HVT is the half value thickness and μ is the linear attenuation coefficient of lead for a specific type of radiation.

What factors can affect the half value thickness for lead?

The half value thickness for lead can be affected by the type of radiation, the energy of the radiation, and the thickness and purity of the lead material. Other factors such as temperature and angle of incidence may also have an impact.

How can the half value thickness for lead be measured?

The half value thickness for lead can be measured using a variety of methods, including direct measurement with a radiation detector, calculation based on the linear attenuation coefficient of lead, or using a half value layer (HVL) phantom and measuring the thickness of lead needed to reduce the radiation intensity by half.

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