Particle physics - absorption length

[JamesJames]

1) What percentage of incident photon radiation passes through 5 mm of material whose absorption coefficient is 0.7 mm^-1? What is an absorption length? I am lost here. Any input ...some mathematical formula for the absoprtion coefficient...would be a great help to me.

2) Why is the mean life of the $$\Delta ^{0}$$ only 10^-23 seconds while that of a $$\Lambda ^{0}$$ is 2.6 * 10^-10 seconds. HINT: think strengths and its relationship to decay rates. Now, I can show how to get 10^-23 seconds but cannot explain the reasoning behind the question.

James

 

[xanthym]

1) What percentage of incident photon radiation passes through 5 mm of material whose absorption coefficient is 0.7 mm^-1? What is an absorption length? I am lost here. Any input ...some mathematical formula for the absoprtion coefficient...would be a great help to me.

The fractional change in transmitted radiation T(x) propagating through material having absorption coefficient $$\alpha$$ is governed by:

$$:(1): \ \ \ \ \frac {dT(x)} {T(x)} = -\alpha dx$$

whose solution for incident radiation $$T(x_0) \ at \ x_0$$ is given by:

$$:(2): \ \ \ \ \color{red}T(x) = T(x_0) exp(-\alpha \Delta x) \ \ \ \ \ \ where \ \Delta x = (x - x_0).$$

For this problem, $$\alpha = (0.7 \ mm^{-1}) \ \ and \ \Delta x = (5 \ \ mm)$$.

The absorption length for a given material is generally defined as the propagation distance through which a factor {1 - e^(-1)} of the radiation is absorbed. Thus, incident radiation T(0), upon propagating through the absorption length, will afterwards have intensity T(0)exp(-1).


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[Duarh]

2) Why is the mean life of the $$\Delta ^{0}$$ only 10^-23 seconds while that of a $$\Lambda ^{0}$$ is 2.6 * 10^-10 seconds. HINT: think strengths and its relationship to decay rates. Now, I can show how to get 10^-23 seconds but cannot explain the reasoning behind the question.
James

The principal decays of the $$\Lambda^{0}$$ are p$$\pi^{-}$$ and n$$\pi^{0}$$. All of these decay products contain only u & d quarks. The principal decay of the $$\Delta^{0}$$ is N$$\pi$$. This is also just u & d quarks. The $$\Lambda$$ contains a strange quark, while the $$\Delta$$ is only u's & d's. So there's a cross-gen interaction involved in the $$\Lambda$$ decays. Now what is the force responsible for cross-gen interactions called? The name suggests why that decay is much less likely.

 

[JamesJames]

Isn' t it the weak force?

 

[Duarh]

Yup - and the weak force is many orders of magnitude weaker than the strong, so weak interactions are waaaay less likely (for rough strength breakup see p. 55 of Griffith's book)

 

[JamesJames]

One question about the absorption length. In class out professor ahd done that exact thing which you did for the radiation length. Does the same thing (THE SAME SOLUTION) apply for absorption length?

James

 

[xanthym]

Yes, radiation length is usually equivalent to absorption length. Both refer to the propagation distance in a given material through which the radiation loses a factor {1-e^(-1)}=(63%) of its incident intensity.


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[JamesJames]

Thanks for all the help. It helped me a lot. Now I have one more question. We got a Weiszacker binding energy curve and two regions were identified as being fission (right side) and fusion (left side). I don' t see how this happens...i.e. why is fission on the right and fusion on the left?

James

 

[Duarh]

I think it's that in the middle of the curve you've got the tighest bonding (lowest potential energy), so that's where all nuclei would "like" to be - to the right of that point, decreasing mass number will yield a lower potential energy, while to the left of that point increasing mass number will yield lower potential energy. A reaction will yield net energy to the environs when going from higher to lower potential energy, so on the left you (the environment) can "gain" energy by fusion, on the right by fission, hence the names of the regions.

Don't take me 100% on my word; I'm taking my first particles class at the moment myself (:) I do have the advantage of taking it from Griffiths himself, though). But I think this is right.