Solve Gamma Ray Scattering from 198Au Incident on Al Slab

[nerdgirl]

Homework Statement

A gamma ray from ¹⁹⁸Au incident normally on a thick aluminum slab travels 1.45 cm before it undergoes a Compton scattering. Calculate the range of angles through which the gamma could scatter and still have at least a 10% probability of re-emerging from the incident slab surface without undergoing another interaction.

Homework Equations

the mass attenuation coefficient for Al at 1.5 cm is .05 MeV
E_$$\gamma$$' = E_y / [1 + (E_$$\gamma$$/E₀) (1-cos$$\vartheta$$)

The Attempt at a Solution

IIn the chart of the nuclides do I use one of the gamma energies or is this even the first formula I should be using?

 

[Jhero]

Hello,

Thank you for your question. To calculate the range of angles through which the gamma ray could scatter and still have at least a 10% probability of re-emerging from the incident slab surface without undergoing another interaction, we can use the formula for Compton scattering:

E\gamma' = Ey / [1 + (E\gamma/E0) (1-cos\vartheta)]

Where E\gamma' is the energy of the scattered gamma ray, Ey is the energy of the incident gamma ray, E\gamma is the energy of the scattered gamma ray, E0 is the rest mass energy of the electron, and \vartheta is the scattering angle.

To calculate the range of angles, we can rearrange the formula to solve for \vartheta:

\vartheta = arccos[(Ey - E\gamma') / (E\gamma/E0 - 1)]

We know that the energy of the incident gamma ray, Ey, is equal to the energy of the scattered gamma ray plus the energy of the recoil electron:

Ey = E\gamma' + E_e

And we also know that the energy of the recoil electron, E_e, is given by:

E_e = E\gamma - E\gamma'

Substituting these equations into our formula for \vartheta, we get:

\vartheta = arccos[(Ey - E\gamma') / (E\gamma/E0 - 1)] = arccos[(E\gamma - E\gamma') / (E\gamma/E0)]

Now, to calculate the range of angles, we need to consider the minimum and maximum values for E\gamma' and E\gamma. The minimum value for E\gamma' is 0, as the scattered gamma ray must have some energy to be able to re-emerge from the slab without undergoing another interaction. The maximum value for E\gamma' is Ey, as the scattered gamma ray can't have more energy than the incident gamma ray.

The minimum value for E\gamma is the energy of the incident gamma ray, and the maximum value is the energy of the incident gamma ray plus the energy of the recoil electron. Therefore, we can rewrite our formula for \vartheta as:

\vartheta = arccos[(Ey - E\gamma') / (E\gamma/E0 - 1)] = arccos[(E\gamma - E\gamma') / (Ey/E0)]

Now