# Gamma-ray atoms displacement

1. Nov 5, 2014

### ch3m

Hi everyone,

For one of my PhD project I am trying to calculate the minimum energy required to displace a carbon atoms with gamma rays.

Knowing the displacement energy (Ed) for C in diamond (30 -40 eV), I've managed to calculate the minimum energy required using electron with the formula:

Ed=2(E+2mc^2)E/Mc^2

Where E is the energy of the electron beam in KeV, m is the rest mass of the electron and M is the carbon mass.

There is any simple way to calculate the same result for gamma?

Sorry for the stupid question but being a chemist I struggle a bit with this kind of stuff.

2. Nov 6, 2014

### Staff: Mentor

You can use momentum conservation. Those 30-40 eV correspond to some momentum, the gamma ray has to provide at least half that value (the other half is gained from re-emission). A reasonable cross-section will need a higher energy, however.

3. Nov 10, 2014

### ch3m

Hi mfb
Thanks for the hint.
Reading couple of paper I found out that for the kind of gamma I am using (Co60) most of the lattice damage will be done by Compton scattering. As a result of that the relation between the energy of the gamma and the recoil electron should be:
E=Eg*(1-cosθ)/[(Ee/Eg)+(1-cosθ)]

Computing the equation, considering Eg=1.48MeV, Ee=0.511MeV and the scattering angle θ=π to get the maximum energy achievable, I obtain an energy for the recoil electron of 1.26MeV which to me sound a bit too much.

Last edited: Nov 10, 2014
4. Nov 10, 2014

### e.bar.goum

That's the right answer! For a gamma of 1.48 MeV, anyway. Co60 has gammas at 1.173 and 1.332 MeV. Where did you get 1.48?

Also, this is the energy of the scattered photon, the energy of the electron is the energy of the photopeak less this energy

If you look at a compton scattering spectrum for 60Co, shown here: http://en.wikipedia.org/wiki/Gamma_spectroscopy#mediaviewer/File:Co60_Spectrum.JPG

You'll see a sharp cutoff at channel ~580 - that's the energy associated with θ=π.

ETA: But! Is this really what you want to know? This calculates the energy imparted to the electron, not the nucleus, which is what I thought you wanted from your OP?

5. Nov 10, 2014

### ch3m

Hi e,bar.goum

I did an error writing up the energy on excel. I do not know 1.48MeV comes from.

"this is the energy of the scattered photon, the energy of the electron is the energy of the photopeak less this energy"
So the energy that I calculated will be the energy of the gamma photon after the scattering not the one acquired by the electron.

From the info that I found on some paper the gamma can displaced atoms via the interaction with the electron in the solid. Am I wrong?

Here the line quoted from an article:
"Gamma rays can displace atoms by first transferring energy to an electron, which transfer energy to a lattice atoms through and electron-atom scattering event." - Kwon, J., & Motta, A. T. (2000). Gamma displacement cross-sections in various materials. Annals of Nuclear Energy, 27(18), 1627–1642. doi:10.1016/S0306-4549(00)00024-4

6. Nov 10, 2014

### Staff: Mentor

Yes this is possible. Then you'll first need the maximal electron energy, and consider the electron/nucleus collision afterwards.

7. Nov 11, 2014

Thanks guys!