Re(2): tritium x-rays -Reply

[FRAMEP@ORAU.GOV]

Rob Gunter wrote:

Hey Rob, let me add a couple of observations here.

>The Radiation Yield (Y) can be calculated using the following 

>Y=(6x10^-4(ZT))/(1+6x10^-4(ZT))

>Where Z is the atomic #; T is the Kinetic E. of the beta in MeV.
>for an average energy of 6keV and a Be encasement, you get:

>Y=(6x10^-4(4*.006)/(1+6x10^-4(4*.006))
> =1.44x10^-5

>Which is the fraction of the 6 keV converted to photons as the Beta
>particle slows down.

True, but this value is for monoenergetic beta/electron particles
whose energy is 6 keV. This does not give the fraction of the H-3
beta energy  converted to photons because tritium betas are not
monoenergetic. Calculating the fraction using the average beta energy
for H-3, as done here, underestimates the average fraction for
tritium because the fraction of the energy converted into
bremsstrahlung increases with the beta energy.

The standard approximation is  ZE/3000 where E is the maximum beta
energy i.e. 0.0186 MeV. See Evan's "The Atomic Nucleus"

This gives (for Be)  4 x 0.0186/3000 or 2.5 E-5, roughly twice the
value you got for monoenergetic 6 keV betas.

>About the average E of the bremsstrahlung, the maximum would be 18
>keV.
> The average E of the Beta is 6 keV, and the bremsstrahlung spectrum
>is approximately flat.

Turner does say the bremsstrahlung spectrum is flat but he means it
is flat for a monoenergetic electron (which emits multiple photons),
not a beta source where the beta energies vary. Unless I'm mistaken,
the spectrum is only flat for monoenergetic electrons hitting thin
targets, something Turner does not say and something that does not
apply in most cases.  Generally we're dealing with thick targets.

>This would make the average E of the photon about 3 keV. 

This answer might be about right (I don't know) but the approach
seems wrong.

The best discussion I've seen on this topic is in Evan's book. What
is nicer in a way is in Fitzgerald's "Applied Radiation Protection
and Control" Vol 2. He actually does example calculations of complete
spectra from basic principles (unlike Evan's who just gives the
equations). Fitzgerald does it for several nuclides. Judging from the
shapes of these things, the average bremsstrahlung photon energy is
heavily weighted towards the low end. Again, based on the shape of
these spectra, I'd say tritium's average bremsstrahlung energy is
below, maybe substantially, below 3 keV.   

This is a complex subject. You must run across Turner in the ORNL
cafeteria, why not have him check this out.  

Best wishes and thanks for bringing up this issue. This is neat stuff
- something I'd like to learn more about.

Paul Frame
Oak Ridge Assoc. Univ.