From: hndcol02@uctvax.uct.ac.za (Colin Henderson, Physics Dept. UCT)
Newsgroups: sci.physics.fusion
Subject: Re: Hide the neutron (or: billiards with feathers)
Date: 13 Aug 92 17:33:02 +0200
Summary: compton scattering unlikely.
Distribution: world
Organization: University of Cape Town

In article <9208120306.AA10493@sleepy.network.com>, logajan@SLEEPY.NETWORK.COM (John Logajan) writes:
> Okay, I think I understand the "virtual neutron" idea now, and how it
> allows the process to occur even though the "trigger" energy seems
> to be insufficient (which I repeat below.)
> 
>  [ .... virtual neutron discussion .... ]

> 
> And now for more speculation.
> 
> Perhaps there is simple "mechanical" movement of D into the vicinity
> of Pd.  We know that D's can be Compton scattered directly (don't we?).
> We know they can be scattered indirectly by electric force "collisions"
> with Compton scattered electrons. [Note: Compton scattering merely
> means that the incident gamma ray induced physical movement in the
> particle, with a resultant deflected gamma with reduced energy.]
> 

O.K.  As Paul Dietz says in a back of the envelope calculation, the
"range" of a virtual neutron will be in the order of fermis (10^-15 m)
Therefor you need pretty nearly contact between the deuterium and
palladium nuclei for the neutron to be transferred.  Essentially what you're
looking at is then the pickup reaction

Pd(d,p)Pd                              (1)

Now the threshold for the T(d,n)4He reaction is some 20 keV. I guess
it's pretty similar for the D(d,n)3He reaction.  The threshold for the
deuteron on palladium is going to be up by the order of Z (=46) for
palladium.  That's because the threshold arises largely from coulomb barrier
penetration requirements.

So by BOTE calc, threshold for reaction (1) is of the order of 1 MeV.
Sorry if this is flogging a dead horse, but that's point (1)

Now for Compton scattering, the other runner in this two horse race.


The photon wavelength shift is given by:

lambda1 - lambda0 = h/mc(1-cos(theta)), where

lambda1 is the scattered wavelength, lambda0 is the initial photon
wlngth, m is the mass of the particle the photon scatters off, and
theta is the angle of scattering.

Converting to energies, and taking the maximum energy transfer at
theta=90 deg, i.e. cos(theta)=0, we have:

hc/E1 - hc/E0 = h/mc

or

1/E1 - 1/E0 = 1/mc^2

For a deuteron, mass m = 1880 odd MeV/c^2, and a E0 = 1.5 MeV gamma,
the new gamma energy E1 will be 1.4988 MeV, i.e. the deuteron will be
given about 1 keV AT MOST, which is insignificant, even for d,d
fusion. So, forget Compton scattering as a mechanism. Nice idea,
though.

> And like a bullet shot into a tank, it is going to bounce around inside

more like a balloon flung by a kid :-)

> the lattice for a while, approaching many Pd nuclei in sequence,
> perhaps until probability smiles on it and it is captured.  (Heh,
> I can even envision a D caught between two Pd's like a ping-pong
> ball between the table top and a descending paddle board -- brrrrt.)

It will lose energy very fast due to collisions with electrons.
 [delete "brrrrt" and insert farting noise :-)]

-- 
Colin Henderson
Physics Dept, UCT, Cape Town, South Africa.
colin@physci.uct.ac.za

------------------natural selection favours paranoia!-------------------