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- Path: sparky!uunet!psgrain!hippo!ucthpx!uctvax.uct.ac.za!hndcol02
- From: hndcol02@uctvax.uct.ac.za (Colin Henderson, Physics Dept. UCT)
- Newsgroups: sci.physics.fusion
- Subject: Re: Hide the neutron (or: billiards with feathers)
- Message-ID: <1992Aug13.173302.202374@uctvax.uct.ac.za>
- Date: 13 Aug 92 17:33:02 +0200
- References: <9208120306.AA10493@sleepy.network.com>
- Summary: compton scattering unlikely.
- Distribution: world
- Organization: University of Cape Town
- Lines: 79
-
- 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!-------------------
-