In reply to Stephen A. Lawrence's message of Tue, 04 Jan 2011 17:30:05 -0500: Hi, [snip] > > >On 01/04/2011 03:16 PM, Jones Beene wrote: >> >> _http://www.journal-of-nuclear-physics.com/?p=338_
Quote: "However, based on the principle of conservation of momentum, as a result of the backlash of this nucleus, the photon energy ? is divided into kinetic energy of this nucleus of large mass (heat) and a photon of low frequency." This is clearly wrong. In two particle systems such as this (nucleus + photon), the energy is divided in inverse proportion to the mass. The mass energy of a Cu nucleus is about 63*1000 MeV or about 63000 MeV. That of the photon only about 5-10 MeV so the photon is much "lighter" (by a factor of at least 6000), and consequently gets nearly all the energy (at least 99.98 %). IOW the wavelength of the photon is barely diminished at all (by no more than 0.02%), and this excuse can in no way be used to explain why the (measured?) energy of the photons is so low. Regards, Robin van Spaandonk http://rvanspaa.freehostia.com/Project.html
