The isospin in a nuclear reaction is conserved. The quantum law is called the conservation of isospin.
On Wed, Apr 8, 2015 at 1:16 AM, Axil Axil <[email protected]> wrote: > I don't get it. 8Be has zero nuclear spin and 4He has zero nuclear spin. > How can a nuclear reaction involving them have huge annular momentum? > > On Wed, Apr 8, 2015 at 12:28 AM, Eric Walker <[email protected]> > wrote: > >> Hi Bob, >> >> The possibility you've been drawing attention to, that the result of the >> decay of the [8Be]* compound nucleus into two 4He nuclei with little linear >> momentum and a great deal of angular momentum makes for an interesting >> thought experiment. Out of curiosity, I calculated the energy that would >> be needed to break up an alpha particle into either tritium and a proton or >> 3He and a neutron, which would be the reverse of these two reactions: >> >> 3He + n → 4He + Q (19.3 MeV) >> t + p → 4He + Q (20.5 MeV) >> >> As I understand it, this implies that angular momentum sufficient to >> produce ~ 19 MeV of centripetal force would be needed to break apart a 4He >> into either 3He and a neutron or tritium and a proton. This suggests that >> a 4He can carry a large amount of angular momentum before it is likely to >> break apart. (I assume the process is probabilistic and that the force >> needed lies along a distribution.) >> >> Further comments inline. >> >> Eric >> >> >> On Tue, Apr 7, 2015 at 1:35 PM, Bob Cook <[email protected]> wrote: >> >> However, I know of know reason why the light nuclei cannot have any spin >>> quantum number--high or low. Any spin quantum is available. >> >> >> Further to the thought experiment, I think we should make a clear >> distinction between two types of "spin" -- there's the actual spinning >> motion of a nucleus (e.g., 4He), and there is the spin state of the >> nucleus. At higher rates of rotation, a heavy nucleus such as an isotope >> of nickel will reconfigure into a higher spin state, presumably through >> deformation. In such a state a photon may be emitted, with the nucleus >> relaxing into a lower spin state. Here my mental model is of neodymium >> magnets spinning around in a clump. When they snap together into a >> lower-energy configuration, a photon is emitted through the movement of the >> magnets as they snap together. The photon is emitted in a direction and >> carries away energy in such a way as to slow the angular movement of the >> spinning nucleus a little (by the amount of energy carried away by the >> photon). The participants involved in such a transition are the nucleons, >> and the energy of the photon that is emitted will correspondingly be in the >> keV or MeV range, which is in the nuclear range. >> >> A light nucleus, such as 4He, does not have a bound excited state. My >> understanding is that it cannot deform under high angular momentum into a >> higher energy state which will emit a photon when it relaxes. The 4He will >> either break apart into lighter constituents under centrifugal forces or it >> will not. But I'm guessing that the actual moment-to-moment velocity of >> the 4He about its axis of motion is in principle a continuous quantity. If >> this is true, perhaps the energy could be released to the environment in >> small amounts. >> >> Where the thought experiment gets interesting is in the supposition that >> you and others have already offered in this thread, that charged body such >> as a 4He nucleus that is spinning at an incredible rate will set up a >> magnetic field. This magnetic field could disturb nearby electrons, >> causing them to emit lower energy photons in the process. >> >> Although I do not see anything special in the 7Li+p to 8Be transition >> that has been proposed (and note Jones's point about the gamma that would >> be omitted in the process), I think the more general notion of the energy >> of a nuclear transition somehow being deposited in angular momentum and >> then released in small amounts is a very interesting one. >> >> >
