How a Lithium Atom Transforms into a Hydrogen Isotope

Mr_Bojingles
Messages
78
Reaction score
0
From what I gather a hydrogen bomb works by exploding a conventional explosive such as TNT which propells fissile Uranium or Plutonium particles at each other which causes fission and this fission causes enough heat to start fusion.

Then I hear lithium is bombarded with by an electrons which cause it to transform into tritium. My question is how does a lithium atom transform directly into a hydrogen isotope??
 
Physics news on Phys.org
2D + 6Li produces the following possible results

3He + 4He + 01n, Q = 2.56 or 1.8 MeV

2 4He, Q = 22.4 MeV

1H + 7Li, Q = 5.0 MeV

7Be + 01n, Q = 3.4 MeV



2D + 7Li produces the following possible results

8Be + 01n, Q = 15.0 MeV

2 4He + 01n, Q = 15.1 MeV

Where Q is the energy (as kinetic energy of the products) released from the reaction.


The second reaction cited by Hurkyl applies to fast neutrons, which would be produced by a fission reaction or D + T reaction. Thermal or slow neutrons are not produced in fission or fusion reactions.

01n (fast) + 7Li -> 3H + 4He + 01n (slow)


Simply suspending an A-bomb inside a vessel of fusion fuel will not work.
The fission trigger and fusion reaction chamber are specially configured, of course.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Back
Top