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1

If Rob Jeffries numbers are right (Here) and it seems as good a baseline as any, you get about 3*10^15th Joules per KG of Neutron Star matter. The Hiroshima bomb was about 6*10^13th Joules, so just 1 KG of degenerate matter, you'd be looking at about 50 Hiroshima bombs released in kinetic energy, in the form of (I think) mostly Neutrons. 1 KG of hydrogen ...


2

Look up Lithium deuteride: a form of lithium hydride where the hydrogen is all deuterium: this will give you most of your answers. The main fusion[1] reaction that lets slip most of the energy and thus the horrendous blast in a fusion bomb is: $$_1^2 D + _1^3 T \rightarrow _2^4 He (3.5{\rm MeV}) + _0^1 n (14.1{\rm MeV})\tag{1}$$ Here I've written $D$ for ...


2

Generally, you are correct, and trying to store gaseous hydrogen for long periods of time without significant losses doesn't work. The usual way around this is use lithium deuteride, a solid compound, as the main fuel in the fusion portion of the device (the secondary). Deuterium/tritium is also used in the fission primary. According to ...


0

What if there's literally only one resulting particle? Suppose two relativistic particles with rest masses $m_1, m_2$ and velocities (in the lab frame) $\vec v_1, \vec v_2$ collide and stick together. Their energy-momentum 4-vectors are $\gamma_i m_i [c, ~\vec v_i] = [\bar m_i ~ c,~ \vec p_i]$ where $\bar m_i = \gamma_i m_i$ is the effective mass and $\vec ...


0

Actually, there are papers that argue that deuterium fusion can appear in planets as small as Jupiter. http://arxiv.org/abs/1506.03793 The argument is that although Deuterium fusion is thought to happen in the order of magnitude of 10 jupiter masses, the fusion of Deuterium can be facilitated by electrons "screening" the protons of Deuterium. If a Deterium ...


11

As dmckee says in his comment - Population III stars have no metals (a tiny bit of lithium and beryllium), but they are not "pure hydrogen stars", they still have the big bang fraction of Helium. Taking the second part of your question first. These "stars" will last for ever. Their final fate is to become a completely degenerate ball of helium, supported by ...


3

First, note that Population III stars are expected to be massive, not tiny, with masses upwards of $10^6\,M_\odot$. The reason for this is due to the Jeans criteria, where the mass follows $$ M_J\propto T^{3/2} $$ In the early universe (~ 1 Myr), the temperature was around 10,000 K; so in a pure-hydrogen environment, no cloud with a mass less than about a ...


0

In your scheme, other processes, such as Coulomb scattering, are much more probable than fusion, so you will have net energy loss, as dmckee noted.



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