# Tag Info

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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 ...

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Yes, the question is basically asking which are the most abundant elements, though it's specifically asking about elements that form molecules and this is presumably why the second most abundant element helium is excluded (helium being chemically unreactive). Big Bang nucleosynthesis produced mainly hydrogen and helium, with small amounts of lithium and ...

-1

The closed orbits are those classical trajectories that can never leave the system and these closed orbits are then correlate with Quantun interference. To under this closed orbit theory read following articles: PRL 58, 1731 (1987) PRA 38, 1896 (1988) PRA 38, 1913 (1988)

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This won't work, though possibly not for the reason you think. High energy protons will go straight through a turbine blade without transferring any significant amount of momentum to it. The LHC uses a seven metre long block of graphite to catch the proton beam if there's a beam dump. Steel has greater stopping power than carbon, but even so a turbine blade ...

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Well there is no reason of why it would not . But really what is the purpose of putting a turbine inside a particle accelerator.That beats the purpose of the accelerator to work without much resistance.If it is to generate energy or something, I doubt it's a viable solution.hydrogen as a gas is very hard to compress so I don't think you can make a solar ...

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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 ...

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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

While there is an origin for the position operator $\vec r$, and therefore the angular momentum $\vec L = \vec r \times \vec p$ has an origin, it is a little glib to say that there is an angular momentum operator. What you can is three operators $\hat L_x,$ $\hat L_y,$ and $\hat L_z$ whose forms are a inspired by the forms of the three components of the ...

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