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In fact hydrogen is an old idea to get a high temperature superconductor, based exactly on the idea of its light mass. The problem is that one has to start from metallic hydrogen, which is a problem on itself. It has not yet been fully experimentally confirmed in the lab. You need pressures of several hundreds of GPa to achieve that (100 GPa is about 1 ...

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Not a proof but: Because the potential only depends on $r$, it can be shown that the wave function is separable into $\Psi=RY$. (This might be shown by David Miller in his online quantum mechanics course at Stanford in his lectures in section 7 on the hydrogen atom.) Then we can say that the $Y(\theta,\phi)$ portion is the spherical harmonics, same as for ...

1

Firstly, you cannot simply 'neglect' $a_0$ (as the *.pdf actually shows). In some test books the following substitution is used: $\rho=\frac{r}{a_0}$, so the $a_0$ doesn't have to be 'carried around'. As your expression for $P(\rho)$ is made of three factors, each factor can be evaluated for extrema individually. $\rho=0$ is an obvious minimum from the ...

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The instructor hasn't thrown out $r^4$. If you do the calculations properly you will get the desired result. What the instructor is telling is that while simplifying and taking commons out for $\frac{\text{dP(r)}}{\text{dr}}=0$ you get,$$(\text r^3)(6-\frac{\text r}{\text a_0})(f(\text r))=0$$which gives, $\text r=0$ and $6\text a_0$ from the first two ...

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Thinking through the problem for water vapour to exist by nature a fixed volume will contain more energy in the form of heat. The heat displaces air molecules around it making the volume lighter as in a hot air balloon. The amount of atoms in the fixed volume in comparison with surrounding air is higher in the water vapour. I see water vapour as heavier in ...

4

When labeling states of the hydrogen atom, one doesn't refer to the z component of the angular momentum, but rather to the total angular momentum. The total angular momentum is positive, but, as you've stated, there are two states for $J=\frac{1}{2}$ with $L=0$, and those are $J_z=\pm\frac{1}{2}$ (Or some linear combination of them) As to why this is, ...

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The core of the Sun is radiative. That means that energy is transported outwards primarily by photons and is stable to convection. This means, to first-order, that the centre of the Sun is not mixed up by convection. As the hydrogen in the core is burned, it forms helium, which has a larger atomic mass. The helium sinks towards the middle and the core fills ...

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In about four billion years the sun will have used up most of the hydrogen in its core and fuse helium into heavier elements. The fusion of helium releases more energy than the fusion of hydrogen. This will cause the sun to expand to, at least, the orbit of Venus and likely the Earth. Both planets will, of course, be vaporized.There was a similar question ...

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For any sample of atoms in a gas they have a range of speeds. The likelihood of any atom having any particular speed is given by a probability distribution called the Maxwell-Boltzmann distribution. But the main point is that low mass atoms are more likely to have high speeds. For hydrogen and helium at temperatures typical of Earth's atmosphere there is ...

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They don't go up and leave earth or its atmosphere. The atoms go up because they are lighter then the other atoms and molecules in the air, but they still have mass. They float over the heavier atoms and molecules. They don't leave earth's orbit or go in space but they go very very high. its all because of the gravity and pressure. Atoms swap places, the ...

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By "consume" we mean "convert into helium." That $6\times10^{11}\ \mathrm{kg}$ of hydrogen is part of the Sun (specifically it is found in the core of the Sun), and it is converted into $6\times10^{11}\ \mathrm{kg}$ of helium. The Sun doesn't need to suck up material from space. Note that this amount of material is miniscule compared to the $2\times10^{30}\ ... 1 In an ideal, theoretical world, we would only have functions from$m$arguments to some space, and we'd say "the partial derivative with respect to the$n^{\text{th}}$argument", with an implicit statement "holding all the other$m - 1$arguments constant." Call this derivative of a function$f(x_1, \dots, x_m)$the derivative$f_{n}$. But, we live in a ... 2 If you have a function$f(x_e,x_p)$then it has the partial derivative$\partial f/\partial x_e$(that holds$x_p$constant and treats the function like a 1d function of$x_e$alone) and similarly has the partial derivatives$\partial f/\partial x_p$(that holds$x_e$constant and treats the function like a 1d function of$x_p$alone). But you can think of ... 0 If you crush hydrogen to very high densities, the Pauli exclusion principle will prevent electrons in the material from occupying the same quantum states. The electrons will obey Fermi-Dirac statistics and they will fully occupy states up to the Fermi energy - which increases as the density gets higher. Presumably in metallic hydrogen what happens is that ... 1 It’s unrealistic to expect most students to able to derive the energy quantisation of a hydrogen atom on the spot. For hydrogen: $$E_n = \frac{-13.6 }{n^2}\rm eV$$ (ignore the minus sign for your problem). For a particle in a 1 D infinite potential well: $$E_n = \frac{n^2h^2}{8mL^2}$$ Set$n = 1$to obtain the energies of both ground states. From the ... 4 The energy is defined as $$E = \frac{p^2}{2m} + V(\vec r)$$ where the first term is the kinetic energy and the second term is the potential energy calibrated so that$V(\vec r)=0$for$|\vec r|\to\infty$. Consequently, you may say that the energy in a given state (an analogy of an orbit in classical physics) is equal to the kinetic energy$T_\infty\$ ...

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The term to look for is Coulomb wave. These wavefunctions are well explained in the corresponding Wikipedia article. Depending on your mathematical background, you should be ready for a bit of a formula jolt, as these wavefunctions rely very intimately on the confluent hypergeometric function. If you want the short of it, then I can tell you that the ...

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