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2

The chemical properties of an element are always determined by the atomic number, that is, the number of protons in the nucleus. All carbon atoms have six protons, all iron atoms have 26, etc. It's the atomic number which is featured prominently in the periodic table, for example. Until the neutron was discovered in 1932, this was fine. After the neutron ...


4

The most common isotope of hydrogen has no neutrons. Other isotopes are deuterium with 1 neutron and tritium, with 2 neutrons. Since virtually all (99.98% according to wiki) naturally occurring hydrogen comes in the no neutron isotope, it seems reasonable that books show a schematic of that one when illustrating hydrogen. As a secondary motivation, the one ...


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It is all over the place, but it's involved. It is normally calculated from the path integral propagator. The most concise source of the radial Green's function you are after is eqn (15) of Grosche 1998, in terms of modified Bessel functions, integral rep, $$ G_l^C(r'',r';E) = \int_0^\infty\dfrac{e^{i e^2s''/\hbar}ds''}{\sqrt{v'v''}} ...


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I dont think we should consider the energy of the electrons. For both para- and ortho- hydrogens, the two electrons are sitting in the same ground state (but with opposite spins) of the hydrogen molecule, so I dont think there is a difference. The energy of a hydrogen molecule arises from its translational, rotational and vibration kinetic energy. At room ...


2

The relativistic generalization of the formula; and the replacement of the electron mass by the reduced mass are clearly two basically independent steps (at least in the leading approximation), and both of them have to be applied to agree with the experiment. The most accurate experimental value of the energy is $-13.59844\,{\rm eV}$, see NIST: ...


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A nice overview of the problem is given in arXiv:1205.3740. I'll summarise the most important points here. Let $d$ be the number of space dimensions. Then the Laplace operator is given by $$ \Delta=\frac{\partial^2}{\partial r^2}+\frac{d-1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\Delta_S\tag{1} $$ where $\Delta_S$ is the Laplace operator on the $d-1$ ...


2

Yes, the conservation of momentum is valid. The photon and the Hydrogen atom acquire equal and opposite momenta when the photon is emitted. However you must be a little careful when you are calculating the momentum as the total energy of the photon and the hydrogen atom is 10.2 eV. This means your equation will be $$ p^2/2m + pc = 10.2 eV$$ You must conserve ...


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The primary difference is that the electrons in metallic hydrogen are nearly completely degenerate. Degenerate electrons cannot be dissipatively scattered and lead to the "metallic" characteristics of extremely high electrical and thermal conductivity. To first order, to produce metallic hydrogen you need to make sure that the electron kinetic energy at ...


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In metallic hydrogen, the protons share an approximately fixed location relative to each other - energetically, a lattice is more favorable than an randomized state. Because of this, the substance is not a plasma - in a plasma, the positive and negative charges both flow freely. Metallic hydrogen only exists at very low temperatures, and very high ...


3

Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $ \psi_{a} $ and $ \psi_{b} $, both of which correspond to energy $ E_{0} $. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent. Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks ...



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