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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 $2\times10^{30}\ ... 4 When labeling states of the hydrogen atom, one doesn't refer to the z component of the angular momentum, but rather to the total angular momentum. The total angular momentum is positive, but, as you've stated, there are two states for$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, ... 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$T_\infty$... 3 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 ... 2 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 ... 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 ... 2 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 ... 2 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 ... 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 ... 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 ... 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 ...