17

Atoms are composed of a positively charged nucleus and an outer shell of negatively charged electrons. When two atoms come into close proximity, their electron shells repel, preventing the atoms from sharing the same space. The "volume" of an object can generally be understood as the total measure of space that is unavailable for other objects to occupy, as ...


17

Quoting from the Nobel lecture of Hans G. Dehmelt (1989): With the rise of Dirac’s theory of the electron in the late twenties their size shrunk to mathematically zero. Everybody “knew” then that electron and proton were indivisible Dirac point particles with radius R = 0 and gyromagnetic ratio g = 2.00. The first hint of cuttability or at least ...


11

The uniform glow is due to ionization and recombination of the residual gas; it's called a glow discharge. At higher pressure, a spark or arc discharge occurs at much higher current density. Fluorescent lamps and neon lamps operate in the glow-discharge region, and high-pressure xenon lamps use an arc discharge. Elements of the residual gas can be ...


11

The figure below shows the situation between configuration for a $p^2$ configuration, terms, levels and states. The word 'state' tends to be used colloquially to mean any of Term, Level or State. The Configuration such as $(1s)^2$ or ...$(2p)^2$ etc. tell us which orbitals are occupied. These are split with electrostatic (Coulomb) coupling to form Terms. ...


7

The atomic orbitals (wavefunctions) $\psi(r,\theta,\phi)$ are comprised of a radial component $R_{n,l}(r)$, as well as an angular component $Y_{l,m}(\theta,\phi)$. These are obtained by separately solving the radial and angular parts of the Schrödinger equation, the details of which can be found in any text. Often these are referred to as the radial and ...


7

When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the $1s$ orbital as: $$ \psi_{1s} = \frac{2}{a_o^{3/2}} e^{-r/a_0} \tag{1} $$ the variable $r$ is the distance from the ...


7

de Brogile explains why orbitals are quantised Strictly speaking de Brogile doesn't prove Bohr's postulates which are mostly wrong. But he did provide an explanation for the most important of Bohr's ideas: electron orbitals are quantised. Bohr's whole model starts with the classical idea that electrons "orbit" a nucleus. But this has several problems not ...


7

The configuration for the free atoms is: Ba - $\mathrm{[Xe]\ 6s^2}$ Br - $\mathrm{[Ar]\ 3d^{10} 4s^2 4p^5}$ S - $\mathrm{[Ne]\ 3s^2 3p^4}$ Si - $\mathrm{[Ne]\ 3s^2 3p^2}$ Thus: Barium has no unpaired electrons. Bromine has one unpaired electron in 4p subshell. Sulfur has two unpaired electrons in the 3p subshell. Silicon has two unpaired electrons in the ...


7

Conceptually you are right as the commenters have mentioned, but since we are on a thread about nitpicking, we might as well go the extra distance. Technically, $r^2R^2$ itself is not a probability but a probability density. In order to get the actual probability, you need to integrate it over a region. The probability of finding an electron between $r = r_1$...


6

You are trying to mix the Bohr model with quantum mechanics. The Bohr model is a semi-classical treatment of the hydrogen atom in which an electron is circling around a proton. Classically, this electron has angular momentum, even in the lowest orbital because it has a well defined position vector $\vec{r}_n$ and velocity $\vec{v}_n$. The classical angular ...


6

Think of the planetary system which inspired Bohr to think about his atomic model. The Earth is rotating around the Sun in a fixed orbit. What keeps the Earth rotating around the Sun? The mechanics of the planetary motion, and other electrical phenomena were very well understood in Bohr's time, so much so that by late 1880 to early 1900 a physics professor ...


5

So for example, the relative molecular mass of water is 18.02 u, and it's almost the same as the molar mass of water, which is 18.015 g/mol. So my question is, how is 18.02 u × (6.022 × 10^23) = 18.015 g/mol? That is not quite correct. The relative molecular mass $M_\mathrm r$ of water is a dimensionless quantity, i.e. the unit for relative molecular ...


5

One principle that goes against the fixed orbit concept is the Heisenberg Uncertainty Principle. Bohr's fixed orbits imply precise knowledge of radial position (the radius of the orbit) and also precise knowledge of the radial momentum (zero), violating the principle for the radial component of position and motion.


5

The morse potential (equation see e.g. wikipedia) basically contains the bond dissociation energy, a "force constant" and the bond length at ground state. It does not give any measurable reality, but is just a mathematical model describing (approximating) the same. The lower part of the curve (usually the ground state and perhaps the first two vibrationally ...


5

The most common and 'classical' method is to use Infra-red, Raman and Microwave spectroscopy to give the frequencies, and equivalently, the gaps between the vibrational and rotational energy levels. Microwaave spectroscopy also give the average bond length as but only as $\langle 1/r^2\rangle$. Normally in a text book a model of the potential energy is ...


5

Your question is historically important. Nobody has the energy to write an answer which would require several pages of an article. In short, it is a can of worms. For almost 5-6 decades chemists and physicists used a different scale for the atomic masses, so the atomic masses in the publications/books used by physicists did not exactly match what the ...


5

The theory for a muon-antimoun "atom" is essentially the same as any hydrogen-like atom, just with masses changed. See this Wikipedia page for the relevant formulae for relativistic (Dirac) and non-relativistic treatments. The non-relativistic energy levels are given by : $$E_n = \frac { \mu c^2 \alpha^2}{2n^2}$$ where $alpha$ is the fine structure ...


5

Your question, Why can't we use normal formula to find effective magnetic moment of tripositive rare earth elements (REs)? is answered excellently by porphyrin' comment (vide supra). Thus, I won't attempt to answer it again except for to mention the calculated values from the equation, $\mu_\mathrm{eff} = g_J \sqrt{J(J+1)}\mu_\mathrm{B}$ (where $\mu_\mathrm{...


5

Basically everything is wrong, unfortunately. $\psi^2$ isn't a mass density, it's a probability density. So there is no point in which you should be getting units of mass. As porphyrin pointed out, the units of $\psi^2$ (and hence $\psi$) will depend on the system which you are looking at. Next, $\psi$ isn't a length amplitude. In fact, it's not an ...


4

The fault in your solution is the part where you assume $\mathrm{\frac{-PE}{2} =KE}$ this is true only if the force follows inverse square law. In this particular case we need to find the force first by using $$F =\frac{-dU}{dr}$$ Which gives us $$F =\frac{-3Ke^2}{r^4}$$ Now this force is central and provides centripetal force $$\frac{mv^2}{r} =\frac{...


4

The latter is the limit of probability per volume, as volume approaches zero, at a particular point in space.


4

Atomic weights in any periodic table are based on the natural abundance of different isotopes of the element. But sometimes the source matters and the results will be different. The most important thing to note here is that creating new isotopes is hard. You can't just "add" neutrons to an atomic nucleus and often when you do it decays to a different ...


4

Thompson observed more than just cathode rays. He experimented with different gases and different pressures in his tubes. When there is a significant amount of hydrogen you get ionisation of the gas and more than one type of charged particle can be generated and deflected (though different conditions are needed to see protons than electrons). Part of the ...


4

To convert between joules (units of energy) and inverse centimetres (conventional units of wavenumber) simply requires multiplication by a factor of $hc$. This is because $$E = h\nu = \frac{hc}{\lambda} = hc\bar\nu$$ Of course, they are not literally equal, but they are physically equivalent.


4

Thomson described the mechanical effect produced by cathodic rays on the first edition of its book Conduction of Electricity Through Gases (1903, pp. 501–502) and again later on the second edition (1906, pp. 629-630). As you mentioned it refers to the Crookes experiment with the paddle wheel in which the electron beam collides with the mill coausing it to ...


4

Atomic number is proportional to the atomic kernel charge. From the classical electrostatics we know that potential around a charge $Q$ is: $$U=-\frac{Q}{4\pi \epsilon_0 r}$$ So it's obvious it the electron energy must depend on this charge. From Schroedinger wave equation we can see the charge is involved: $$ \psi_{n\ell m}(r,\theta,\varphi) = \sqrt {\left ...


3

The energy difference between $n=2$ and $n=3$ in Bohr's model is \begin{equation} \Delta E=13.6\ \mathrm{eV} \times \left(\frac{1}{4}-\frac{1}{9}\right) = 1.88\ \mathrm{eV} = 3\times10^{-22}\ \mathrm{kJ} = 181.4\ \mathrm{kJ/mol} \end{equation} Thus, for this excitation to occur, a single energy quantum (photon) with an energy of $1.88\ \mathrm{eV} = 3\...


3

For degenerate eigenstates any (normalized) linear combination of them will be again an eigenstate of same energy. This in fact means that there are infinitely many of them. So where does the 5 (in case of $d$-orbitals) come from, when we actually have infinitely many? It means we can select at most 5 of them, which are orthogonal (or more generally ...


3

Imagine a universe that is not expanding, and contains only one large star and one small planet. It is technically possible to put the two objects into orbit no matter how far apart they are. At some very large distance, the gravitational attraction is so small that it is nearly negligible, but if given infinite time you will be able to trace the orbits for ...


3

Wavenumbers are a unit of energy. Given that $1\mathrm{eV}$=$8100\mathrm{cm^{-1}}$, and the relations $E=h\nu$ and $c=\lambda\nu$, where $h$ is Planck’s constant, $c$ is the speed of light, and frequency $\nu$ and wavelength $\lambda$, you should be able to solve this with some simple algebra. Once you calculate the wavelength, try to find where it falls on ...


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