# Tag Info

18

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 ...

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 ...

16

Now that's a mildly non-trivial observation. Why would they be equal, really? Let's say a particle with mass $m$, charge $q$, and initial velocity $v$ enters an area of length $L$ where an electric field $E$ starts to deflect it sideways. This is a clear example of uniformly accelerated motion, and its laws are well known: $x=vt,\;y={at^2\over2}$, where the ...

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. ...

9

I'd like to copy the answer by John Rennie to a similar question at Physics.SE, since it's much better than current answers here IMHO. Although it mostly speaks about binding energy, the same principle applies to the quanta of excitation energy. The mass of a hydrogen atom is $1.67353270 \times 10^{-27}$ kg. If you add the masses of a proton and electron ...

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 ...

6

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 $\mu_\mathrm{... 6 Does electron mass decrease when it changes its orbit? Essentially yes. If you add the mass of a free proton and a free electron you'll get a greater mass than that of a hydrogen atom. The mass difference will be equivalent to 13.6 eV which is the ionization energy of hydrogen. Now for any "practical" chemistry experiment the assumption is that ... 5 One principle that goes against the fixed orbit concept is the Heisenberg Uncertainty Principle. Bohr's fixed orbits imply precise knowledge of radial position (the radius of the orbit) and also precise knowledge of the radial momentum (zero), violating the principle for the radial component of position and motion. 5 The morse potential (equation see e.g. wikipedia) basically contains the bond dissociation energy, a "force constant" and the bond length at ground state. It does not give any measurable reality, but is just a mathematical model describing (approximating) the same. The lower part of the curve (usually the ground state and perhaps the first two vibrationally ... 5 The most common and 'classical' method is to use Infra-red, Raman and Microwave spectroscopy to give the frequencies, and equivalently, the gaps between the vibrational and rotational energy levels. Microwaave spectroscopy also give the average bond length as but only as$\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 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 ... 5 Very nice question. I also had the same question. These orbitals are nothing more than functions (wavefunctions). They are solutions to Schrodinger's equation, a second order differential equation. So you solve the equation and the solutions you get are determined by$4$numbers also called quantum numbers$n, l, m_l, m_s$. A shell consists of all orbitals ... 5$E_n$represents the energy of the$n$th electronic state relative to the ground state (i.e. the ground-state electronic energy$E_0 = 0$). Often, electronic states are very high in energy and the excited-state energies$E_1, E_2, \cdots$are far greater than$k_\mathrm{B}T$, such that$\mathrm{e}^{-E_n/k_\mathrm{B}T} \approx 0$. Under these conditions, the ... 5 Pauli's exclusion principle is a consequence of electrons being indistinguishable fermions. Fundamental particles are indistinguishable in that two of the same type differ only in a few properties but otherwise behave identically (compare that say to two apples - they are always in principle distinguishable because they differ in so many ways). Fermions ... 5 The question is misguided by the obsolete idea of shells and subshells as separated exclusive regions of the space around the nucleus. All orbitals largely overlaps and statistical distribution of electron density, (aside of being in the particular orbital) depends on the nucleus charge and overall electron configuration. It can be said, that distribution of ... 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 ...

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