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It is a very interesting question, but comparing a combustion spectrum with an atomic emission one is like comparing apples and oranges. A flame is a luminous gas phase chemical reaction where the hydrogen atoms are combining with oxygen atoms. It is a chemiluminescence phenomenon. A discharge tube emission is an atomic emission. You posted a picture of a ...

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 ...

18

The volumes of the nuclei are negligible compared to the atomic radii (like a "pea in a soccer/football/cricket stadium"). The higher the nuclear charge, the closer the electrons to the nucleus in a given shell. Source: https://www.mdpi.com/1422-0067/3/2/87/htm Helium has the smallest atomic radius.

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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

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 ...

12

These orbitals represent the angular part of the wavefunction. The solution obtained directly from solving the Schrödinger equation produces equations containing complex numbers so cannot be drawn on normal $xyz$ axes and are hard to visualise. The angular part of the wavefunction is given by functions called Spherical Harmonics these usually are given 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. ...

10

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 ...

9

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 ...

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 ...

8

The value of the energy in the Bohr model is zero when the quantum number is infinity because that is the limiting value of the Coulombic potential at large distances, and because the electron is assumed bound to the nucleus (the atom is stable), which constrains the value of the total energy. The energy of a stationary hydrogen-like atom is described by ...

8

Although density may be a relatively easily measurable property of solid materials, it may not suggest the most fundamental relationship between mass and volume for the elements. Molar volume (the volume required to contain $6 \times 10^{23}$ atoms of each the elements) is pictured in the graph below (WebElements): It is amazing that ~100 elements can be ...

8

Although I liked James Gaidis's answer, I do not agree with some of arguments because they are all parts of one or more continuous trends. For instance, look at the melting points and boiling points trends of alkali metals and other trends as illustrated in the following table: $$\begin{array}{l|ccc} \hline \bf{\text{Physical property}} & \ce{Li} &&... 7 When we solve the Schrodinger equation for the hydrogen atom we general make the simplifying assumption that the proton stays fixed and the electron moves in the potential of the fixed positive charge. So when we write, for example, the 1s orbital as:$$ \psi_{1s} = \frac{2}{a_o^{3/2}} e^{-r/a_0} \tag{1} $$the variable r is the distance from the ... 7 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 This excerpt is from an article in J Chem Ed (J. Chem. Educ., 1961, 38 (6), p 297 DOI: 10.1021/ed038p297) describing the contents of Markovnikov's 1870 paper in Liebig’s Annalen (translated into English, I guess): So this was before the discovery of the electron, but as you can see from the diagrams, the concept of chemical structure, atoms and valence was ... 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 There is a misconception here. A p orbital is a 3D-function, and these functions don't have shapes, they have values at any point in space. If you describe an electron distribution with one of these functions, you can plot contours at a chosen value, and these contours have shapes. For example, you could choose a contour level such that an electron has a 90% ... 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 ...

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