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26

A particle's rest mass never changes. It's mass is a natural constant, and one of the numbers which uniquely identifies it (like its spin). On the other hand, the invariant mass of the atomic system does increase as the electron becomes excited, bringing the atom into a higher energy state. In that sense, the atom (not the electron) gets "heavier" because of ...


26

This is really an extended comment to Geoffrey's answer, so please upvote Geoffrey's answer rather than this. The mass of a hydrogen atom is $1.67353270 \times 10^{-27}$ kg. If you add the masses of a proton and electron together then they come to $1.67353272 \times 10^{-27}$ kg. The difference is about 13.6eV, which is the ionisation energy of hydrogen ...


17

No. The decay products of a certain particle are not equivalent to its constituents. This is evident especially in the context of fundamental particles: quarks can decay into other particles, but that does not mean that a quark is not elementary (see my answer to this question). Nuclei are made of neutrons and protons, which in turn consist of quarks and ...


15

Graphene is only transparent because it is very thin (one atom thick). If it absorbs 2% per layer then just a few hundred layers would absorb almost all light and that would still be a very thin sheet of graphite. The question should be why does graphene absorb so much light compared to diamond which really is transparent? A simplified answer is that ...


11

No. The atoms are protons, electrons and neutrons. The fact that neutrons beta decay into a proton + electron + electron antineutrino does not mean that neutrons are made of a proton and electron and a neutrino.


9

This is the plum pudding model of the atom Left: Expected results: alpha particles passing through the plum pudding model of the atom undisturbed. Right: Observed results: a small portion of the particles were deflected, indicating a small, concentrated positive charge. There are no electrons and no outer shells, quantum mechanics was yet to come ...


8

The $\alpha$ particle, attracted by the electrons on the outer shell of the pudding, orbits nearly parabolically around the atom, causing the near-180 degree deflection angle seen. This wouldn't happen because of momentum conservation. It was reasonably established in 1909 (when the gold foil experiment was done) that electrons were light, so if an ...


7

If you want to prove an isolated atom is spherically symmetrical, you could proceed by showing the sum of the probabilty functions (wave functions squared) of each orbital results is a spherically symmetrical distribution. Certainly s-orbitals are spherically symmetrical. The sum for an entire subshell of orbitals (such as all three 2p orbitals) is ...


6

Is the Coulomb potential also used to solve the hydrogen atom in relativistic quantum mechanics? Yes, the Coulomb potential is there in the solution of the hydrogen atom with the Dirac equation, which is formulated in the relativistic framework. Now it is time to specialize to the hydrogen atom for which $$\frac{V}{\hbar c}=-\frac{Z\alpha}{r}$$ ...


6

You say: The Coulomb potential comes from classical electrodynamics but actually the Coulomb potential is predicted by quantum electrodynamics as a low energy limit. Quantum field theory describes the interactions between charged particles as the exchange of virtual particles, and it's not immediately obvious that it would lead to an inverse square ...


4

I have read on science forum that electron in orbital s has no angular momentum and would fall into nucleus, so hydrogen atom would not be possible. In this sentence you are encapsulating the reason quantum mechanics was "invented". The planetary like theory of the Bohr model imposed the "not falling", quantized stable orbits with a minimum binding ...


3

Quick answer : electrons are a liquid, but you can replace them with fake electrons which behave like they were in a gas, so both formulations are correct and meaningful. Now for a longer answer. Indeed, the valence electrons in (most) metals form a state which is closer to a liquid than to a gas. This liquid of interacting electrons is called a Fermi ...


3

No, electrons do not pop in and out of existence. They do, however, switch between localized and delocalized states: When the atom is undisturbed, the electron will be in the energy eigenstates of the system, the so-called atomic orbitals. These states are not sharp position states, the position expectation is "smeared out" over a region around the nucleus ...


3

The principle quantum number is denoted $n$ because it is a 'natural number', $n=1,2,3,4....$. The secondary azimuthal quantum number (also known as orbital angular momentum) is denoted $\ell$ through its association with angular momentum (typically denoted $L$).


3

I assume the biggest factor is the thickness. Graphene is a layer of carbon one atom thick. Light is absorbed/reflected by the top layers of a material and if you make any material into a layer one atom thick you'll find it increases transparency a lot. The thing that is special about graphene is that it forms bonds in a 2D layer where most materials ...


2

You are right to think that $4$ quantum numbers completely describe the state of an electron within an atom. I'll explain each in turn. $n$ is related to the energy of the electron, i.e. the eigenvalue of the Hamiltonian operator. It's called the principal quantum number since we usually think of the energy as the most basic quantum quantity. We use the ...


2

The magnetic quantum numbers $m$ give you the projection of the angular momentum $L$ in units of $\hbar$ on one specified axis, normally the $z$- axis. So it's not one $m$ for every axis, but multiple $m$'s for one axis. For the $P$-state for example - that is $L=1$ - the projection on the $z$-axis can take on the values $$\hbar,0,-\hbar~,$$ so ...


2

The $p$ orbitals that you refer to are actually certain linear combinations of the ones we usually use with the magnetic quantum number, i.e., the wave=functions are of the form $$\begin{eqnarray}p_z &=& p_0\\ p_x &=& \frac{1}{\sqrt{2}} \left(p_1 + p_{-1} \right)\\ p_y &=& \frac{1}{i\sqrt{2}} \left( p_1 - p_{-1} ...


2

The answer to this question depends on your interpretation of quantum mechanics. According to the Copenhagen interpretation, the interpretation that is taught standardly in undergraduate quantum mechanics courses, the answer is no. The wavefunction simply describes the probability of finding an electron in a given location. When we are not directly ...


2

Quantum mechanics tells us the probability for an electron in a certain orbital to be in a certain position or momentum state. Inverting this to find the probability for the electron to be in a certain orbital given its position or momentum state requires some a priori idea of what orbitals are the most probable. This is stated succinctly by Bayes' theorem: ...


2

The Rydberg electron - the electron in the high n level - is highly polarizable and very weakly held. The binding energy of the electron is very small. A small electric field will distort the wavefunction of the Rydberg electron so that it spends more time on one side of the atom than the other. The result is the formation of dipole. Rydberg electrons in ...


2

No, it is an approach called semiclassical quantum Mechanics. They mix the Schrödinger equation in which quantum particles interact with the classical potentials, to see what the calculations does. And it happens that it gives good results. So even if flawed, it is still predictive enough to be considered good. For instance, it predicts the fine structure of ...


2

The "anomalous" Zeeman effect is the name for the Zeeman effect with more than three - symmetrically distributed – sublines of the original lines. Without the spin, one would expect 3 lines in the pattern only because the transition may change a single type of the spin $S$ only, by $\pm 1$, roughly speaking, from the spin of the photon. Quantum mechanically ...


1

I suggest that we need 'electron gas' for metals instead of 'electron liquids' because electrons in metals move much more like particles of gas than particles of liquid. The big difference between gas and liquid is that gas fills its container and the particles can all travel throughout the volume of the gas and (for an ideal gas) the spacings between ...


1

According to Grand Unification Theory, protons can decay into electron (even at low energy; just the probability is very low). It doesn't mean you can replace proton with electron.


1

The "rotating particle" picture does not work for $L=0$, unfortunately. Both the electron and the nucleus are waves who create a standing spherically symmetric wave of their relative motion.


1

Electron has mass and momentum, but we don't know anything about how the electron "moves" between the various points within that orbital. Stated another way, we can observe that the probability of finding the particle around the nucleus has a particular form (the s-orbital), and the "motion" if it could be defined for a wavelike dispersed electron in an s ...


1

Actually, you have rediscovered the flaws of Rutherford's atomic model . Now, in order to solve this, Bohr proposed postulates, one of them is: Electrons while moving in a certain orbit cannot radiate energy and can only take pre-ascertained orbits. What he said is based on the fact that Energy can't be radiated continuously. It is discrete in nature. ...


1

According to Quantum Mechanics, the electron actually has a wave function described by the Schrodinger Equation and it has different energy levels and its energy levels are quantized. At the most stable state the ground state ($n = 1$, $l = 0$, $m_l = 0$), it does not radiate energy, it only radiates energy when going between levels from higher to lower ...


1

Take for example the $n = 13$ line on the graph, which intercepts the $y$ axis at $650\text{cm}^{-1}$. Converting $650\text{cm}^{-1}$ to a wavelength by taking the reciprocal and dividing by $100$ gives the corresponding wavelength as $\lambda = 15.38\mu\text{m}$, and converting to an energy using $E = hc/\lambda$ gives $E = 1.292 \times 10^{-20}\text{J}$ or ...



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