New answers tagged

1

Case 1 ( Magnetic field is constant)No, while it can change velocity DIRECTION, it cant change the magnitude of velocity. That is, since force due to magnetic field is always perpendicular to direction of propogation, we infer that the speed of a moving charged particle always remains constant in a magnetic field.Case 2 (Varying magnetic field)Situation is ...


3

The simplest answer, if I did not misunderstood your question, is to adiabatically compress the gas, both pressure and temperature will raise.


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If you take a bottle of gas and carry it with you on a supersonic plane, then the molecules will go much faster without the temperature changing. If you let pressurized gas flow through a well-designed nozzle (De Laval nozzle), the gas will accelerate to supersonic velocity (i.e., faster than the original thermal speed of the molecules) while the ...


2

You can't have strong enough electric fields to tear the proton away from the nucleus but it is really a very subtle thing and the inability is just "by a little bit". The strongest electric field that may exist is given by the Schwinger limit. In $\hbar=c=1$ units, the field is $m_e^2 /q_e$. Once you reach this value, electron-positron pairs start to be ...


3

It isn't really appropriate to describe Gabrielse and Peil's experiment as an artificial atom as it's completely different to an atom. It is just a system with an electron moving in response to a potential. The system is described by the Schrodinger equation just like an atom is, and like an atom it has quantised energy levels (called Landau levels and well ...


2

The property of Caesium that makes it such a stable oscillator is the lone electron in its $6s$ orbital. All other electrons in the lower energy levels take a symmetrical electron configuration and leave the $6s$ as an "outsider". The spin of the Caesium nucleus can cause a so-called hyperfine transition in that $6s$ electron which has a very specific ...


2

Every atom, including cesium-133, emits (or absorbs) electromagnetic waves (light or its generalization to invisible colors) when the electrons jump from one state in the atom to another. The electromagnetic radiation is a periodic process in which the electric (and similarly magnetic) fields at a given point of space behave as $$ E = E_0 \cdot \cos (2\pi f ...


0

Your Hamiltonian is much the same actually. If I write $e^{i\omega t}~=~cos(\omega t) + i sin(\omega t)$ the quoted driving term is $$ H_d = \epsilon \left(a(cos(\omega t) + i sin(\omega t)) + a^\dagger(cos(\omega t) - i sin(\omega t))\right) $$ $$ = \epsilon \left((a + a^\dagger)cos(\omega t) + i(a - a^\dagger)sin(\omega t))\right) $$ You can check that ...


2

The chemical properties of an element are always determined by the atomic number, that is, the number of protons in the nucleus. All carbon atoms have six protons, all iron atoms have 26, etc. It's the atomic number which is featured prominently in the periodic table, for example. Until the neutron was discovered in 1932, this was fine. After the neutron ...


4

The most common isotope of hydrogen has no neutrons. Other isotopes are deuterium with 1 neutron and tritium, with 2 neutrons. Since virtually all (99.98% according to wiki) naturally occurring hydrogen comes in the no neutron isotope, it seems reasonable that books show a schematic of that one when illustrating hydrogen. As a secondary motivation, the one ...


0

In the second example, only the 3d shell is not completely filled. Using the aufbau principle to fill this shell with the 7 valence electrons leaves three unpaired electrons resulting in a total electron spin of $S=3/2$, so you arrive at a quarted state. The electron orbital angular momenta of the five possible states in the d orbital are -2,-1,0,1, and 2. ...


0

I found the answer I was looking for in Chemistry S.E. As you can see it is not that impossible to have a picture of what you calculate. http://chemistry.stackexchange.com/questions/51568/what-is-the-reason-why-protons-and-electrons-do-not-collide/51576#51576


-1

Possibly it can be proven in this way. Let $H_0$ denote the unperturbed Hamiltonian, and let $D_z = \sum_{i=1}^N z_i $ denote the $z$-component of the electric dipole. Define $$ H (f) = H_0 - f D_z . $$ Let $E_g(f)$ be the ground state energy of $H(f)$. First, apparently, $E_g$ is an even function of $f $. Let $|\psi(f)\rangle $ be the ground state of ...


1

No, there can't be atoms like that, at least not in the real world. In the magnetic case, diamagnetism means that the magnetic susceptibility may be negative (so the permeability may be lower or higher than in the vacuum, the magnetic susceptibility may have both signs). But in the electric case, the electric susceptibility is always positive, and the ...


1

You are unlucky, because the microworld of electrons nuclei, atoms and molecules has been studied with mathematical models for over a hundred years and it is not open to hand waving hypothesis of the type: I would say electrons are very tiny containers of energy, which can contain between a minimum and a maximum of such energy, depending on how much energy ...


6

To combine some of the comments that have already been stated, plus some extra remarks, into something more formal: The second process described in the OP is not usually called ionization (even though the end result is an ion), but rather electron capture. In general, though, this cannot happen exactly as stated for an isolated atom $A$ capturing a free ...


0

How does the matter-radiation system, on its own, goes over to the Blackbody distribution? Evolution towards equilibrium (in macroscopic sense) happens when the system matter + radiation is isolated, for example if a piece of matter is inside a cavity that slows down leakage of energy out of the system. For example, a piece of coal in a well reflecting ...


-1

There are 5 types of atomic scattering: Elastic Inelastic Fluorescence Photoelectric Compton scattering The net change in charge depends on the type of scattering. Ionization of an atom is most likely to happen through Compton scattering via X-ray production. However, wave/particle duality results in an implicit ambiguity between an electron or a ...


28

You're correct and the video is mistaken. In fact, if cesium atoms were constantly oscillating between the two hyperfine states, cesium beam clocks wouldn't work at all! In its simplest form, a cesium beam clock uses a magnet to separate a stream of atoms into two streams based on their hyperfine state; one state is selected to continue down the tube to be ...


8

Yes, they really are oscillating between two different states (not simply driven in one direction), but as you suspect they are not oscillating at the reference frequency. Rather than "just" sending radiation at the atoms to absorb, they also interact with an oscillating magnetic field (which is at the reference frequency). This field spurs some of the ...


32

The definition for the cesium clock is: 9192631770 cycles per second is frequency of the radio waves which cause maximum resonance, a physically measurable condition, in the cesium atoms. This corresponds to a particular tuning of the radio. Keeping it tuned provides the reference frequency cited.


0

By E=−Z^2RE/n2 where RE is the Rydberg energy As n increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level< Or EPE = 1/4πε( Qproton Qe-) /r, As r increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level< Thanks to everyone that helped !< I beg to differ on the above explanation provided by ...


1

By E=−Z^2RE/n2 where RE is the Rydberg energy As n increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level Or EPE = 1/4πε( Qproton Qe-) /r, As r increase, EPE becomes less -ve(i.e. more +ve) , indicating higher energy level Thanks to everyone that helped !


5

The potential energy stored in a two like charge system will increase with decrease in distance between them. While for a two unlike charge system, the potential energy decreases with decrease in distance (means potential energy gets liberated if they come close), accounting for increase in attraction. In the equation, you provided, the potential energy ...


8

The energy in a level $n$ is given by $$E = - \frac{Z^2 R_E}{n^2} $$ where $R_E$ is the Rydberg energy ($R_E = 13.6\mathrm{eV}$). Therefore, greater $n$ means lower energy (in absolute value), i.e., the electron is less bounded.


1

Why can't there be any continuous energy band in an atom? This is the basic reason quantum mechanics had to be invented. Once the existence of positive and negative charges was discovered, Maxwell's equations when solved for a planetary model of a central positive charge and an orbiting negative one, are completely unstable, in contrast to the ...


0

An electron in an atom is in a bound state. Since it is a bounded particle, can analyse the problem with particle in a one dimensional potential box. Consider a one-dimensional crystal lattice with lattice constant $L$. We assume that the particle is free to move within this distance and cannot go outside. So we have two potential barriers that tend to ...


2

This is well explained on the basis of sub shell electronic configuration. But first, let's look it by the concept of shells alone. See in the example of calcium, it has $20$ electrons. Of course the outermost shell can accommodate $18$ electrons. If it goes like $2,8,10$ then the outermost shell contains 10 electrons. The stable state is either having an ...


4

In the Schrödinger treatment of the Hydrogen atom, the excited states are true eigenstates, that is, their lifetime is infinite. This means that, if the Hamiltonian is $$ H=\frac{P^2}{2m}+\frac{\alpha}{r} $$ then $\tau=\infty$ for all the energy levels. What this description misses is the interaction of electrons with the electromagnetic field, that is, ...


1

Your intuition is correct. The error is in your calculation of the ground state energy without interactions. There are two electrons, each with energy $-4\,{\rm Ry}$, for a total of $-8\,{\rm Ry}$. The repulsion raises the energy to the experimental value.


2

Yes, the conservation of momentum is valid. The photon and the Hydrogen atom acquire equal and opposite momenta when the photon is emitted. However you must be a little careful when you are calculating the momentum as the total energy of the photon and the hydrogen atom is 10.2 eV. This means your equation will be $$ p^2/2m + pc = 10.2 eV$$ You must conserve ...


0

This answer is written from perspective that the atoms are indistinguisable and that you mean the 100-density matrix, not a reduced one. The main conclusion is that these systems can be distinguished if multiple atoms can be measured simultaneously. Assuming indistinguisable atoms, there is something in the definition of your system 2, which can lead to ...


2

Before talking about entropy, we need to discuss what possible states an atom can be in. I will start by the most general case that consists in considering a single-atom gas in a 3D box. In that case, the microstate of the atom is described by: The definite linear momentum states $| \textbf{k} \rangle$ of the atom (that are eigenvectors of the hamiltonian) ...


0

Compton scattering occurs when a photon impacts an atom, and is reflected with less energy, and at an angle. The larger the angle, the larger the share of the photons energy is absorbed. This proves that photons have a particle nature and posses quantized energy. Imagine a golf ball hitting a pole and getting deflected. If the pole is more massive, there ...


6

When a high energetic photon (like the gamma or X ray photon) hit a charged particle like an electron, due to inelastic collision, the photon loses some energy and the electron get scattered. The energy lost by the photon will be equal to the energy gained by the scattered electron. This process of inelastic scattering of electron by a photon is called ...


5

The entropy of a single atom does not make sense per se, unless you specify the preparation. The entropy of a single isolated atom, fixed at a point, is indeed not defined – the entropy is, after all, a property of an ensemble not of a system. The entropy of an ensemble of isolated atoms prepared at a specific energy, on the other hand, is well defined (this ...


0

Does it go from one state to another via straight line or it makes smaller and smaller orbits till it reaches the next orbit The electron is an elementary particle and as such, at the level of atoms, it can only be mathematically modeled by a quantum mechanical probability distribution. These spatial probability distributions for the electron of the ...


0

You can define velocity as $v=\frac{\sqrt{\langle \hat{p}^2 \rangle}}{m_e} $. For ground state $v=Z\alpha$.


1

As phrased, this question suggests suggests a very Bohr-like image of electrons. This answer is intended to nudge you in the direction of a quantum model, but is by no means complete. Honestly, there's no way to understand QM without delving deep into the math, and even then its difficult to make a math <-> reality correspondence. Firstly: quantum ...


0

Yes you can do that; the space and spin parts just have to have opposite symmetry characteristics so that the total wavefunction is antisymmetric.


0

Let's first write the density matrices. Assuming that the first atom is the excited one, we can write $$ \rho_{total}=\left|2p\right>\left<2p\right| \otimes\left(\bigotimes^{100}_{i=2}\left|1s\right>\left<1s\right| \right) $$ while in the second case we have $$ \rho_{total}=\bigotimes^{100}_{i=1}\left[{\frac{1}{ ...


3

Let's think about a system that has a two-fold degeneracy for some given energy level. That is, two states $ \psi_{a} $ and $ \psi_{b} $, both of which correspond to energy $ E_{0} $. An example would be a spin-1/2 particle with a Hamiltonian that is spin-independent. Now imagine that when we apply a perturbation, H', to the system, the degeneracy breaks ...



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