Tag Info

New answers tagged

0

Bockaste.K(1974), MEAN-VALUES OF POWERS OF RADIUS FOR HYDROGENIC ELECTRON ORBITS. Physical Review A. 9(3): p. 1087-1089, DOI: 10.1103/PhysRevA.9.1087. This is the standard ref and has a table.


0

When two atoms approach each other, there comes a point where it is no longer possible to describe the system as "two atoms". Rather, it becomes "one molecule". This is a gradual transition. When the atoms are far away from each other, they behave like separate atoms: one wave function for the atom on the left and an identical atom on the right having ...


0

This is an attempt at hand-waving over an effect that is not really explained in the statement you gave. In a periodic (lattice) arrangement of identical atoms, the wavefunctions of electrons can also be described as having the same periodicity. In fact, it is the best (in the sense of most practically useful) description for this sort of system. Then your ...


2

Anyway, the conjecture is that, possibly if the intensity of the light is strong enough, we can still get photoelectric effect with $\omega < \omega_c$? The frequency treshold for the photoelectric effect is observed with light of common intensities; even very low intensity light will work. There are ways to ionize atoms without such high-frequency ...


1

By absorbing more than one photon an electron can still be detached. However, the probability will be small. On the other hand, using a focussed intense laser beam, the situation improves. In fact, isolated atoms can be ionised in this way. The process is called multi-photon ionisation and has been verfied experimentally.


0

The number of absorption or emission lines depends on the number of photons involved in the process, not the number of electrons. There could very well be a case where a single photon transitions between two different states that involve changes in quantum numbers for two electrons. In an atom, the electrons are often tightly coupled, so it helps to think of ...


1

The ground state wave function must have a symmetric orbital wave function. Wrong (or not even wrong). The total wave-function (spin and space) must be anti-symmetric under particle interchange. But in HF we need a Slater determinant wave function. So how to handle the problem? Use a Slater determinant. There is no problem with this approach. ...


1

There is no problem. In the framework of Hartree Fock approximation you will treat the nuclei as a potential focus (not really like a particle). The particles involved are just electrons, so the wavefunction is antisymmetric under a permutation of their coordinates.


3

Suppose you start with a proton and an electron separated by a large distance. The mass of this system is just $m_p + m_e$. Now let the proton and electron fall towards each other under their mututal electrostatic attraction. As they fall they will speed up, so by the time the proton and electron are about one hydrogen atom radius apart they're moving with ...


1

Because E = mc^2, B/c^2 = m. The binding energy ends up adding mass to the system of the atom.


0

Of course the photon will have an amount of energy and that is radiation energy which if defined by Planck's equation. Radiation energy is also related directly to wavelength,frequency and wavelength number.The higher the energy, the less is the wavelength ( all this from Planck's equation). What Heisenberg equation is about, does not include the energy.It ...


1

It is possible to set up experimental conditions which allow you to take a ordinary photograph of a single atom in visible light. This works by building an electromagnetic trap (in particular a magneto-optical trap); getting some atoms in there; shaking them gentle to boil atoms away until only one remains; then exciting it to fluoresce. The linked article ...


0

If you mean that you want to observe some physical material using only visible light (e.g. through an insanely well-crafted light microscope), and discern where the invividual atoms in the material are, then no, I don't believe that it is possible. The reason is that you can usually only see objects that are roughly as small as the wavelength of the light ...


2

We see things because some light gets bounced off them, and this "bounced" light is due to electrons jumping from higher energy states to lower energy states. That is not the way we "see" things. Photons getting absorbed and re-emitted is not the most probable interaction of photons with atoms, particularly of visible light. Photons interact with the ...


4

This is could be considered chemistry question and might receive a more interesting response on Chem.SE. But it's probably fine here. According to ChemWiki at UC Davis, "Atomic size gradually decreases from left to right across a period of elements. This is because, within a period or family of elements, all electrons are added to the same shell. However, ...


2

As you said each line in the spectrum represents a photon of a specific wavelength. The photon is simply the form of energy released when an electron transitions from a higher energy level to a lower one. As the precise energies of the energy levels are discrete (as opposed to bring energy bands, so to speak) the possible differences between them are also ...


0

So, one by one: how different radiations can be produced due to the excitation and relaxation of its one and only electron? The excitation of the electron (by photoelectric absorption of a photon, or by Compton scattering) is to different energetic levels. In stars, at the huge temperatures the electron of H can be raised to any high energy level of ...


0

Okay, so even if you have one electron, there are lots of different energy states it can be in. (If you're still at the most basic level, think that there are a lot of allowed "orbits" of different energies.) Photons can be absorbed or emitted if they have an energy equal to a difference between two energy states. If an atom has a lot of electrons, then ...


0

A classical is Bransden, Joachain - Physics of Atoms and Molecules, a one-thousand tome covering a lot of basic material. Another good book is Atkins, Friedmann - Molecular quantum mechanics. This one is at a slightly lower level than Bransden, and also contains reviews of basic concepts of quantum mechanics (it's pretty much self-contained). It has a lot ...


0

Some good books on atomic physics are: The classic "Atomic physics" by Max born. Atomic Physics by J Foot. The Feynman's lectures volumes. Introductory Nuclear Physics by K S Krane. Moreover there is a series of "very short introductions" books, those are good too for a pre insight on a subject of interest.


1

The answer to your question comes from math, not from physics. Consider the wave equation in 1+1 dimensions: $$\partial^{2}_{x}\phi(x,t) - \partial^{2}_{t}\phi = 0$$ One obvious way to solve this is to convert the PDE in two variables into two ODEs in one variable. This is done by assuming that the solution can be written as some sum $$\phi(x,t) = ...


1

The quantum numbers serve to enumerate the solutions to the time-independent Schrodinger equation (Energy eigenstates) which are also eigenstates of the angular momentum (squared) operator and have a definite component of angular momentum in some direction labeled $ z $. When solving the problem, it just turns out to be handy to label the solutions with the ...


80

This entirely depends on what you mean by "see". Let me start of by noting: As per my knowledge, atoms are small beyond our imaginations No. Atoms are quite big compared to certain other things we play around with, like its constituents (protons, electrons) in particle accelerators. The size of atoms is of the order 0.1 nanometres (of course, there is ...


0

why doesn't the electron fall into the nucleus if it is emitting radiation (accelerating charge emit radiation) . i have come to know they emit E=h*frequency waves, or does it get any energy? If the electron orbiting the nucleus is in a stationary state, i.e., a state of fixed energy as defined by a solution to the stronger equation $$ \hat H \Psi = ...


0

The easiest way to see that $p^4$ is spherically symmetric is to view it in momentum space. If you apply $p^4$ to a momentum eigenstate $|p\rangle$ the result clearly only depends on the magnitude of the momentum vector of the state and not on its orientation, so if $R$ is a rotation operator we have \begin{equation} p^4 R|p\rangle = Rp^4|p\rangle ...


0

The following refs show that the eq.2 is used in Universities : IMO the eq.1 is not reducible to the eq.2 . In page 21 of Atomic structure lecture (Cambridge) $$\Delta\ E= \frac{ m_e c^2}{2} \left (\frac{Z\alpha}{n}\right )^4 n\times \frac{8}{3}\ g_N \frac{m_e}{m_p} \frac{1}{2}\begin{Bmatrix} I\\ -I-1 \end{Bmatrix} \propto m_e$$ Splitting of the ...


-1

all d-block elements will have their last electron in 'd' orbital. thus even if it is in the 5th period it has no 5th shell as it first fills the 'd' block in 4th shell



Top 50 recent answers are included