# Tag Info

## New answers tagged atomic-physics

1

Your coloured object is absorbing light, i.e. light is changing into mechanical energy, while the atoms in the Bunsen burner are emitting light, i.e. mechanical energy is changing into light. If you have, for example, sodium atoms in a flame those atoms are continuously colliding with air molecules. The velocities of the air molecules are a function of ...

1

In Bunsen burner atoms get heated up which means they absorb energy, and go to some excited states (or even get ionized). And then they de-excite in any way they can which means radiative decay is dominant, as there is almost no other way to dissipate energy. There is no crystal lattice, and compared to luminescence rate collisions are infrequent. So yes - ...

1

So, I assumed a pair of electron and proton to behave as a dipole Classically, this is correct. It would behave as a dipole. But in quantum mechanics we don't talk about a localized electron, but about orbitals. This is because particles have an associated wavefunction $\psi(t)$, which tells you the probability of finding your particle at a particular ...

0

Here is a representation of the hydrogen atom energy levels. It displays the availabe solutions of the Schrodiner equation for an atom composed of a proton in the nucleus and an electron existing in their mutual potential. Systems stay in the minimum energy state, and for the single electron of hydrogen the minimum energy state is the n=1 state and the ...

0

I thought, if it was a possibility, then electron would constantly need to lose energy, whenever excited, at last, it would collapse into the nucleus. You seem to have forgotten that when the electron is excited, it gets energy, which is then released when it emits it. So it wouldn't collapse because energy absorption and emission are balanced.

1

Here in case of electron, it has already emitted absorbed energy as quanta. So, is it that electron losses some energy other than the energy absorbed from the source, to come down to ground state. I thought, if it was a possibility, then electron would constantly need to lose energy, whenever excited, at last, it would collapse into the nucleus. But, this ...

1

Based on some of your comments, I think what might be tripping you up is the first statement you started with: From the Bohr's atomic model, it is clear that electron can have only certain definite energy levels. and ...If suppose, we assume electron losses total energy, electron can't stay in any particular shell, as it would not have that ...

1

Reading your comment in reply to mine, I understood what you wanted to ask. This is where so many people are confused. Since we start middle school chemistry, we are taught about electron orbits which are like concentric circles. Everyone innately assumes that the 1st orbit is inside the 2nd orbit, which is inside the 3rd and so on. It isn't like that! ...

0

Even if an electron gives up energy (and the quantum rules require certain energy states in a bound system), the nuclear repulsive forces (not electromagnetic) would keep it from getting "near" to the nucleus itself. Take a look at the kind of speeds particles are accelerated to in CERN or LINAC to allow them to collide with a nucleus.

2

I will put in my two cents: The Bohr model as such can be saved by postulating standing waves for the electrons. The contrast with the Schrodinger formalism lies not only, as observed by others , that the solutions of the Schrodinger equations are more accurate and can be generalized to complicated potentials, but that the Bohr model is only one step higher ...

1

Here is my answer to this very difficult issue. In my opinion the elementary Schroedinger approach does not solve the problem of radiation. The electron still radiates when it changes its energy level and this process is not described in the elementary Schroedinger model based on the Coulomb potential only. Experiments prove that every level is not stable, ...

2

In Bohr's theory the smallest possible orbital angular momentum is $\hbar$. The measured value is $0$. On the other hand the picture developed by solving the (time independent) Schödinger equation reproduces the energy levels from Bohr's model and gets the minimum angular momentum and the angular momentum step size right (it also fives you the quantization ...

0

A potential well in the Schrodinger equation will produce energy levels similar to the energy levels of the hydrogen atom A nucleus with two protons to be neutral will need two electrons. These will be accommodated in the lowest two energy levels: because of the spin statistics of electrons they cannot occupy the exact same energy level. A nucleus with ...

2

I don't think you really understood the outcome of the discussion Is uncertainty principle a technical difficulty in measurement?. This has nothing to do with our ability to perform a precise measurement. The position and momentum of the electron simply does not exist simultaneously in a definite state. It is like trying to measure the exact day that winter ...

2

The reason Bohr's theory is considered surpassed is that Heisenberg and Schroedinger developed more powerful theories, in which Bohr orbits do not play major role. Bohr's theory works nice only for few-electron systems, like hydrogen atom or ion Li$^{2+}$. For more complicated systems like the molecule of water H$_2$O it is difficult to see how to generalize ...

2

So, is there any reason for overruling the idea of fixed orbit? or is there any thing wrong in my opinion about the concept, if so please explain, so that I would not proceed with that wrong thinking? An insurmountable problem with the Bohr atom is that one has two charged particles orbiting around each other. Electromagnetism was an exact science at ...

3

Unlike electrical or magnetic field, which acts differently on particles of different charge, for the gravitational field there is equivalence principle, which means that electrons and nuclei would experience the same acceleration due to gravity. There is, of course, the overall shift of energy level if the atom is observed from the place with the different ...

3

Your question is asking you to calculate the de Broglie wavelength of the electron, which can be obtained from its momentum $p$ via de Broglie's relation $$\lambda=\frac h p$$ where $h$ is Planck's constant. Regarding units: You should work consistently with your units. You can work with the kinetic energy in Joules, $h$ in J s and $p$ in kg m/s. But ...

3

As Mitchell says in his comment, this is related to the uncertainty principle. The uncertainty principle states that if you have some system with a position $x$ and a momentum $p$ then there is an uncertainty in the position, $\Delta x$, and an uncertainty in the momentum, $\Delta p$, related by Heisenberg's uncertainty principle:  \Delta x \Delta p ...

0

Muonic atoms should be stable in electron-degenerate matter (white dwarf material) as long as the Fermi energy is more than $m_\mu - m_e$. This is more or less exactly a analogy with neutron stability in the nucleus where the the protons are effectively in a degenerate state. Any answer has to forbid electrons (which isn't going to be possible as they share ...

2

I'm not sure what you mean by "collapse", but if I interpret that as "no hydrogen is formed" or "the electron is not captured", then 2 things can happen: 1) Elastic electron-proton scattering: the electron and proton just "bounce" off each other under some angle theta. By observing the cross section of the scattering versus the theta angle it was shown that ...

1

depends on the energy of the electron. For low energies, a bound state will be formed due to electromagnetic interaction between the two. In the case of higher energy, the proton can be transformed into a neutron.

1

Your diagram of 2 H -> He is extremely misleading. First, He has Neutrons and can't be made from just 2 H. It need deuterium and tritium and it's a multi-step process involving first making deuterium via $\beta^{+}$ decay. Also, your question and diagram seem to imply that there is a formula describing the curve you show but there isn't. Your curve is ...

2

Not really an answer, but rather some organized comments. First, you may become disappointed but the trully fundamental laws, as we know them today, are not written in terms of force laws. Even though the concept of force is still present in Physics, it is not used in the way it was before and which seems to be the way you are thinking about them. Force is ...

1

What laws (formulas) govern the fundamental forces of nature? None. Columb's law and Newton's law of gravitation classical explanations of electrostatics and gravitation,respectively. But the are no analogus formulas for fundamental interactions. They must be described in the context of Quantum field theory. QFT is to complex to give you a ...

0

Very intuitive. No maths. There is an excited state with a symmetrical probability distribution and no e/m dipole moment. There is a ground state (or less excited state) also with no dipole moment. There is a tiny probability that the excited state electron will be in the ground state that allows both states to be present at the same time producing a finite ...

Top 50 recent answers are included