2
$\begingroup$

With the development of quantum mechanics, it was found that the orbiting electrons around a nucleus could not be fully described as particles, but needed to be explained by the wave-particle duality. In this sense, electrons do not orbit the nucleus in the sense of a planet orbiting the sun, but instead exist as standing waves. The lowest possible energy an electron can take is therefore analogous to the fundamental frequency of a wave on a string. Higher energy states are then similar to harmonics of the fundamental frequency.

I got a question here, if electrons exist as standing waves (matter waves), can they get superposed by non-mechanical wave such as electromagnetic wave? My book says that, the wavelength of an electron with mass $9.11\cdot10^{-31}Kg$ and moving with the velocity of $10^6m/s$ has the wavelength of about $7.28\cdot10^{-10}m$. The wavelength associated with the moving electron is said to be of the same order of magnitude as of $X$-rays, which can be easily measured. If assume that electron gets superposed with electromagnetic radiation, I thought by superposition, electron would loss its existence as standing wave (I didn't get any source to support this view, if it is wrong, please explain).

Actually atoms get incident by electromagnetic radiation at every instant, if we assume electrons to get superposed by electromagnetic wave, electron can't exist as stationary wave, but according to schrodinger model, electron is said to exist as stationary wave. I don't know whether I have gone wrong any where or is it that electron can't exist as stationary wave? if any is the case please explain.

standing wave
[standing wave]

$\endgroup$
4
$\begingroup$

Several concepts are tangled up in your question.

The electrons in their orbitals are a stable solution to a potential problem, the potential supplied by the charges in the problem. The orbitals describe the probability of finding the electron in a position (x,y,z) around the nucleus.

How can one see the electron orbital? In this link you can see that it is not a simple matter, but it has been done, by disturbing the electron with photons and getting statistically the original location.

Once one has seen/measured the electron's location the solution of the potential no longer holds for this free electron. The probability of finding it can be described by a traveling plane wave, until it is caught by another potential and radiates down to a ground state or an energy level that was empty in an ionized atom.

Actually atoms get incident by electromagnetic radiation at every instant, if we assume electrons to get superposed by electromagnetic wave, electron can't exist as stationary wave, but according to schrodinger model, electron is said to exist as stationary wave.

A photon hitting a bound ( probability standing wave) electron can expel it from the atom . A photon of appropriate energy can kick it up to a higher energy level , the photon disappears and the electron falls back to ground state by emitting that special spectral line. It is true that atoms, and this means the electron cloud about the nucleus, are continuously hit and interacting with photons, but the grand majority of photons do not have enough energy even to ionise the atoms, let alone free the electron.

The photons in our environment are bounded in energy to first order by the black body radiation of the earth and by the spectrum from the sun which supplies the energy to the surface of the earth. The high tail from the sun which can ionize atoms and thus is dangerous is mostly cut off by the atmosphere, our shelters and clothes.

So the only two outcomes on atoms from the "superposition" you imagine are either change of energy level and disappearance of photon, or ionization of atom and change in energy of photon. This last does destroy the standing wave probability function of the electron to a plane wave one.

A free electron hit by a photon can either scatter elastically or inelastically, but a free electron is no longer described by a probability standing wave, just by a plane wave propagating ( again , a probability wave), which is also a solution of Schrodinger's equation in the absence of a potential.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.