What's the (Electron) Frequency, Kenneth?

[gareththegeek]

I was wondering what frequency a ground state hydrogen atom's electron should have?

I read somewhere that it has 13.6eV of energy but I think this is the energy required to release it from orbit. I tried to examine the 13.6eV value to check if it seemed correct using the following method. Can anyone tell me where I've gone wrong? It gave me a value of 14.5nm for the Bohr radius (apparently its meant to be more like 0.05nm)

$$E = h\nu$$

$$\nu = \frac{c}{\lambda}$$

Assuming quantum number n=1 then circumference of circular orbit =wavelength

$$\lambda = 2 \pi r$$

$$E = \frac{hc}{2 \pi r}$$

$$r = \frac{\hbar c}{E}$$

If I feed in the value E = 13.6eV I get r ~ 14.5nm (which is wrong).

Thanks,
G

 

[gareththegeek]

I think I see the problem, 13.6eV is the energy required by the electron (in addition to its rest state energy) in order to free the electron from orbit about the nucleus.

Taking Bohr radius to be equal to 5.2917720859x10^-11 we have

$$E = \frac{\hbar c}{r}$$

$$E_{rest} \approx 3731eV$$

$$E_{free} \approx 3731+13eV = 3745eV$$

This seems like a disproportionately large amount of energy but I shall continue

$$\nu = \frac{E}{h}$$

$$\nu_{rest} \approx 9.02x10^{17}hz = 902 Phz$$

$$\nu_{free} \approx 9.06x10^{17}hz = 906 Phz$$

Is that right?

 

[f95toli]

No
The electron is a particle, and particles do not have "frequencies".
So the question does not really make sense.

 

[gareththegeek]

Thanks for the response.

I was under the impression that de Broglie hypothesised that electron might exhibit both wave and particle nature and used the electron as a wave to explain the discreet energy levels held by electrons. The electron would only orbit in integer multiples of its wave length so as not to destructively interfere with itself. The integer multiple being the principal quantum number.

Can you point me in the direction of an article which explains why the above is incorrect?

G

 

[f95toli]

Well, de Broglie wasn't exactly wrong. But today most people would say that his ideas were a bit naive. Remember that he was working at the time when quantum mechanics was still in its infancy (he was one of the people who helped create QM).

Anyway, the modern view is that an electron is "something" which exhibits both wave-like and particle-like behaviour. Note also that this is true for everything, not only electrons. "Classical" particles in the Newtonion sense do not exis.

But, getting back the original question. You can't really think of an electron as having a "wavelength" when it is in a bound state in an atom (although it can of course be described by a wavefunction).

 

[gareththegeek]

So isn't the wavefunction just a probability field for where the wave can be found? In a hypothetical world couldn't we take the frequency of an electron following the most probable orbit given by the wavefunction and say that it is comparable to the above calculations? Or am I on the wrong path there?

Thanks

 

[gareththegeek]

I noticed this thread https://www.physicsforums.com/showthread.php?t=294836"

So my mistake is to use equations for a massless particle on a massive particle.

$$\lambda = \frac{h}{mv}$$

$$\nu = \frac{v}{\lambda}$$

$$\lambda = \frac{h}{mv} = \frac{v}{\nu}$$

$$\nu = \frac{mv^{2}}{h}$$

$$E = mc^{2}$$

so for a photon where v = c

$$E = h\nu$$

However the electron cannot have a v = c so my original calculations were wrong.

Is this correct?

 

[f95toli]

Yes, your original calculations were wrong.
But that is sort of irrelevant since your original question is "wrong" as well.

Yes, the wavefunction has -to keep it simple- to do with where you are most likely to find the electron; but there is no "frequency" associated with the wavefunction. If you want you can think of the electrons as being "smeared out" in a ring around the nucleus so nothing is oscillating (and even the p- d- etc shells where there IS an angular dependence the wavefunction is static, no time dependence means no frequency).

Also, E=hv is, is as you say only valid for massless particles.

 

[Matterwave]

To obtain the Bohr radius by assuming 1 full deBroglie wavelength of the electron must constitute the circumference of the orbit, you can try:

From the fact that 1 wavelength will fit in the circumference of 1 circular orbit:
$$\lambda=\frac{h}{m_e v}=2\pi r$$

$$r=\frac{\hbar}{m_e v}$$

From properties of circular orbits (e is elementary charge, k is the Coulomb constant):
$$m_e\frac{v^2}{r}=\frac{ke^2}{r^2}$$

$$r=\frac{ke^2}{m_e v^2}$$

Combining equations:
$$r=\frac{\hbar}{m_e v}=\frac{ke^2}{m_e v^2}$$

$$v=\frac{ke^2}{\hbar}$$

Finally, plugging v back in:
$$r=\frac{\hbar^2}{m_e ke^2}$$

This is indeed the Bohr radius.
If you plug in all the accepted values, you should get:
$$r=5.3\times 10^{-11}$$This model of the hydrogen atom is very elegant imo. So elegant, it's a shame it's not quite right.

 

[Matterwave]

I forgot to get to your original question. If we model your hydrogen atom this way (indeed, it is not quite right), then the electron is whizzing around at ~2 million m/s (plug in the values yourself to check).

We can take the period of revolution to be:
$$T=\frac{2\pi r}{v}$$

Then, the frequency of revolution is:
$$\nu=\frac{v}{2\pi r}=\frac{ke^2}{2\pi\hbar a_0}=\frac{ke^2}{ha_0}$$

Where the a_0 is the Bohr radius.
Plugging in the values, I get $$\nu=6.6\times 10^{15} Hz$$

Indeed then, if we took this model of the atom at face value, we should expect the electron to be always emitting UV rays...which it obviously isn't! If the electron really were spinning around like this, it would lose energy in the form of EM radiation (since it's always accelerating), and it would crash into the nucleus.

The electron doesn't do this.

 

[gareththegeek]

Fascinating, thanks. That has cleared up some of the material I have read online. Taking the wavefunction model does the first orbital's "average" (or, rather, most probable) radius come out to being comparable to the Bohr radius?

As for the wavefunction not having a frequency doesn't it derive from

$$\Psi(x, t) = e^{i(\varphi(x, t))}$$

$$\varphi$$ being the phase of the wave using the angular frequency as one argument?

$$\varphi(x, t) = kx - \omega t$$

where $$\omega$$ is the angular frequency and

$$E = \omega \hbar$$ is used?

 

[SpectraCat]

Fascinating, thanks. That has cleared up some of the material I have read online. Taking the wavefunction model does the first orbital's "average" (or, rather, most probable) radius come out to being comparable to the Bohr radius?

As for the wavefunction not having a frequency doesn't it derive from

$$\Psi(x, t) = e^{i(\varphi(x, t))}$$

$$\varphi$$ being the phase of the wave using the angular frequency as one argument?

$$\varphi(x, t) = kx - \omega t$$

where $$\omega$$ is the angular frequency and

$$E = \omega \hbar$$ is used?

The phase of an eigenstate does indeed have a time-dependent oscillation with a "frequency" given by E/hbar. However, I am almost certain that the phase of a wavefunction is not an observable quantity in physics, so this frequency doesn't have the meaning you seem to want to ascribe to it. There are experimental observations, such as quantum beat patterns, that can be understood in terms of the relative phase difference between different quantum states, but AFAIK there is no way to measure the absolute phase of a single eigenstate.

 

[f95toli]

As for the wavefunction not having a frequency doesn't it derive from

No, it comes from the fact that the wavefunction is static in this case; it is not changing in time (the fact that the phase has time-dependence does not change the fact that it is a static solution).
Also, note that the word "wavefunction" does not imply something that looks like waves in e.g. water; it just refers to the fact that the SE belongs to a class of equations known as wave-equations.
The angular part of the solutions in this case are spherical harmonics whereas the radial parts are polynomials (Laquerre polynomials); neither is"wavy".

 

[gareththegeek]

I guess I am confused by how quantum mechanics began (and I have only read and tried to understand early quantum mechanics at this point) by examining electrons &c using wave-particle duality and deriving the equations from treating particles as waves but went on to a point where frequency doesn't really exist or has no physical significance.

As for the idea that the wave function is static (non-time dependant); could I in any way think of this as analogous to a standing wave or am I yet again on the wrong path :P

Thanks.

 

[PhilDSP]

Then, the frequency of revolution is:
$$\nu=\frac{v}{2\pi r}=\frac{ke^2}{2\pi\hbar a_0}=\frac{ke^2}{ha_0}$$

But just to be clear there's no relationship there to the "matter wave" that de Broglie postulated for the electron which is much, much higher in frequency and was experimentally verified many times over.

To clarify further, de Broglie's theory is actually a bit more sophisticated than most physicists at the time realized. And the situation may be similar today even. He did not postulate that the electron can be evaluated as either a wave or a particle (depending on the observational technique) but rather as both wave and particle simultaneously. If you disregard that, his concept doesn't quite work.

 

[conway]

The angular part of the solutions in this case are spherical harmonics whereas the radial parts are polynomials (Laquerre polynomials); neither is"wavy".

They're actually pretty wavy.

You can argue if you like that the electron isn't a wave, but the functions that describe it's distribution are wave functions.

 

[SpectraCat]

No, it comes from the fact that the wavefunction is static in this case; it is not changing in time (the fact that the phase has time-dependence does not change the fact that it is a static solution).

I don't understand the above ... the probability distribution of an eigenstate is time-independent, the wavefunction certainly isn't. The phase is time-variant, and you can't just ignore it. If there were no time-dependence of eigenstates, then you could create arbitrary new eigenstates with any energy you like simply by making arbitrary linear combinations of non-degenerate eigenstates, and of course that is not possible. Such creations are solutions to the time-dependent SE, but they are not eigenstates (solutions to time-independent SE).

Also, note that the word "wavefunction" does not imply something that looks like waves in e.g. water; it just refers to the fact that the SE belongs to a class of equations known as wave-equations.
The angular part of the solutions in this case are spherical harmonics whereas the radial parts are polynomials (Laquerre polynomials); neither is"wavy".

Gotta disagree with that last statement too ... Laguerre polynomials are oscillatory functions with nodes ... that seems like a pretty good description of "wavy" to me. The same argument holds for the spherical harmonics ... in fact, spherical harmonics are classical descriptions of surface waves on spherical particles.

 

[SpectraCat]

I guess I am confused by how quantum mechanics began (and I have only read and tried to understand early quantum mechanics at this point) by examining electrons &c using wave-particle duality and deriving the equations from treating particles as waves but went on to a point where frequency doesn't really exist or has no physical significance.

The phase of the wavefunction definitely has physical significance, even if it is not a directly measurable quantity (see my response to f95toli above).

As for the idea that the wave function is static (non-time dependant); could I in any way think of this as analogous to a standing wave or am I yet again on the wrong path :P

Thanks.

You are definitely on the right path ... a wave-function is a standing wave ... however, even in the classical case, standing waves have time-dependent characteristics (such as phase).

 

[f95toli]

I don't understand the above ... the probability distribution of an eigenstate is time-independent, the wavefunction certainly isn't. The phase is time-variant, and you can't just ignore it.

Yes, you are right that the phase varies in time and of course that means that the wavefunction is strictly-speaking time-dependent. However, remember that the OP was about the "frequency" of the electron; a time dependent phase does in itself not mean that an electron is really "oscillating" in any way; so from this point is view it is a static problem.
Also, I was trying to keep things as simple as possible, so I didn't really see the point in making the distinction between the wavefunction and the probability distribution at this "basic" level.

Btw, yes Laguerre polynomials are oscillatory functions but I would not consider them to be "wavy" since they are not periodic, And again, I don't think you could assign a single "frequency" to a phenomena described by a Laguerre polynomial even if it WAS time dependent.

 

[SpectraCat]

Yes, you are right that the phase varies in time and of course that means that the wavefunction is strictly-speaking time-dependent. However, remember that the OP was about the "frequency" of the electron; a time dependent phase does in itself not mean that an electron is really "oscillating" in any way; so from this point is view it is a static problem.
Also, I was trying to keep things as simple as possible, so I didn't really see the point in making the distinction between the wavefunction and the probability distribution at this "basic" level.

Btw, yes Laguerre polynomials are oscillatory functions but I would not consider them to be "wavy" since they are not periodic, And again, I don't think you could assign a single "frequency" to a phenomena described by a Laguerre polynomial even if it WAS time dependent.

Fair enough ... I just wanted to help the OP understand that the fact that non-degenerate eigenstates have phases that oscillate at different frequencies, and that is really the only sensible way to talk about "frequencies" of bound states IMO. I certainly agree that the electron cannot be oscillating spatially in a classical sense, for the reasons already presented in this thread (i.e. radiation, relaxation, crashing into the nucleus, panic, mayhem, etc. ).

 

[Matterwave]

But just to be clear there's no relationship there to the "matter wave" that de Broglie postulated for the electron which is much, much higher in frequency and was experimentally verified many times over.

To clarify further, de Broglie's theory is actually a bit more sophisticated than most physicists at the time realized. And the situation may be similar today even. He did not postulate that the electron can be evaluated as either a wave or a particle (depending on the observational technique) but rather as both wave and particle simultaneously. If you disregard that, his concept doesn't quite work.

Actually, why isn't there a relationship? If we look at the deBroglie relation:

$$\lambda=\frac{h}{p}=\frac{h}{mv}$$

$$\nu\lambda=v$$

One could see that simply setting lambda to be 2*pi*r, we get the exact same formula for both frequencies. The circumference of the circle is exactly lambda, and the velocity we use in both relations is obviously the same (just the velocity of the electron).

We see from the above relations that:
$$\nu=\frac{mv^2}{h}$$

When I plugged in the values for this, I got exactly what I should have got before of 6.6*10^15 Hz. Incidentally, I found my previous calculation was wrong by a factor of ten. I should have previously stated a frequency of 6.6*10^15 Hz instead of 6.6*10^16 Hz (just a mis-calculation on my part).

Indeed, if the electron flying around 1 orbit must complete 1 wavelength, then there can be no other frequency for this system.

How would you find the frequency of the matterwave some other way?

 

[f95toli]

Indeed, if the electron flying around 1 orbit must complete 1 wavelength, then there can be no other frequency for this system.

But the electron is not "flying around", you can't think of an electron as a discrete particle in this case. The problem is that you are trying to apply classical physics to a situation where it does not apply.

 

[Matterwave]

I'm not applying classical physics to a situation where it doesn't apply. I'm just going through with the implications of a specific atomic (orbital) model.

The implications within the model itself should be consistent, even if the model is not reflective of nature.

I have mentioned several times that this model is not the correct model.

No where have I said that the electron is ACTUALLY orbiting the nucleus at such speeds and at such frequencies.

 

[LostConjugate]

I have to agree with conway. The electron may have particle properties but it has wave nature. Even though the wave function is in an infinite dimensional LVS and is not an observable it still shows that all observables have to be described by wave mechanics. So something wave-like is happening under the hood.

I can't say that the electron is made up of any sort of observable waves, but I can say for sure that it is no particle. I know I already said this but how can the ground state of the electron in a hydrogen atom have zero angular momentum if the electron is a particle?

 

[zincshow]

... but I can say for sure that it is no particle. I know I already said this but how can the ground state of the electron in a hydrogen atom have zero angular momentum if the electron is a particle?

Why not? lots of things have zero angular momentum?

 

[PhilDSP]

"Also, note that the word "wavefunction" does not imply something that looks like waves in e.g. water; it just refers to the fact that the SE belongs to a class of equations known as wave-equations.
The angular part of the solutions in this case are spherical harmonics whereas the radial parts are polynomials (Laquerre polynomials); neither is"wavy".

Gotta disagree with that last statement too ... Laguerre polynomials are oscillatory functions with nodes ... that seems like a pretty good description of "wavy" to me. The same argument holds for the spherical harmonics ... in fact, spherical harmonics are classical descriptions of surface waves on spherical particles.

Maybe the difference is that the wave is not a "traveling" wave but one that purely undergoes dissipation. As an analogy it would be more of a wake or a shadow than a wave. There's at least one chapter in Maxwell's enormous treatise where he says that all movement of charge is accompanied by the counter-balanced movement of opposite charge, in this case supplied by the medium - whatever that might be.

 

[PhilDSP]

We see from the above relations that:
$$\nu=\frac{mv^2}{h}$$

When I plugged in the values for this, I got exactly what I should have got before of 6.6*10^15 Hz.

Okay, I think you're right. I had skimmed the post and thought that the frequency you were calculating was the frequency of the orbit.

One thing that seems a mystery is whether an electron in a low energy state in a simple atom, especially the hydrogen ground state, stays intact with a definite spatial center but moves chaotically or whether it dissolves in some sense to become an energy cloud. Before that question can be answered we may need a better model and understanding of exactly what an electron is comprised of.

 

[gareththegeek]

You are definitely on the right path ... a wave-function is a standing wave ... however, even in the classical case, standing waves have time-dependent characteristics (such as phase).

Yes, you are right that the phase varies in time and of course that means that the wavefunction is strictly-speaking time-dependent.

These two observations seem to be at odds.

a time dependent phase does in itself not mean that an electron is really "oscillating" in any way

So what does the phase mean in this case?

Many thanks!
G

 

[f95toli]

These two observations seem to be at odds.

Not really, remember that the wavefunction does not necessarily represent anything "physical" in QM (whether or not wavefunctions "exist" depends on what interpretation of QM you prefer); it is not until we take the absolute value of the wavefunction squared that we get "real" (or at least measurable) properties of the system. When you do this here the phase -and all time dependence- disappears.

Note that this is what you would expect, the "in" data for this problem is a particle in an electrostatic potential, so there is no way you could get a time dependent solution where something undergoing spatial oscillations (strictly speaking you get a stationary state solution to the SE).

So what does the phase mean in this case?

In this case I would say it does not really represent anything physical. There ARE cases in QM where phase IS important, but this is not one of them.

 

[SpectraCat]

Not really, remember that the wavefunction does not necessarily represent anything "physical" in QM (whether or not wavefunctions "exist" depends on what interpretation of QM you prefer); it is not until we take the absolute value of the wavefunction squared that we get "real" (or at least measurable) properties of the system. When you do this here the phase -and all time dependence- disappears.

Note that this is what you would expect, the "in" data for this problem is a particle in an electrostatic potential, so there is no way you could get a time dependent solution where something undergoing spatial oscillations (strictly speaking you get a stationary state solution to the SE).



In this case I would say it does not really represent anything physical. There ARE cases in QM where phase IS important, but this is not one of them.

Yup ... that's about the size of it!

 

[gareththegeek]

it is not until we take the absolute value of the wavefunction squared that we get "real" (or at least measurable) properties of the system. When you do this here the phase -and all time dependence- disappears.

The squared value of the wavefunction represents a particular measurement and a measurement must be taken at a certain time, so wouldn't the time dependence be implied? The measurement shows the state of the system at a given time, reducing the probability of any other (mutually exclusive) state to zero.

Is this correct? (I am sorry if I am being slow. I am trying my utmost to understand all that is said here and it is a little complicated ;p).

 

[f95toli]

The squared value of the wavefunction represents a particular measurement and a measurement must be taken at a certain time, so wouldn't the time dependence be implied? The measurement shows the state of the system at a given time, reducing the probability of any other (mutually exclusive) state to zero.

Is this correct? (I am sorry if I am being slow. I am trying my utmost to understand all that is said here and it is a little complicated ;p).

No, there is no reference to a particular measurement in the formalism. What you get is simply the expectation value, there is no time dependence.

If you were to e.g. drive the atom with an electromagnetic field you could start to induce transitons between levels, in this case the problem become time-dependent which in turn means that you also get time-dependent expectation values (look up e.g. "Rabi oscillations").

 

[gareththegeek]

So do you mean that the expectation value is the probability of that measurement regardless of time because the electron is in a fixed pattern about the nucleus?

 

[f95toli]

Sort of, but this is a general "feature" of all systems of this type.

The same thing is true in e.g. the artificial quantum wells that are used as semiconductor lasers; the levels are stationary so nothing changes over time, which is why the frequency of the laser is -to first order- constant.