# What's the (Electron) Frequency, Kenneth?

1. Mar 3, 2010

### 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

2. Mar 3, 2010

### 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?

3. Mar 3, 2010

### f95toli

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

4. Mar 3, 2010

### 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

5. Mar 3, 2010

### 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).

6. Mar 4, 2010

### 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

7. Mar 4, 2010

### gareththegeek

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?

Last edited by a moderator: Apr 24, 2017
8. Mar 4, 2010

### 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.

9. Mar 4, 2010

### 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.

10. Mar 4, 2010

### 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.

Last edited: Mar 4, 2010
11. Mar 4, 2010

### 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?

12. Mar 4, 2010

### SpectraCat

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.

13. Mar 4, 2010

### f95toli

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".

14. Mar 4, 2010

### 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.

15. Mar 4, 2010

### PhilDSP

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.

16. Mar 4, 2010

### conway

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.

17. Mar 4, 2010

### SpectraCat

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).

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.

18. Mar 4, 2010

### SpectraCat

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

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).

19. Mar 4, 2010

### f95toli

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.

20. Mar 4, 2010

### SpectraCat

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. ).