# Frequency of an Electron

My question is very simple. If frequency is defined as the cylces per unit time, Then what is meant by "Frequency of an Electron" ? If the rotation of electron around a nucleus is considered then, Which phenomena we consider for a free electron i.e: electron in a force field (unstable alone electron).

EDIT: Is "Frequency of an Electron" a experimental quantity?

My teacher told me how to calculate the frequency of an electron. We started from finding Energy of Electron then difference in energy, then we get this equation according to Bohr's Radius of Hydrogen Atom and his postulates: i.e:

$$f = \frac{z^2e^42\pi^2m}{h^3} (\frac{1}{n_1^2} - \frac{1}{n_2^2})$$

Where:

• $z =$ Atomic number
• $e =$ Charge on Proton
• $m =$ mass of electron
• $h =$ Plank's constant
• $n =$ Orbit number

From the last part of my equation, I am confused. The $n_1$ and $n_2$ shows that, that frequency will be the frequency of energy? or electrons?

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I don't think the term "frequency of an electron" has any intrinsic meaning. You would need to consider the context to work out what it means. Can you give us a link to the document where you found the phrase? –  John Rennie Feb 12 '13 at 16:32
Hi @devWaleed: Instead of flagging your question for deletion as you did, you can delete it yourself. –  Qmechanic Feb 12 '13 at 16:39
@JohnRennie I thought, "Frequency of an Electron" is a property or quantity which is available. But if you say there is no such thing, Now I am clear now. -Thanks. –  devWaleed Feb 12 '13 at 16:41
@devWaleed: Do you mean the Compton frequency of the electron or the Rydberg frequency of the Hydrogen atom? –  Qmechanic Feb 12 '13 at 17:08
Frequency is a physical thing, but is hard for our feeble minds to interpret. If psi = e^(i(kx - wt)), then the electron will oscillate through time and space, and will oscillate in the complex plane. Impossible to visualize, but if you combine electrons with different such phases, one can calculate and observe physically destructive and constructive interference occurring, after one looks at the observable quantity, the amplitude. –  Mew Feb 12 '13 at 23:29

Since you used the tag , I imagine you mean the frequency $f$ that corresponds to an electron's energy $E$ via Planck's relation, $$E=hf,$$ where $h$ is Planck's constant. That is a valuable question and nothing to get picked on for. After all, if the electron is a wave with wavelength and so on, it surely has a frequency, right?

It turns out that this frequency is not very easy to measure. The reason for this is that the electron "wave" is usually complex-valued. That is, the thing that oscillates is a complex number $\psi=a+ib$, usually called its wavefunction. The real and imaginary parts of this wavefunction "rotate" into each other: $\psi$ will be real, then imaginary, then negative real, then negative imaginary, then real again, and so on and so on, in a continuous fashion. The frequency you're asking about is the frequency at which this happens.

Unfortunately, we are only ever able to directly measure the modulus of $\psi$, i.e. quantities of the form $|\psi|^2=a^2+b^2$, and this is constant even though $a$ and $b$ are oscillating. Schemes to try and measure $\psi$ in some (indirect) way are some of the most interesting measurements in quantum mechanics.

In this case there is a second problem which is also quite interesting, and it is the fact that only differences in energy can have physical meaning. Thus to ever measure the frequency$\leftrightarrow$energy of a particle, then we need to compare it with a second particle with a different frequency$\leftrightarrow$energy, and then measure the difference in frequencies$\leftrightarrow$energies. This will be present as a "beat" in the wavefunction, as we add together two complex numbers that are rotating at different frequencies, and it is in principle possible (though damned hard!) to measure.

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I am not sure I understand your question clearly, but here are some ideas trying to cover as many cases as possible:

For the electron in the first Bohr orbit in the hydrogen atom: The frequency of its rotational motion is the number of times it will rotate around the proton in one second, and it isabout

$f=6.58\times 10^{15} s^{-1}.$

In a uniform magnetic field: For an electron that has entered a uniform magnetic field of flux density B, depending on the speed $v$ of the electron, the magnetic field can put it into a circular orbit with frequency that can be found using these two equations

$Bev=\frac{mv^2}{r}$

which is the balance equation between the magnetic and centripetal forces, and

$v=2\pi fr$

which is from the circular motion of the electron at uniform speed $v$. These two lead to the equation

$f={\frac {Be}{2\pi m}}$.

For an electron in a piece of wire: carrying an electric current of frequency 50Hz say, it means that the electron oscillates at 50Hz (i.e. goes back and forward, and it does this 50 times per second.)

For a free electron: The frequency is of quantum mechanical nature. It relates to the the wave function of the electron

$\psi (x)=u(p)e^{i({\bf p.r}-Et)/h}$.

Note that in the above equation $E/h$ is the frequency of rotation of the phasor (the exponential part), it does not mean the electron goes back and forward so many times per second. So the larger the energy the larger the frequency of rotation of the phasor, hence of the wave function of the electron. For a relativistic electron, the energy is

$E=c\sqrt{p^2+m_o^2c^2}$

so that the frequency is given by

$f=c\sqrt{p^2+m_o^2c^2}/h$,

hence the origin of the more general part $\hbar\omega t$ of the phasor (in the wave function) representing an electron.

I hope this helps.

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ok, you understood my question what I was asking but the real question is that, If a particle is vibrating at 50 herts, It means it will go back and forth 50 times in a second. Then what is meant by frequency of electron? Does electron vibrate? or its revolving around the nucleus is considered as its frequency? –  devWaleed Feb 12 '13 at 17:35
@devWaleed Well, still I am not sure I understand your question completely, but I have edited my answer to cover as many possibilities as possible, and you need to decide which one of these fits your real question. –  JKL Feb 12 '13 at 19:32