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I understand that the frequency is the number of waves that pass a place in a given amount of time. And it is shown like this:

enter image description here

But, the frequency of waves are always shown above as if the source of wave (let’s say it is light) is continuously emitting waves. What i mean is, isn’t there a time elapse between the individual pulses of waves.

For example light emitted from an hydrogen atom, the frequency of that light would be shown as above. But when each time the electron of hydrogen atom goes to an higher state and a lower state, i think there must be a time elapse. And if you were not to take that into account, wouldn't you get the wavelength of that wave source longer.

enter image description here

Note: My English language is not very good, so if i made mistakes that made the question harder to understand, I apologise.

John Rennie
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Tuna
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    What do you mean by "time pass"? Between what parts of the wave? – Emilio Pisanty Oct 28 '17 at 11:54
  • It is probably when photon comes handy – Alchimista Oct 28 '17 at 12:02
  • @EmilioPisanty can you see the uploaded picture sir? – Tuna Oct 28 '17 at 12:05
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    I think the OP tries to ask how a finite light pulse can have a frequency. He claims that frequency is a property of an infinite wave, while it is also associated with pulses. – eranreches Oct 29 '17 at 09:16
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    @eranreches Not exactly. I think the second picture in the question explains the question but looks like it doesn’t. What i mean is when each pulse of light wave is emitted, isn’t there a time between each pulses. In my clap example, if you were to calculate the wavelength of the sound wouldn’t you get a longer wavelength than the original wavelength since you are also counting the time between each pulse of sound wave. – Tuna Oct 29 '17 at 09:59
  • @Tuna You will have more then one frequency. Try to rephrase your above question so it will be more clear and reopen. I would suggest removing the pictures and writing something like 'Suppose a light source of some wavelength blinks...'. Does this fit your question? – eranreches Oct 29 '17 at 10:09
  • @eranreches I edited the question some. Can you please now take a look. If it is still unclear, i think i will close the question because i cannot make it more clear. I’ll have to find it by myself. – Tuna Oct 29 '17 at 10:16
  • The transition of the hydrogen atom from the excited to the ground state takes quite a long time compared to the frequency of the light, so it emits a wave that is long compared to the wavelength. As a rough guide the decay takes around a nanosecond, and the frequency of the light is around $10^{15}$ Hz, so the light emitted in a single decay contains about a million cycles. – John Rennie Oct 29 '17 at 11:58
  • @JohnRennie Thank you. So how can we calculate that hydrogen’s light frequency? Do we take that decay into account? – Tuna Oct 29 '17 at 12:05
  • @Tuna: the decay doesn't produce a single burst of light in the way a hand clap produces a single burst of sound. In effect the atom oscillates between the excited and ground states and each oscillation produces one cycle of the light. So the decay produces what looks like a portion of a regular EM wave, and as I indicated above for a hydrogen atom that portion includes about a million wavelengths of the wave. – John Rennie Oct 29 '17 at 12:40
  • This will probably be too technical for you, but Emilio Pisanty describes this very elegantly in his answer to Is there oscillating charge in a hydrogen atom? – John Rennie Oct 29 '17 at 12:40
  • Thank you for your answer. I did not know that each oscillation produced one cycle of light. I think i got the picture now. – Tuna Oct 29 '17 at 12:55
  • I disagree with this closure - the question is perfectly clear to me. – Emilio Pisanty Oct 29 '17 at 12:55

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Your understanding of what constitutes a 'pulse' is flawed. If you excite an atom, then the emitted radiation will indeed be confined in time, forming a pulse, but this will take the form of a larger envelope that encases the individual oscillations at the resonance frequency.

A typical example looks like this:

Mathematica graphics

You have an initial ramp-up phase, caused by whatever is bringing the population from the ground to the excited state, followed by an exponential decay as the atom emits its energy as electromagnetic radiation. (Why an exponential? because the Einstein-coefficient formalism tells you that the rate of emission is proportional to the population in the excited state.)

Within that envelope, however, you have individual oscillations of the electromagnetic field, and those are a good map of the charge oscillations within the atom as it winds down. The frequency of the emitted light is (the inverse of) the time between successive maxima of this oscillation.

It is important to note that the diagram above has parameters that are somewhat exaggerated to be able to draw things. For most normal atomic emission lines, the pulse is much longer (compared to the carrier frequency $\omega_0$) than what I've drawn: the pulse envelope decays with a lifetime $\tau$ of a few nanoseconds, whereas the period of the internal oscillations is about a few femtoseconds, i.e. the envelope will typically fit several million oscillations of the carrier. This means that when you're calculating the frequency of the emission, as an initial approximation, you can just forget about the presence of the envelope.

If you do want to include it, however, then the usual approach is to add it as the imaginary part of the oscillation frequency: the electric field of the signal can incorporate both the trigonometric oscillations and the exponential decay as a single dependence if you rephrase it as $$ E(t) = \mathrm{Re}\bigg[E_0 e^{-i\left(\omega_0 -\frac{i}{\tau}\right)t}\bigg], $$ with a single complex frequency $\omega = \omega_0 -i/\tau$. However, that's probably too technical for you at this stage: for all intents and purposes, you just calculate the frequency by looking at the internals of the pulse, but if you do want to include the envelope, then this is how you do it.

Now, in many experiments you will excite the atom, look at the radiation, then wait a bit, and then re-excite it to produce a second pulse of radiation, wait some more, re-excite, and so on. This will then produce a train of pulses, but their separation is not called their frequency - it is called the repetition rate of the source.

Emilio Pisanty
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