# Why does photoelectric emission not depend on intensity?

1. Dec 30, 2014

### PWiz

I understand that intensity is power per unit time and that $I = 2π^2f^2x^2_oρv$ for regular waves (I don't know if the formula applies to E.M. waves or not). What I don't understand is why electrons are only released when the electromagnetic radiation incident on the metal surface is beyond the threshold frequency.
I read the equation $E = ħf$ and understand that the energy per photon must be greater than the work function energy for the photoelectric effect to take place. When the frequency increases, the photon energy increases, but why does the photon energy not increase with an increased amplitude of the E.M. wave? Even if I ignore the rate at which energy is transferred and look on the fact that a single photon only interacts and transfers energy to a single electron, the equation (taking the atoms/electrons to be harmonically oscillating due to the wave) $E = <KE> + <PE>$ becomes $\frac{1}{2} m (2πf)^2 x^2_o$ , and since the only energy carriers of E.M. waves are photons, their energy depends on $m$,$f$ and $x_o$ too. So the energy transferred by a photon to an electron should clearly depend on the amplitude of vibration of the oscillating electron which released the particular photon while coming down to ground state (or any other lower energy configuration) as well, which clearly isn't the case, since any electromagnetic wave, regardless of its amplitude, doesn't cause photoelectric emission in a metal unless it's above the metal's threshold frequency.
Sorry but I'm still a little new to quantum mechanics, the transition from classical wave theory isn't very easy

2. Dec 30, 2014

### jonny23

intensity increase number of photons...not their energy..... one photon can eject only one electron...
so number of electron increase but no effect on kinetic energy......
energy is only proportional to frequency ..

3. Dec 30, 2014

### Bystander

Fair enough.
From where does that model/mental picture come? The energy comes from the difference in energies of the two states.

4. Dec 30, 2014

### jonny23

the energy transfer from light to electron is based on quantum theory in which photon is a discrete packet of energy with each photon having energy E=hf....
energy transfer is due to vibration in classical EM theory but not in quantum theory...

5. Dec 30, 2014

### PWiz

So what exactly determines the amplitude of the E.M. wave? $\delta \ E = E_2 -E_1$ , but the energy of an electron in a particular energy state depends entirely on its kinetic/potential energy. The energy difference between the two states corresponds to a difference in the kinetic/potential energy of the electron when it jump back down to a lower level,emitting a photon. The difference between the kinetic and potential energy eventually depends on the amplitude of electron oscillation in the two states, does it not? The oscillatory amplitude determines the energy level the electron will occupy for maximum atomic stability, so the photon energy, directly or indirectly, depends on the amplitude of vibration of the electron which emitts it. (at least that's what I can make of it, and therefore I refrained from applying the intensity formula onto E.M. waves at the beginning)

6. Dec 30, 2014

### jonny23

if you consider light as wave the energy is dependent on AMPLITUDE of vibration
but in light as a particle (quantum) energy depends on frequency..
since photoelectric effect was explained by particle nature so we take E=hf

7. Dec 30, 2014

### PWiz

@Jonny So there is no quantitative difference between the photons of a high amplitude E.M. wave and low amplitude one with the same frequency?

8. Dec 30, 2014

### Bystander

It does not. There is no oscillation between states or energy levels.

9. Dec 30, 2014

### jonny23

@PWiz the only quantitive difference will be number of photon per unit area....

10. Dec 30, 2014

### vanhees71

No, that's a often used claim in the popular-science literature (and unfortunately in some introductory chapters of otherwise good textbooks on quantum mechanics, which should never start the discussion with photons, because as massless spin-1 quantum fields these are the most complicated objects to discuss).

The photo-effect is completely explainable with semi-classical theory, i.e., you can treat the bound electron as quantum and the electromagnetic field as classical. Then you do first-order time-dependent perturbation theory to precisely get out what Bystander said in #3.

Photons can be adequately treated only with relativistic quantum field theory, i.e., QED. It's not possible to explain it correctly without the rather involved mathematics necessary to formulate QED.

11. Dec 30, 2014

### PWiz

@Bystander Okay just look at this example (made up, a tad bit long, but stick with me please):
An electron is at an energy level $K$. It is oscillating harmonically. Its Kinetic+Potential energy is therefore given by the formula $E=2π^2f^2mx^2_o$ . Let's say due to thermal heating, its amplitude of oscillation doubles to $2x_o$ . To maintain atomic stability, the electron jumps to a higher energy level $S$(It's frequency of oscillation stays constant). After a while, it jumps down to the lower energy level $K$, emitting a photon. The difference between the two energy levels is therefore, $6π^2f^2mx^2_o$ and so this is the energy that the emitted photon possesses. If the thermal heating effect (due to a collision with another electron for example) instead raised its oscillatory amplitude to $4x_0$, then the electron would jump to another higher energy level $Z$. When it goes back to the energy level $K$, the energy difference $\delta \ E$ would be $30π^2f^2mx^2_o$ . Clearly, the energy level an electron occupies depends on its oscillatory amplitude as well . The amount of increase in the oscillatory amplitude can dictate which energy level the electron should jump to, and can hence influence the value of $\delta \ E$, or the difference between the energy levels. The differences between any two particular energy levels stays fixed, but by making the electron jump to different energy levels by changing it's amplitude only, we can have different energy changes. When the electron goes back to a lower energy level, a photon will be emitted. So the oscillatory amplitude of the electron should affect the energy the emitted photon possesses. So an electron with a sufficient amplitude of vibration should cause photoelectric emission by raising the photon energy to the work function energy level of the metal on which the wave is incident, which doesn't really happen.

12. Dec 30, 2014

### Bystander

You've wandered off in the weeds on line 2. We're not dealing with microscopic charged marbles attached to springs --- delete that picture from your mind. There is nothing useful to be gained from such a model.

13. Dec 30, 2014

### PWiz

@Bystander Oh my gosh you just erased the thing on which I was basing my whole understanding off My A levels Physics book says that the electrons and protons in the atomic nuclei together can be approximated to harmonically oscillating systems with the centripetal force of attraction being represented by a hypothetical spring. Is this approximation generally unsuitable for understanding the quantum mechanics behind such phenomena? If so, what formula gives the energy that an electron possesses at any particular energy level? (Average energy taking into consideration KE and PEs)

14. Dec 30, 2014

### vanhees71

With a harmonic oscillator you have no photo effect, because the harmonic oscillator has only bound states. Of course you can absorb and emit photons by exciting/deexciting an electron in a harmonic-oscillator potential. The photo effect however describes the process of absorbing a photon by a bound electron, kicking it out of its potential into a scattering state, i.e., the electron flies away from the atomic nucleus it was bound to before the impact of the electromagnetic wave.

15. Dec 30, 2014

### Bystander

When you were in 2nd grade you couldn't subtract a larger number from a smaller. In 3rd, you could, but weren't allowed to divide 2 by 3. In 4th you got to divide 2 by 3 but couldn't take square roots of negative numbers.
For the moment, I would recommend that you stick with the "model," but understand that it will be replaced at some point with something that will answer the questions this current model can't handle.