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### Significance of the photoelectric effect

Posted: Fri Oct 19, 2018 11:16 pm
Could someone explain the significance of the photoelectric effect, and how to solve problems using
E(photon) - E(energy to remove e-) = E (excess) = EK (e-) = 1⁄2 mv2?

### Re: Significance of the photoelectric effect

Posted: Fri Oct 19, 2018 11:23 pm
The photoelectric effect is significant because it demonstrates that light has particle-like qualities. It established that we can consider light as photons (packets) of energy where one photon interacts w/ one electron and each photon must have sufficient energy to remove each electron. This is why, in the original photoelectric experiment, increasing the intensity of the light didn't excite the electrons in the metal surface.

### Re: Significance of the photoelectric effect

Posted: Fri Oct 19, 2018 11:53 pm
I believe the photoelectric effect's significance stems from showing how light rejects the traditional wave model, and how light is instead perceived as photons/quanta of energy . According to the wave model, if the amplitude increases then the intensity of the light wave should increase and eject more electrons. However, this hypothesis was proven incorrect. It wasn't amplitude that increased the number of electrons ejected, but rather the number of photons (the # of photons is proportional to the intensity of light). The effect shows how photons truly interact with electrons on a metal surface.

There are many ways to use such a formula, so it would vary on the problem and the data it gives you. For example, a problem could give me the energy of the photon and the velocity of the ejected electron. Then it could ask me to calculate the threshold energy needed to eject the electron. I would then plug in the values as so: E(photon, known) - 0.5(mass of electron, known from formula sheet)(velocity of ejected electron, known)^2 = (threshold energy). Or maybe the problem would not give me the photon's energy, but its frequency instead. Then I would have to use the E=h*(frequency of electron) to calculate its energy and then plug it into the formula and solve.

Other clarifying notes:

Here is a shortened interpretation of the effect from our textbook:
1) An electron requires a certain threshold energy (which varies by the identity of the metal) to be ejected by a photon of a certain frequency.

2) Once the threshold is reached, the electron is ejected immediately.

3) The electron's kinetic energy is directly related to the frequency of radiation used.

To clarify the nature of photons: one photon interacts with only one electron, increased intensity (or # of photons) increases the number of electrons ejected, and an increase of frequency/decrease of wavelength causes an increase in kinetic energy of the electron ejected (not the number of electrons ejected).

### Re: Significance of the photoelectric effect

Posted: Sat Oct 20, 2018 1:09 pm
The photoelectric effect is significant because it showed that light has both wave-like properties and particle-like properties. The finding was that increasing the intensity of the light would not result in electrons being emitted from the metal surface. Instead, you have to increase the frequency/decrease wavelength to meet and/or exceed the work function (threshold energy of the metal surface) to eject electrons. Once you meet this threshold energy, however, increasing the intensity of the light would result in more electrons being emitted (2x the intensity = 2x the ejected electrons)!

The resulting equation demonstrates the conservation of energy throughout the experiment. Incoming light energy E(photon)=hv. The work function is the threshold energy of the metal surface to eject electrons, usually given in Joules. If the energy of the photon exceeds the threshold energy, then the ejected electron will have excess energy, aka kinetic energy as its emitted. So, the greater the photon's energy is than the threshold energy, the more kinetic energy the electron will have once its ejected.