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So from today's lecture, from my understanding as opposed to normal waves, intensity of a photon merely indicates number of photons bombarding a given object but not necessarily correlative to its energy (which instead relies on frequency/wavelength) which is against the wave model, which usually indicates intensity does equal energy. What property of photons makes this so?
Yes, the photoelectric effect showed that photons don't always behave as waves. That's part of what made studying atoms at the quantum level so mysterious, light behaved as a wave and as a particle. This became known as the "wave-particle duality" of light. Later on, we would in fact learn that all matter behaves as waves. It's just the large mass that makes it less observable.
Because of the particle quality of a photon, the interaction between photons and electrons occur one to one, that is the energy from only a single photon is transferred to one electron. If the energy of this single interaction does not meet the work function threshold, the electron will not be ejected. If light solely behaved like a wave, as was previously theorized, increasing intensity would proportionally increase the energy of light and allow for a change in electron ejection to occur even if wavelength is kept constant.
Ruby Richter 4G wrote:Does that mean that if the energy of a photon is greater than or equal to the energy needed to eject an electron, then increasing intensity results in more ejected electrons?
Yep! Basically because at that point, increasing intensity means you increase the number of photons with that equal or higher energy, which can then eject more electrons from the metal.
Yes, in lecture, Dr. Lavelle's slides said that "if the energy per photon is greater than/equal to the energy needed to remove an electron, then increasing the light intensity results in more ejected electrons."
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