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Physics is so annoying =.= (1 Viewer)

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So im studying "Ideas to Implementation" and i went over my notes and found this:
"Photons of light below the threshold frequency did not carry sufficient energy, so no photoelectrons could be emitted regardless of the intensity of light"
Can someone help me out? Is it just saying that frequency does not affect the intensity of light being shone onto the metal?
 

LeightonChops

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Yea, increasing intensity (photon amount) just results in more electrons IF IT IS ABOVE THE THRESHOLD FREQUENCY, but i dont know what that results in, im gonna ask my teacher. Increasing frequency of the light that is shone will increase the KINETIC ENERGY in removing the electrons from the surface => higher voltage (I THINK)
 
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fl0

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If the photon of light is not above the threshold frequency (the frequency required for photoelectrons to be emitted - hence for photoelectric efffect to occur) it doesn't have sufficient energy because E = hf. The associated energy the photon has must be greater than the work function that binds the electron to the metal surface. Intensity of the light just means there is more photons, so even if there was high intensity but each individual photon doesn't have high enough frequency (hence energy) then photoelectrons cannot be emitted since there is not enough energy to release electrons from the surface of the metal. If the photon did have a frequency greater than the threshold frequency and the intensitiy of that light was increased, that just means the photoelectrons making up the current would be increased in size. If the frequency of the photons of the incident light was increased, that would mean the photoelectrons would be emitted with greater energy hence velocity.

Hope that helps a bit!
 

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