Help on KE of photoelectron emmision

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The discussion revolves around calculating the kinetic energy (KE) of photoelectrons emitted from an aluminum surface when exposed to 180nm light, with a work function of 4.2 eV. The maximum KE of the fastest emitted photoelectrons is calculated to be 2.69 eV using the equation KE max = hf - W. For the slowest emitted photoelectrons, the minimum KE is determined using the threshold frequency, which corresponds to the work function energy. The cutoff wavelength is found by calculating the threshold frequency and converting it to wavelength. Overall, the participants clarify the equations needed for these calculations and confirm the approach to finding the minimum KE and cutoff wavelength.
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'light of wavelength 180nm falls on an aluminium surface. the work funct is 4.2eV.
what is the KE of the fastest and slowest emmited photoelectrons?
Also the cutoff wavelength for the Aluminium?
i only need the right equations an a little guidance as i myt as well be a complete novice.

i have worked out what i think is the max KE as 2.69eV

using KE max = hf - W (w as the work funct)
 
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The second part asks for kinetic energy of the slowest emitted electrons. What do you imagine is the slowest speed an electron can be emitted? The cutoff wavelength is just that wavelength that provides the same energy of the work function.
 
so i use the work funct equation an work backwards to get the threshold freq an then the cut off wavelength?

and for the min KE use the threshold freq?
 
exactly! :smile:
 
thank you so much!
 
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