Photoelectric effect: maximum and minimum kinetic energy at 180nm

[carus88]

'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)

 

[Kurdt]

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.

 

[carus88]

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?

 

[Kurdt]

exactly!

 

[carus88]

thank you so much!