The work function for potassium is 2.24 ev. if potassium metal is illuminated with light of wavelength 350 nm , find:

a) the maximum kinetic energy of the photoelectrons at this wavelength.

b) the cut off wavelength for potassium

Photo electric effect.Some guidance notes which may help ...
The energy of a photon of wavelength lambda is E=hc/lambda where h is Plank's constant, c is the speed of light, and lambda is the wavelength.The units of eV (electron volts) and the value of Planck's constant probably have to be determined. Calculate E=hc/lambda as hx3x10^8/(350x10^-9) and compare that to 2.24eV (expressed in consistent units which could be Joules)If the 350nm photon energy is higher than 2.24eV, then the difference should be the kinetic energy of the ejected photo electron.
if a photon has equal to or less than 2.24 eV work function, then it won't be ejected. That wavelength can be found by equating the work function to the photon energy using the equation E=hc/lambda2.24eV=hx3x10^8/(lambda_cut_off)

If the incoming light photon has an energy less than the work function of the metal, then the metal will not emit a photon.
This is the photo-electric effect, explained by Einstein who got a Nobel prize for this, and also for relativity. The point was, it seems, that it didn't matter how intense the light source was, it was the energy of photons within the light source, expressed by their wavelength, that mattered. 
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Ithink the answers is b i hope it my friend message me if i’m wrong and i’ll change my answer and fix it for you