The threshold frequency of a potassium metal plate is 4.35×1014 Hz. Determine the wavelength of a photon required to reach 2.50 eV of the maximum kinetic energy of an emitted electron.
Options:
A. 2.88×102 nm
B. −4.96×102 nm
C. 9.50×10−21 nm
D. 6.69×102 nm
E. 2.02×1015 nm
From what I understand, I used the equation KE=hν+hν0 and replaced ν with cA because wavelength and frequency are inversely proportional. Then I plugged in the given components, ν0 and a fraction of KE to solve; however, I don't understand how to approach this as I'm only given 2.5eV(4.005×10−19J) of maximum KE, not an actual value for KE.
