The work function for tungsten is 4.58 eV . Find the threshold frequency and wavelength for the photoelectric effect to occur when monochromatic electromagnetic radiation is incident on the surface of a sample of tungsten. Find the maximum kinetic energy of the electrons if the wavelength of the incident light is 200 nm .

The work function of a metal phi , the wavelength of the incident photon lambda required to eject an electron with the maximum kinetic energy K_(max) is expressed by Einsteins photoelectric equation.
K_(max)= (hc)/lambda - phi
The threshold wavelength lambda_t , or lowest energy wavelength necessary to free an electron occurs when the K_(max)=0 .
0=(hc)/lambda_t - phi
(hc)/lambda_t = phi
lambda_t=(hc)/phi=((4.136*10^-15 eV*s)(2.9998*10^8 m/s))/(4.58 eV)= 271 nm
Using the relation lambda*f=c we can now solve for the frequency.
f_t=c/lambda_t=(2.9998*10^8 m/s)/(271 *10^-9 m)=1.11*10^15 Hz
If the wavelength of the incoming photon were 200 nm, then  K_(max) can be found by:
K_(max)=E-phi=hf-phi=(hc)/lambda-phi
K_(max)=hc/(200*10^-9 m)-4.58 eV
K_(max)=0.380 eV
https://physics.info/photoelectric/