# 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`