[Lynda92]

The threshold wavelength for photoemission of electrons from potassium is 564nm.Calculate the maximmum velocity of the photoelectrons that will be emitted when the metal is irradiated by light with a wavelength of 300nm.

the answer in the book was 8.25x10^5 m/s however my answer was 3091 m/s.I can't see where I've gone wrong, so could someone please help me!my workings out are below.thankyou!

the equations I used were 1/2mv² =hv-ɸ, E=hv and c=λv

first I worked out the workfunction:
at a wavelength of 564nm, hv=ɸ

E=hv= h(c/λ) =6.626x10^-34 x (2.997x10^8/564x10^-9)=3.52x10^-19 J

therefore ɸ=3.52x10^-19J
 
then the energy of the light at 300nm  = 6.622x10^-19J

then I worked out the kinetic energy
1/2mv^2=6.622X10^-19J -3.52x10^-19J=3.102x10^-19J

v^2= 6.203x10-19/(39.1 x (1.660x10-27kg))

V=3091m/s



                  






    

[rabolisk]

v^2= 6.203x10-19/(39.1 x (1.660x10-27kg))

V=3091m/s

Everything is right until that point. Remember that you're trying to solve for the velocity of the electron(s) that will be emitted.

[Lynda92]

v^2= 6.203x10-19/(39.1 x (1.660x10-27kg))

V=3091m/s

Everything is right until that point. Remember that you're trying to solve for the velocity of the electron(s) that will be emitted.

Oh right yeah, I was meant to use the mass of the electron ( 9.109 x 10^-31 kg) instead of potassium.
Thank-you Rabolisk!