in a photoelectric effect experiment, a beryllium surface (work function= 8.01 x 10^-19 J) is bombarded with photons from a 125 nm source of electromagnetic radiation.
a.) what is the energy of the incident photons?
b.) what is the maximum speed of electrons emitted from the berylliums surface?

Question

in a photoelectric effect experiment, a beryllium surface (work function= 8.01 x 10^-19 J) is bombarded with photons from a 125 nm source of electromagnetic radiation. 
a.) what is the energy of the incident photons?
b.) what is the maximum speed of electrons emitted from the berylliums surface?

cuanchi · Answer

a) $$E= hc/\lambda $$
h= 6.626 x 10^(-34) Js
c= 3.0 x 10^8 m/s
lambda= 125 x 10^(-9) m

anonymous · Answer

so maximum speed?

cuanchi · Answer

with the difference of energy, the energy of the photon - work function you will have the energy of the electron released. 
you have lambda = h/ mv ; E= hc/lambda
v=c/mE
v= velocity
c= 3.0 x 10^8 m/s
m= mass of an electron =9.10938291 x 10^(-31) kilograms
E = energy (J)  kg m^2/s^2