Alaser beam of total power 1 mw and wavelength 365 nm is incident on a clean piece of metal in vacuum. the metal has a work function of 3.0 ev. find the maximum possible number of photoelectrons that could be emitted from this metal per second, and the maximum possible kinetic energy of the emitted electrons in ev. how do your answers change if the laser wavelength is doubled to 730 nm?

1.84* 10 ^15 electrones /s for 365 nm. Maximum energy kinetic, 0.4 eV.

0 electrons/s for 730 nm,

Explanation: In order to solve this problem we have to take into account the following relationship for the the power is laser beam.

Power= Energy/time so we can considerer a numeber electrons/time* energy of each photon the we have the power of photon flux given by:

P=n*hν where n is the number of electrons per second and h the Plack constant.

The phton energy is equal h*c/λ;  where c is the speed of light

for  λ = 365 nm  we have E=1240/365 nm=3.4 eV ( 1eV=1.6 * 10^-19J)

so n= P/hν=1*10^-3/3.40 *1.6*10^-19 =1.84 * 10^15 electrons/s

The maximun kinetic energy of photolectrons is given by:

hν-W= Ek where W is the work function

then Ek=(3.4-3)eV= 0.4 eV

Finally, for the case of  λ =730 nm, the energy for this photon is

E=1240/730=1.7 eV

so with this energy it is not possible to obtain any photoelectrons because this energy is lower that the work fuction for this material.

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