Photoelectric work function for cesium at 2.14eV?

a) What is the max kinetic energy of electrons ejected from the surface of cesium by light of wavelength: lambda = 546nm?
b) What is the maximum speed of the electrons?
c) What is meant by the stopping potential? Describing the effects on photocurrent?

1 Answer
Apr 10, 2018

Below

Explanation:

a) #T_(max) = h nu - phi = (h c) / lambda - phi#

For photon energy in electronvolts, there is a shortcut:

#E_(eV)=(1.2398)/(lambda [ mu m])#

#implies T_(max) = (1.2398)/(0.546) - 2.14 = 0.131eV#

b) From #T = 1/2 m v^2#, #v_(max) = sqrt( (2 T_(max) )/(m_e) )#

# = sqrt( (2 times 0.131 times color(red)(1.6 * 10^(-19)) )/(9.1*10^(-31)) )#

The figure in red, the charge on an electron, converts eVs to Joules

#implies v_(max) = 214, 630 \ ms^(-1) approx 215 kps#

c) The stopping potential, which is applied across the plates in the experiment, is designed to prevent electrons that are emitted from the lit plate, from moving to the collecting plate, as the electric field opposes the photo current.

Once the current is zero, we can therefore equate stopping potential to the max kinetic energy of electrons that were otherwise reaching the collecting plate. This is on the basis that the work done by the field retarding electrons equals that max kinetic energy.

#e V_o = T_(max)#

Here the stopping potential is:

#V_o = (0.131 eV)/e = 0.131 V#