What frequency of light is needed to eject an electron from a metal whose threshold energy is #"201 kJ/mol"#?

1 Answer
Nov 2, 2016

#5.04 xx 10^14# #"Hz"# or #"s"^(-1)#


This sounds like it's about the photoelectric effect.

The binding energy #phi# is the energy that would have needed to be supplied by an incoming light source in order to remove a desired number of electrons from the surface (in this case, #"1 mol"# of electrons).

It is also defined as the threshold that needs to be overcome in order to eject a specific number of electron(s). Hence, its associated frequency IS the threshold frequency.

http://www.cyberphysics.co.uk/

If you supply more energy than the required threshold to eject a specific quantity of electrons, the difference in the supplied energy and the binding energy gives the remaining kinetic energy that goes into the ejected electrons.

Therefore, the equation becomes:

#bb(stackrel("leftover kinetic energy")overbrace(K_e) = stackrel("supplied")overbrace(E_"photon"))# #-bb()# #bb(stackrel("required threshold")overbrace(phi))#

#= bb(hnu - hnu_0)#

where #nu_0# is the threshold frequency.

Based on your question, all you need is the threshold frequency #nu_0#, which means:

#phi = hnu_0#

#color(blue)(nu_0) = phi/h#

#=# #[(201 cancel"kJ")/cancel"mol electrons" xx cancel"1 mol electrons"/(6.022xx10^(23)) xx (1000 cancel"J")/cancel"1 kJ"]/[6.626xx10^(-34) cancel("J")cdot"s"]#

#= color(blue)(5.04 xx 10^14# #color(blue)("Hz")# or #"s"^(-1)#