What is the ionization potential for an electron going from #n = 2# to #n = 3# in #"eV"# for hydrogen atom?

1 Answer
Jun 22, 2017

About #"1.89 eV"# is released for the absorption, or absorbed for the emission process between #n = 3# and #n = 2#. That means it takes about #1.89# #"eV"# of work to accelerate one electron through a one potential difference of one volt.

To be accurate, I don't think that's an ionization potential... that's really the energy used to excite the atom (ionization potential is for removing electron(s) altogether).


I think you're asking about the change in energy going from #n = 3# to #n = 2# in the hydrogen atom? That can be found from the Rydberg equation:

#DeltaE = -"13.6 eV" (1/n_f^2 - 1/n_i^2)#

where:

  • #"13.6 eV"# is the Rydberg constant in electron volts (or the negative of the hydrogen atom ground state energy).
  • #n_i# is the initial state's principal quantum number #n#.
  • #n_f# is the final state's principal quantum number #n#.

So, if you wanted the final state to be #E_3#, then:

#DeltaE = color(blue)(E_3 - E_2) = -"13.6 eV" (1/3^2 - 1/2^2)#

#=# #color(blue)("1.89 eV")#