# 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

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 **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

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

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