How much energy is involved in ionizing hydrogen atom at #n = 1# to #3#?
1 Answer
I think you are asking the amount of energy input needed to bump the electron up from
Either way, the idea is to use the Rydberg equation (or some form of it), which describes the difference in energy for electronic transitions in the hydrogen atom:
#DeltaE = -R_H(1/n_f^2 - 1/n_i^2)#
It just so happens that
ELECTRONIC TRANSITION ENERGY
In this case, we have
#color(blue)(DeltaE) = -"13.61 eV"(1/3^2 - 1/1^2)#
#=# #color(blue)(ul"12.10 eV")#
In
#12.10 cancel("eV") xx (1.602 xx 10^(-19) "J")/(cancel"1 eV") = color(blue)(ul(1.938 xx 10^(-18) "J"))#
This should make physical sense. Electronic excitations (to higher
Also, this number should be somewhat familiar; the electron has a charge of
IONIZATION ENERGIES
And if you somehow mean, what are
#E_1 = -"13.61 eV"/1^2 = -"13.61 eV"# (at#n = 1# )
#E_2 = -"13.61 eV"/2^2 = -"3.40 eV"# (at#n = 2# )
#E_3 = -"13.61 eV"/3^2 = -"1.51 eV"# (at#n = 3# )
From Koopman's approximation theorem, we say that the ionization energy from an orbital containing one electron is approximately the same magnitude as the energy of that orbital, because it should take the electron past the
So, the ionization energies would then be:
#"IE"(n = 1) = color(blue)(ul(+"13.61 eV"))#
#"IE"(n = 2) = color(blue)(ul(+"3.40 eV"))#
#"IE"(n = 3) = color(blue)(ul(+"1.51 eV"))#