So I did the problem without such liberal rounding, and it came out to about 1.11*10^-7.
11.26 eV = 1.6*10^-19 C
Thus, 11.26 x 1.6*10^-19 = 1.8*10^-18 C (the energy of the ionization)
E=hf
1.8*10^-18 = 6.63*10^-34 x f
f = 2.71*10^15
v=λf
3*10^8 = λ x 2.71*10^15
λ = 1.11*10^-7 m
but that isn't 110 nm, right? If we put that number in nanometers, it'd be .011 nm, right??