Sulfur vapor is analyzed by photoelectron spectroscopy (PES). Measurements determine that photoelectrons associated with the 1st ionization energy of sulfur move with de Broglie wavelength λ=5.091 A˚. What is the maximum wavelength (in meters) of radiation capable of ionizing sulfur and producing this effect

I guess there is an error. it is 2.091, not 5.091...

I'm having problems with this as well. I did all the calculations but still don't have a correct answer for it :(

i'm having problems in the same question but i have λ=3.591 not 2.091 or 5.091...

that makes use for this problem but with λ = 3.091

that equation use

how do you calculate? i do not understand

what concept to be applied for measuring maximum wavelength

Please what is the formula?

lambda=4.591 angstroms for me

what is the answer?