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?
equation please
equation?
4.372281
How is this calculated? someone to give the little formula
Dear participants of EDX 3.091,
You are being warned not to cooperate during the examination, the punishment for such breaking the rule will be cancellation of the certificate for the whole students currently enrolled.
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Jiskha is not part of the MITx platform!!! How the honor code of MITx could be applied to an user of other system?
It obviously applies to the user, no matter what system they use to try to cheat.
p=h/lambda(meters) ,
E= p^2/2m + 1.659*10^-18,
LambdaMax=hc/E