Asked by amar chauhan

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
Answered by Anonymous
I guess there is an error. it is 2.091, not 5.091...
Answered by ghgl
I'm having problems with this as well. I did all the calculations but still don't have a correct answer for it :(
Answered by kk
i'm having problems in the same question but i have λ=3.591 not 2.091 or 5.091...
Answered by hh
that makes use for this problem but with λ = 3.091
Answered by hh
that equation use
Answered by bigga
how do you calculate? i do not understand
Answered by lkp
what concept to be applied for measuring maximum wavelength
Answered by a
Please what is the formula?
Answered by drd
lambda=4.591 angstroms for me
Answered by guru
what is the answer?
Answered by chelo
equation please
Answered by b
equation?
Answered by Anonymous
4.372281
Answered by chelo
How is this calculated? someone to give the little formula
Answered by MIT EDX 3.091
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.

Regards.
Answered by c
Jiskha is not part of the MITx platform!!! How the honor code of MITx could be applied to an user of other system?
Answered by Ms. Sue
It obviously applies to the user, no matter what system they use to try to cheat.
Answered by c
p=h/lambda(meters) ,
E= p^2/2m + 1.659*10^-18,
LambdaMax=hc/E
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