Asked by selly

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=2.091 A˚. What is the maximum wavelength (in meters) of radiation capable of ionizing sulfur and producing this effect?
Answered by kerry
could somebody just post the solution process or post the formula? thanx a lot
Answered by amar chauhan
E=hv
E=E(A)-B
lambda=hc/lambda
Answered by kerry
what does E=E(A)-B mean?
Answered by kerry
if i solve it by last equation i get something with 2.xxxxx*10^-7
could this be the right solution?
Answered by kerry
helloooo? please somebody answer...
now i have a bunch of solutions here but i cant figure out the right one and
Answered by chelo
Which equation using?. help me thanks
Answered by selly
Can someone post the right equation please?
Answered by c
p=h/lambda(meters) ,
E= p^2/2m + 1.659*10^-18,
LambdaMax=hc/E
Answered by can u help me?
Is the answer 116016636.521?
Answered by can u help me?
No...is the answer0.116016636521?
Answered by PLEEEEEASE HEEEELP MEEEEE
is the answer 0.116016636521?
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