chemistry
posted by selly on .
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?

could somebody just post the solution process or post the formula? thanx a lot

E=hv
E=E(A)B
lambda=hc/lambda 
what does E=E(A)B mean?

if i solve it by last equation i get something with 2.xxxxx*10^7
could this be the right solution? 
helloooo? please somebody answer...
now i have a bunch of solutions here but i cant figure out the right one and 
Which equation using?. help me thanks

Can someone post the right equation please?

p=h/lambda(meters) ,
E= p^2/2m + 1.659*10^18,
LambdaMax=hc/E 
Is the answer 116016636.521?

No...is the answer0.116016636521?

is the answer 0.116016636521?