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