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?
11 answers
could somebody just post the solution process or post the formula? thanx a lot
E=hv
E=E(A)-B
lambda=hc/lambda
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?
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
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
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?