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=3.591 A˚. What is the maximum wavelength (in meters) of radiation capable of ionizing sulfur and producing this effect?
5 answers
How is this calculated? someone to give the little formula
Look for Lecture S6E4, its the same, the difference is that they are asking you for lamda in the exam instead of the velocity in the lecture. Hope that this helps.
Remember
E(photon) = E(electron)+E(Binding)
Look also for deBroglie wave.
E(photon) = E(electron)+E(Binding)
Look also for deBroglie wave.
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
(what is p?
6 answers