Question

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=3.091 A˚. What is the maximum wavelength (in meters) of radiation capable of ionizing sulfur and producing this effect?

Answer 1

help pleasee

Answer 2

The de Broglie wavelength is related to the momentum p of the electron:\[\lambda_{\rm DB} = \frac{h}{p}\]where h is the Planck constant. Next, you'll need the fact that the momentum and energy E of the electron are related:\[E = \frac{p^2}{2m }\]Finally, you'll need the fact that the energy of a photon is related to its frequency:\[E = h \nu\]and the fact that frequency and wavelength of a photon are related to the speed of light c:\[c = \lambda \nu\]