Question

Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 58.4 nm.
Energy of a photon of this light is 21.2 eV.
Write an equation that shows the process corresponding to the first ionization energy of Hg.
the kinetic energy of the emitted electrons is measured to be 10.75 eV. What is the first ionization energy of Hg, in kJ/mol?
With reference to the figure, determine which of the halogen elements has a first ionization energy closest to that mercury.

Answer 1

E(Joules) = hc/wavelength.
I would convert 21.2 eV to joules. h is Planck's constant, c is the speed of light in m/s, and wavelength is in meters.

Answer 2

p=h/lambda(meters) ,
E= p^2/2m + 1.659*10^-18,
LambdaMax=hc/E

Answer 3

Hg(g) -> Hg+(g) + e-
1010kg/mol
I