One way to measure ionization energies is ultraviolet photoelectron spectroscopy

Question

One way to measure ionization energies is ultraviolet photoelectron spectroscopy (UPS, or just PES), a technique based on the photoelectric effect. In PES, monochromatic light is directed onto a sample, causing electrons to be emitted. The kinetic energy of the emitted electrons is measured. The difference between the energy of the photons and the kinetic energy of the electrons corresponds to the energy needed to remove the electrons (that is, the ionization energy). Suppose that a PES experiment is performed in which mercury vapor is irradiated with ultraviolet light of wavelength 98.1 nm.

Explanation / Answer

Hey!

This question needs more information, but I give you the sample of how you can solve this in case you have the following questions:

Energy of a photon of this light in eV?

E = h * c /

Where,

E = energy (J)

h = Plank’s constant 6.626 x10^-34 J s

c = speed of light 3.0 x 10^8m/s

= wave length (m) ------> 98.1 nm = 98.1 *10^-9 m

Ephoton = 2.02 * 10^-18 J = 12.6 eV = 2.02 * 10^-21 kJ

2) Equation of the process corresponding to the first ionization energy of Hg

Hg (g) ---> Hg+ (g) + e-

3) The first ionization energy of Hg, in kJ/mol, if you have the kinetic energy in eV:

Kinetic energy (kJ) = (Kinetic energy in eV / 6.241509·10^18) * (1/1000)

First Energy of ionization (kJ/mol) = Ephotons - Ekinectic (kJ) / 1 mol

Hope this helps!

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.