Question

In two-photon ionization spectroscopy, the combined energies carried by two different photons are used to remove an electron from an atom or molecule. In such an experiment a copper atom in the gas phase is to be ionized by two different light beams, one of which has wavelength 327 nm. What is the maximum wavelength for the second beam that will cause two-photon ionization? Please answer in nanometers

Hint: The ionization energy of copper is 745.4 kJ/mol please

Answer 1

Ionization energy = 745.4 KJ/mol

= 7.454*10^5 J/mol

= 7.454*10^5/(6.022*10^23) J/photon

= 1.238*10^-18 J/photon

Lets find the energy of 1st photon

Given:

lambda = 3.27*10^-7 m

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(3.27*10^-7 m)

= 6.079*10^-19 J

Use:

Energy of 2nd photon = Ionization energy - energy of 1st photon

= 1.238*10^-18 J - 6.079*10^-19 J

= 6.301*10^-19 J

use:

E = h*c/lambda

6.301*10^-19J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda

lambda = 3.155*10^-7 m

Answer: 3.16*10^-7 m or 316 nm