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 rubidium atom in the gas phase is to be ionized by two different light beams, one of which has wavelength 795 nm. What is the maximum wavelength for the second beam that will cause two-photon ionization? Hint: The ionization energy of rubidium is 403.0 kJ/mol ____ nm???

Answer 1

Given:

lambda = 7.95*10^-7 m

Find energy of 1 photon first

Given:

lambda = 7.95*10^-7 m

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(7.95*10^-7 m)

= 2.5*10^-19 J

This is energy of 1 photon

Energy of 1 mol = energy of 1 photon * Avogadro's number

= 2.5*10^-19*6.022*10^23 J/mol

= 1.506*10^5 J/mol

= 150.6 KJ/mol

This is energy of 1st photon.

So,

energy of 2nd photon = total energy - energy of 1st photon.

= 403.0 KJ/mol - 150.6 KJ/mol

= 252.4 KJ/mol

Given:

Energy of 1 mol = 2.524*10^2 KJ/mol

= 2.524*10^5 J/mol

Find energy of 1 photon first

Energy of 1 photon = energy of 1 mol/Avogadro's number

= 2.524*10^5/(6.022*10^23)

= 4.191*10^-19 J

This is energy of 1 photon

use:

E = h*c/lambda

4.191*10^-19J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda

lambda = 4.743*10^-7 m

lambda = 474.3 nm

Answer: 474.3 nm