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 beryllium atom in the gas phase is to be ionized by two different light beams, one of which has wavelength 235 nm. What is the maximum wavelength for the second beam that will cause two-photon ionization?

Hint: The ionization energy of beryllium is 899.4 kJ/mol

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Answer 1

Given:

lambda = 2.35*10^-7 m

use:

E = h*c/lambda

=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(2.35*10^-7 m)

= 8.459*10^-19 J

This is energy of 1 photon

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

= 8.459*10^-19*6.022*10^23 J/mol

= 5.094*10^5 J/mol

= 509.4 KJ/mol

This is energy of 1st photon

Now use:

Total ionization energy = energy of 1st photon + energy of 2nd photon

899.4 = 509.4 + energy of 2nd photon

energy of 2nd photon = 390.0 KJ/mol

Given:

Energy of 1 mol = 390 KJ/mol

= 3.9*10^5 J/mol

Find energy of 1 photon first

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

= 3.9*10^5/(6.022*10^23)

= 6.476*10^-19 J

This is energy of 1 photon

use:

E = h*c/lambda

6.476*10^-19J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda

lambda = 3.069*10^-7 m

= 307 nm

Answer: 307 nm