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
aluminum atom in the gas phase is to be ionized by
two different light beams, one of which has wavelength
395 nm. What is the maximum wavelength for the
second beam that will cause two-photon ionization?
Hint: The ionization energy of aluminum is
577.6 kJ/mol
nm?
Find energy of 1 photon first
Given:
lambda = 3.95*10^-7 m
use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(3.95*10^-7 m)
= 5.032*10^-19 J
This is energy of 1 photon
Energy of 1 mol = energy of 1 photon * Avogadro's number
= 5.032*10^-19*6.022*10^23 J/mol
= 3.031*10^5 J/mol
= 303.1 KJ/mol
This is the energy of 1st photon
Use:
Total energy = energy of 1st photon + energy of 2nd photon
577.6 = 303.1 + energy of 2nd photon
energy of 2nd photon = 274.5 KJ/mol
Given:
Energy of 1 mol = 2.745*10^2 KJ/mol
= 2.745*10^5 J/mol
Find energy of 1 photon first
Energy of 1 photon = energy of 1 mol/Avogadro's number
= 2.745*10^5/(6.022*10^23)
= 4.558*10^-19 J
This is energy of 1 photon
use:
E = h*c/lambda
4.558*10^-19J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda
lambda = 4.361*10^-7 m
= 436 nm
Answer: 436 nm
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