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 gallium atom in the gas phase is to be ionized by two different light beams, one of which has wavelength 403 nm What is the maximum wavelength for the second beam that will cause two-photon ionization? Hint: The ionization energy of gallium is 578.8 kJmol nm Submit Answer 5 question attempts remaining

Answer 1

IE = 578.8 KJ/mol
= 578.8 / 6.022*10^23 KJ/atom
= 9.611*10^-22 KJ/atom
= 9.611*10^-19 J/atom

Find the energy supplied by 403 nm:

Given:
lambda = 4.03*10^-7 m

use:
E = h*c/lambda
=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(4.03*10^-7 m)
= 4.933*10^-19 J

Find energy carried by 2nd wavelength:
E = IE - E1
= 9.611*10^-19 J - 4.933*10^-19 J
= 4.678*10^-19 J

use:
E = h*c/lambda
4.678*10^-19J =(6.626*10^-34 J.s)*(3.0*10^8 m/s)/lambda
lambda = 4.249*10^-7 m
lambda = 425 nm
Answer: 425 nm