Question

Photoelectron Spectroscopy utilizes the photoelectron effect to measure binding energy of electrons in molecules and solids, by measuring the kinetic energy of the emitted electrons and using the relation in problem 9.3 between kinetic energy, wavlength, and binding energy. One variant of photoelectron spectroscopy is X-Ray wavelength is 0.2nm, calculate the velocity of electrns emitted from molecules in which the binding energies are 10eV, 100eV and 500eV?

Answer 1

wavelength (l) = 2 nm = 2 x 10^-9 m

With binding energy = 10 eV

KE = hc/l - binding energy

     = (4.14 x 10^-15 x 3 x 10^8/2 x 10^-9) - 10

     = 611 eV = 611/6.24 x 10^18

     = 9.79 x 10^-17 J

KE = 1/2.mv^2

9.79 x 10^-17 = 1/2 x 9.11 x 10^-31 x v^2

velocity of ejected electron (v) = 1.47 x 10^7 m/s

With binding energy = 100 eV

KE = hc/l - binding energy

     = (4.14 x 10^-15 x 3 x 10^8/2 x 10^-9) - 100

     = 521 eV = 521/6.24 x 10^18

     = 8.35 x 10^-17 J

KE = 1/2.mv^2

8.35 x 10^-17 = 1/2 x 9.11 x 10^-31 x v^2

velocity of ejected electron (v) = 1.35 x 10^7 m/s

With binding energy = 500 eV

KE = hc/l - binding energy

     = (4.14 x 10^-15 x 3 x 10^8/2 x 10^-9) - 500

     = 121 eV = 121/6.24 x 10^18

     = 1.94 x 10^-17 J

KE = 1/2.mv^2

1.94 x 10^-17 = 1/2 x 9.11 x 10^-31 x v^2

velocity of ejected electron (v) = 6.52 x 10^6 m/s