Question

Total: 100 points Return Homework 3 on Wednesday, November 13, at the beginning of class. Show your calculations to receive full credit. In the interest of time, do not type the formulas, handwrite them instead. Always present your results with the appropriate number of significant figures. 1. Photoionization is one of the ionization methods employed in the chromatography-mass spectrometry interface. Photoionization uses intense vacuum ultraviolet radiation from a source such as a 118nm laser. La. (15 points) Why does the vacuum ultraviolet photoionization with the 118nm wavelength radiation produces less molecular fragmentation than the 70eV electron ionization method? Hint: compare the energies of the two ionization methods. Show your calculations to receive full credit. leV = 1.602 x 10- J; Plank constant: h=6.63 x 10- Js; speed of light in vacuum: c = 3 x 10 m/s.

Answer 1

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Inm = 10 m. -7m 110nm= 118810 9 = 1.18x10 ENERGY OF ULTRAVIOLET PHOTOMONIZATION DATE WAVELENGTH A = 118nm = 1.18 x 10- m. -34 ENERGY = 2 x C = 6.626810 Is x 3x10 m/s 1. 18410 -7 m 19:978 x 10-26 - 16:846 x 100J 1.18X10 7 or 1.6846 X 10-18 ] ENERGY Electron Tenization method Energy = 70 eV 1 ev = 1.602 x 10 19 J To ev = = 70 81.602 X10-19 J 112.14 X10-19 J = 1.1214 x10-17 J Slepte EnERLY of ULTRAVIOLET Photo ionization 1.6846810-10 I ( Electron ionization methode L 1.1214 X 10-17 J So Electron ionization method due to high Energy Produce more fragment