Kellie K.
asked • 09/18/22What is the ionization energy for an atom (in kilojoules per mole) if the ejected electron has a velocity of 4.29x10^5 m s^-1
when ionized with a 307.5mn wavelength laser. The mass of the electron is 9.1094 x10^-31 kg. (to one decimal place with answer)
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1 Expert Answer
Gwo J. answered • 09/26/22
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The kinetic energy of the ejected electron:
K.E. = (1/2)mv2 = (1/2)(9.1094×10--31Kg)(4.29×105 m/s)2
= (8.38x10--20 Joule/atom)(6.022×1023 atoms/mole)(1KJ/1000J) = 50.5 KJ/mole
The incoming energy of 307.5nm wavelength laser
= hc/λ = [(6.626×10-34J•s)(3×108m/s)/(307.5nm)(1m/109nm)]
= (6.464×10-19 J/photon)(6.022×1023photons/mole)(1KJ/1000J) = 339 KJ/mole
The ionization energy
= Incoming Laser energy - Kinetic energy of the ejected electron
= 339 - 50.5 = 288.5 KJ/mole
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Douglas B.
First determine the energy of the electron using KE = 1/2mv^2, where m = the mass of the electron (look it up), and v = velocity of the electron. Don't forget to SQAURE THE VELOCITY OF THE ELECTRON! Then determine the energy of the laser using c = frequency times wavelength, where c = 3x10^8 m/s. Then subtract this from the KE of the electron and this should give your the ionization energy of removing the electron with the KE left over. Good luck09/26/22