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sami23
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Cesium is often used in "electric eyes" for self-opening doors in an application of the photoelectric effect. The amount of energy required to ionize (remove an electron from ) a cesium atom is 3.89 electron volts (1 eV = 1.60 x 10-19 J). Show by calculation whether a beam of yellow light with wavelength 5230Å would ionize a cesium atom.
I converted wavelength: 5230*1x10^-10 = 5.23*10^-7m
Then I calculated the energy by using hc/[tex]\lambda[/tex] :
(6.62*10^-34 * 3*10^8)/(5.23*10^-7) = 3.797*10^-19 J
I also calculated 3.89eV*(1.6*10^-19) = 6.224*10^-19 J energy required to ionize Cesium
How can I tell the beam of yellow light ionizes the Cesium atom?
I converted wavelength: 5230*1x10^-10 = 5.23*10^-7m
Then I calculated the energy by using hc/[tex]\lambda[/tex] :
(6.62*10^-34 * 3*10^8)/(5.23*10^-7) = 3.797*10^-19 J
I also calculated 3.89eV*(1.6*10^-19) = 6.224*10^-19 J energy required to ionize Cesium
How can I tell the beam of yellow light ionizes the Cesium atom?