The binding energy = #3.23xx10^(4)kJ.mol^(-1)#
The KE of the emitted electron will be equal to the energy of the X-ray photon - the binding energy of the electron:
#(1)/(2)mv^2=hf-E#
We need to get everything into standard units:
#c=flambda#
#f=(c)/(lambda)= (3xx10^(8))/(0.938xx10^(-9))=3.198xx10^(17)s^(-1)#
#(1)/(2)mv^2=990xx1.6xx10^(-19)=1.584xx10^(-16)J#
#E=hf-(1)/(2)mv^2#
#E=(6.63xx10^(-34)xx3.198xx10^(17))-(1.584xx10^(-16))J#
#E=(2.12xx10^(-16))-(1.584xx10^(-16))J#
#E=5.36xx10^(-17)J#
#E= 5.36xx10^(-20)kJ#
The Avogadro Constant #L=6.02xx10^(23)mol^(-1)#
#E=5.36xx10^(-20)xx6.02xx10^(23)kJ.mol^(-1)#
#E= 3.23xx10^(4)kJ.mol^(-1)#