Question #d8ab4

1 Answer
Feb 16, 2015

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)#