Question

Question #d8ab4

Answer 1

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)`