Answer to Question #226248 in Atomic and Nuclear Physics for Fazi

Question #226248
Q.4
(a). Derive the cohesive energy for ionic crystals. Also explain the significance of Madelung’s constant using suitable diagram
1
Expert's answer
2021-08-16T17:46:21-0400

Cohesive energy of ionic solid

F=Fatt+FrepF=F_{att}+F_{rep}

F=ArmBrnF=\frac{A}{r^m}-\frac{B}{r^n}

Fatt=Ae24πϵr2F_{att}=\frac{Ae^2}{4\pi\epsilon r^2}

Frep=BrnF_{rep}=\frac{B}{r^n}

Work done by system of atom equal to negative of potential energy

W=P.E,P.E=WW=-P.E,P.E=-W

PE=Ae24πϵr+BrnPE=-\frac{Ae^2}{4\pi\epsilon r}+\frac{B}{r^n}

Equilibriumr=r0r=r_0

PE=-W=U(r)=min

Differenciate U(r) at r=r0r=r_0 W.r.t. to r is zero

dU(r)drr=r0=0\frac{dU(r)}{dr}|_{r=r_0}=0

ddr(Ae24πϵr2)+ddr(Br0n)=0\frac{d}{dr}(-\frac{Ae^2}{4\pi\epsilon r^2})+\frac{d}{dr}(\frac{B}{r_0^n})=0

Ae24πϵ(1r02)nBr0n1=0\frac{Ae^2}{4\pi\epsilon}(-\frac{1}{r_0^2})-nBr_0^{n-1}=0

Ae24πϵr0α1n=Br0n\frac{Ae^2}{4\pi\epsilon r_0}\alpha\frac{1}{n}=\frac{B}{r_0^n}

Now potential energy

Ae24πϵr0+Ae24πϵr0×1n=PE-\frac{Ae^2}{4\pi\epsilon r_0}+\frac{Ae^2}{4\pi \epsilon r_0}\times\frac{1}{n}=P E


P.E=Ae24πϵr0(11n)P.E=-\frac{Ae^2}{4\pi \epsilon r_0}(1-\frac{1}{n})

Unit are ev and Jules

Cohesive energy ionic solid per molecule per ion

PE=Ae24πϵr0(11n)×12PE=-\frac{Ae^2}{4\pi \epsilon r_0}(1-\frac{1}{n})\times\frac{1}{2}


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