Physics, asked by smsk7956, 1 day ago

Calculate the cohesive energy per ion-pair of LiCl crystal using following information. Madelung constant = 1.748 Li+ and Cl- spacing = 2.57 Å Ionization Energy of Li = 5.4 eV Assume that the repulsive potential energy is negligibly small. ​

Answers

Answered by adarsh9876543210987
0

ask the question in pdf formste please

Answered by adventureisland
2

The repulsive potential energy is small is -4.002eV.

Explanation:

V=-\frac{\alpha .e^{2}}{4\pi e_{0}.r_{0}} (1-\frac{1}{n} )

V=-\frac{\alpha .e^{2}}{4\pi e_{0}.r_{0}}

let's see what ew get,

e=1.602.10^{-19}   E_^{0}=8.854.10^{-12}  r_{0}=0.257.10^{-9}  \alpha =1.748

V=-\frac{\alpha .e^{2}}{4\pi e_{0}.r_{0}}

V=-1.569.10^{-18}

v:=\frac{v}{1.602.10^{-19}}

V=-9.793

the cohesive energy is ,

V_{ion} :=5.4   V_{affin=-3.61}

E_{coh}=V+(V_{ion} +V_{affin}  )

E_{coh}=-8.003 eV

E_{cohesive}=\frac{E_{coh}}{2}

E_{cohesive}=-4.002eV.

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