Question

Consider eight of the ions Cl? and Na+ in a crystal lattice of common salt. The ions are located at the vertices of a cube measuring 2.82

Answer 1

At every corner you will observe that one anion is surrounded by 3 cations and every cation is surrounded by 3 anions, also every anion has next adjacent atom as anion and every cation has adjacent atom as cation

also an anion has diagonal atom as cation and a cation has diagonal atom anion

So taking Cl for force consideration at origin we have

also charge on every atom is electronic charge which has value= e=1.6x10^-19C

Now Force due to adjacent atoms, Fa= K*((e²/d²)i +(e²/d²) j +(e²/d²) k) =(K.e²/d²)*(i+j+k)

Now force due to next adjacent atoms, they will be at a distance of ?2*d from Cl atom

Hence, Fn,a = K*(-(e²/2d²)*?2i -(e²/2d²)*?2 j -(e²/2d²)*?2 k) [we have taken coponent in x, y z of forces by all anions and adding them will lead to above)

=-(K.e²/?2d²)*(i+j+k)

Now force due to diagonally opposite cation, Fd = K*((e²/3d²)*(1/?3)i +(e²/3d²)*(1/?3)j +(e²/3d²)*(1/?3) k)

[taking component in x, y andz direction]

=(K.e²/3?3d²)*(i+j+k)

Thus net force on anion, F= Fd+Fa+Fn,d =(K.e²/3?3d²)*(i+j+k)-(K.e²/?2d²)*(i+j+k)+(K.e²/d²)*(i+j+k)

magnitude,|F|= (K.e²/d²)*((1/3?3) -(1/?2)+1)*|(i+j+k)|

=0.485*1.732*9x10⁹x(1.6x10^-19)²/(2.82x10^-10)²

=2.434x10^-9 N