Question

Figure represents a crystal unit of cesium chloride, CsCl. The cesium atoms, represented by open circles are situated at the corners of a cube of side 0.40 nm, whereas a Cl atom is situated at the centre of the cube. The Cs atoms are deficient in one electron while the Cl atom carries an excess electron.



(i) What is the net electric field on the Cl atom due to eight Cs atoms?

(ii) Suppose that the Cs atom at the corner A is missing. What is the net force now on the Cl atom due to seven remaining Cs atoms?

Answer 1

(i) As given Cl atom is at centre of the cube and Cs atoms are at corners of the cube. So, the Cl atom is equidistant from each Cs atom. Thus, electric field will be cancelled due to opposite Cs atoms.

Therefore, net electric field on the Cl atom due to eight Cs atoms is zero.

Final answer: Zero

(ii) If eight Cs atoms are situated at the corners of the cube, then net electric field at the centre of the cube will be zero.

So, the vector sum of the electric field due to charge on A + other corner charges = 0

${\overrightarrow{E}}_A+{\overrightarrow{E}}_{sevencharges}=0$

${\overrightarrow{E}}_A=-{\overrightarrow{E}}_{sevencharges}=\frac{e}{4\pi {ϵ}_0{r}^2}$

(As given Cl atom has one excess electron and Cs atom deficient by one electron)

The net force on Cl atom is only due to charge diagonal opposite to point A.

$F=e{\overrightarrow{E}}_{sevencharges}$

$F=\frac{e^2}{4\pi {ϵ}_0{r}^2}$

𝑟 = Distance between Cl and Cs atom

$r=\frac{1}{2}\left(\sqrt{3}(0.40\times {10}^{-9})\right)m$

(half of the diagonal of the cube)

$r=0.3464\times {10}^{-9}m$

So, $F=\frac{(9\times {10}^9){(1.6\times {10}^{-19})}^2}{{(0.3464\times {10}^{-9})}^2}$

$F=1.92\times {10}^{-9}N$

(Directed from A to Cl)


Final answer: $1.92\times {10}^{-9}N$ (Directed from A to Cl)