In crystals of the salt cesium chloride, cesium ions Cs + form the eight corners

Question

In crystals of the salt cesium chloride, cesium ions Cs + form the eight corners of a cube and a chlorine ion Cl –is at the cube’s center. The edge length of the cube is L = 3.5 x 10 –10 m. The Cs + ions are each deficient by one electron (and thus each has a charge of +e), and the Cl– ion has one excess electron (and thus has a charge of –e). (a) If one of the Cs+ ions is missing, the crystal is said to have a defect; with a reference potential at a point infinitely far away from the crystal, what is the net potential at the location of the missing Cs + ion? (b) How much energy is required to assemble this crystal of seven Cs + ions and a Cl –ion at the center? Note that the computation may be tedious but it is straight forward.

Explanation / Answer

Imagine a close cube and at each corner +e charge and at the center -e charge. If one is missing at any corner, we need to find the potential due to other charges. Since potential is a scalar, we need to just add up the potential due to other.

3 +e charges are edge lenght distance to the corner. i.e v1 = 3kq/L (here, k=1/4 * pi* epsilon

3 +e chrage are the diagonal length (side surfaces)i.e v2 = 3 kq/(sqrt(2)*L)

1 +e charge is diagonal to cube i.e v3 = kq/(sqrt(3) * L)

1 -e charge at half diagonal to cube i.e v4 = - 2* kq/(sqrt(3) * L)

Net potential = v1+v2+v3+v4 = kq/L (3 + 3/sqrt(2) + 1/sqrt(3) - 2/sqrt(3)) = (9 * 109 * (1.6 * 10 -19 ))/(3.5 * 10 -10 ) (3+2.12-0.58) = 18.67 V

The total potential energy required is sum of potential energies at the center.

U = 8 * k(+e)(-e)/ half-diagonal distance = - 2 * (8 * k e2 ) / (sqrt(3) * L) = - 60.8 * 10 -19 J

Carefully go through calculations again and convert the answer the form you require.

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