(Model answers will be posted shortly after the due date.)

Question 2.12

Calculate the force of attraction between a Ca²⁺ ion and an O^2- ion, the centres of which are 1.25 nm apart.

Answer 2.12

This just involves plugging numbers into the formula

F = q₁q₂/(4 pi epsilon r²)

The numbers to plug in are: q1 = q2 - 3.2 * 10^-19 coulombs, r=1.25 * 10^-9 m and epsilon = 8.85 * 10^-12F/m, giving the result F = 5.89 * 10^-10 N .

Question 2.13

The net potential energy between two adjacent ions may be represented as

E= - A/r + B/rⁿ (2.11)

Calculate the bonding energy E₀ in terms of the parameters A, B, and n, using the following procedure:

1. Differentiate E with respect to r and set the resulting expression equal to zero.
2. Solve for r₀, the equilibrium interatomic spacing, in terms of A, B and n.
3. Obtain an expression for E₀ by substituting r₀ into Equation 2.11.

Answer 2.13

Differentiating and setting equal to zero gives us

Arⁿ⁺¹=nBr²

from which we deduce that

r₀ = (nB/A)^(1/(n-1))

and

E₀=-A/r₀ + B/r₀ⁿ

Question 2.16

The net potential energy E_N between two adjacent ions is sometimes represented by the expression

E = - C/r + D exp(-r / p) (2.12)

where r is the interionic separation, and C, D and p are constants depending on the specific ions.

1. Derive an expression for the bonding energy E₀ in terms of the equilibrium ionic separation r₀ and the constants D and p, using the following procedure:
   1. Differentiate E with respect to r and set the resulting expression equal to zero.
   2. Solve for C, in terms of D, p and r₀, the equilibrium interatomic spacing.
   3. Obtain an expression for E₀ by substituting for C in Equation 2.12.

2. Now derive another expression for E₀ in terms of r₀ and the constants C and p, using similar means.

Answer 2.16

dE_N/dr=C/r₀²-(D/p)e^-r₀/p=0

Hence C = (Dr₀²/p)(e^-r₀/p)

and E₀=(-De^-r₀/p)(r₀/p+1)

Similarly, D = Cp/(r₀²e^-r₀/p)

and hence

E₀=(C/r₀)(p/r₀-1)

Question 3.2

What is the difference between a crystal structure and a crystal system?

A crystal structure is any regular arrangement of atoms, ions, or molecules. Very many different crystal structures exist.

A crystal system is a set of crystal structures having similar values for the six lattice parameters; for example, the cubic crystal system is one in which the first three lattice parameters all take the same value. There are seven crystal systems, each corresponding to a number of crystal structures.

Question 3.11

Some hypothetical metal has a simple cubic crystal structure. If its atomic weight is 74.5 grams/mol and its atomic radius is 0.145 nanometres, calculate its density.

Answer 3.11

The unit cell of a simple cubic crystal structure has an edge length of 2R, where R is the atomic radius. A unit cell contains one atom. So the density of this solid is 74.5 amu per (2 * 0.145)³ cubic nanometres.

1 amu is 1.66 * 10^-27kg, so this density is equal to:

74.5 * 1.66 /(2 * 0.145)^3 = 5070 kg/cubic metre

(This looks like the right order of magnitude; the density of water is 1000 kg/cubic metre, while the density of iron is about 7800 kg/cubic metre; so this is a metal that's a bit less dense than iron.)

Question 3.14

Niobium has an atomic radius of 0.143 nm and a density of 8.57 grams per cubic centimeter. Determine whether it has a FCC or BCC crystal structure.

Answer 3.14

Assume it has an FCC structure. The unit cell of an FCC crystal contains 4 atoms, and the length of side of the unit cell is a = 2^1.5R, where R is the atomic radius. So the density is 4 atoms in a volume of 2^4.5R³. Looking up the atomic weight of niobium in the back of the book, we find it is 92.9 amu. So the density would be:

92.9 * 4 * 1.66 * 10^-27/(2^4.5 * 0.143³ * 10^-27) kg/cubic meter

= 9322 kg/cubic metre.

This is higher than 8570 kg/cubic metre, which is the actual density, so our initial assumption of an FCC structure must be wrong. We should recalculate assuming a BCC structure. However, rather than do this, we note that the packing fractions for FCC and BCC structures are 0.74 and 0.68 respectively. So if niobium is BCC, its density should be 9322 * (0.68/0.74) = 8566 kg/cubic metre, which matches the official value.

Question 3.48

1. Derive planar density expressions for the BCC (100) and (110) planes in terms of the atomic radius R.
2. Compute and compare planar density values for these same two planes for molybdenum (which has an atomic radius of 0.136 nm).

Answer 3.48

I didn't clarify the distinction between planar density, which is measured in atoms per square metre, and fractional planar density, which is a dimensionless number, until Monday's lecture; so I'll accept either quantity as an answer to this question.

For the BCC unit cell, the length of cell side is a, where 3^0.5a = 4R, and R is the atomic radius. The 100 plane of the BCC structure is a square, side length a, containing 1 atom (a quarter of an atom from each of the four corner atoms). So the planar density is 1 atom per a² square meters, or 1 atom per ((16/3)R²) square meters. The fractional planar density is pi R^2/((16/3)R²) square meters, or (3 pi / 16), or 0.589.

The 110 plane of the BCC structure is a rectangle, a by 1.414 a, and it contains 2 atoms. So the planar density is 2 atoms per (1.414 * (16/3) * R^2) square meters, or 1 atom per 3.77R^2 square meters. The fractional planar density is pi R^2 / 3.77 R^2 = 0.833.

So we can conclude right away that the 110 plane is more densely packed for any BCC crystal, including molybdenum. Specifically, the planar density of the (100) plane is 1.0137 * 10 ¹⁹ atoms per square metre, and the planar density of the (110) plane is 1.433 * 10¹⁹ atoms per square metre.