Answers to questions on superconducting metals & X-ray crystallography

Answers to Superconducting metals & X-ray crystallography questions.

1. (a) i. The number of closest neighbouring atoms surrounding each atom.
ii. The number of coordinate bonds formed by the ligand(s) to the transition metal ion.

(b) I. Simple cubic cell, number of atoms 1, coordination number 6.
II. Body-centred cubic unit cell, number of atoms 2, coordination number 8.
III. Face-centred cubic unit cell, number of atoms 4, coordination number 12.

2. Volume of one unit cell = (3.613 x 10^-10)³ = 4.716 x 10^-29 m³ = 4.716 x 10^-23 cm³
Mass of one atom of copper = 63.55 ÷ 6.02 x 10²³ = 1.056 x 10^-22 g
Each FCC unit cell contains four atoms
Density = mass ÷ volume = 4 x 1.056 x 10^-22 ÷ 4.716 x 10^-23 = 8.96 g cm^-3

3. (a) Type 1 as there is a sudden change to superconductivity.

(b) x is the critical temperature below which the material is superconducting

(c) Below the critical temperature the Cooper pairs of electrons act as a single entity and are able to move freely through the superconductor.

(d) As it still has resistance at zero Kelvin it is a normal conductor.

4. (a) Atomic mass of gold = 196.97 g mol^-1
Volume of one mol of gold = 196.97 ÷ 19.32 = 10.195 cm³ = 1.0195 x 10^-5 m³
Since the atoms only occupy 74% of available space the volume occupied by one mol of gold atoms = 74/100 x 1.0195 x 10^-5 = 7.544 x 10^-6 m³
Volume of one gold atom = 7.544 x 10^-6 ÷ 6.02 x 10²³ = 1.253 x 10^-29 m³
(Radius of a gold atom)³ = (1.253 x 10^-29 x ¾ ) ÷ π = 2.991 x 10^-30 m³
Radius of a gold atom = 1.44 x 10^-10 m

(b) This is about 11% bigger than the covalent radius of 1.30 x 10^-10 m.

Answers to Superconducting metals & X-ray crystallography questions