5.2 Problems 295
5.10 The energy of interaction of two atoms a distancerapart can be written as:
E(r)=−a
r+
b
r^7
whereaandbare constants.
(a) Show that for the particles to be in equilibrium,r=ro=(7b/a)^1 /^6
(b) In stable equilibrium, show that the energy of attraction is seven times
that of the repulsion in contrast to the forces of attraction and repulsion
being equal.5.11 In Problem 5.10, if the two atoms are pulled apart, show that they will separate
most easily whenr=(28b/a)^1 /^6.
5.12 Let the interaction energy between two atoms be given by:
E(r)=−A
r^2+
B
r^8
If the atoms form a stable molecule with an inter-nuclear distance of 0.4 nm
and a dissociation energy of 3 eV, calculateAandB.5.13 Lead is a fcc with lattice constant 4. 94 A. Lead melts when the average ampli- ̊
tude of its atomic vibrations is 0. 46 A. Assuming that for lead the Young’s ̊
modulus is 1. 6 × 1010 N/m^2 , find the melting point of lead.
5.2.3 Metals ...........................................
5.14 Take the Fermi energy of silver to be 5.52 eV.
(a) Find the corresponding velocity of conduction electron.
(b) If the resistivity of silver at room temperature is 1. 62 × 10 −^8 Ωm estimate
the average time between collisions.
(c) Determine the mean free path. Assume the number of conduction electrons
as 5. 86 × 1028 m−^3.5.15 Find the drift velocity of electron subjected to an electric field of 20 Vm−^1 ,
given that the inter-collision time is 10−^14 s.
5.16 Aluminum is trivalent with atomic weight 27 and density 2.7g/cm^3 , while the
mean collision time between electrons is 4× 10 −^14 s. Calculate the current
flowing through an aluminum wire 20 m long and 2 mm^2 cross-sectional area
when a potential of 3 V is applied to its ends.
5.17 Given that the conductivity of sodium is 2. 17 × 107 Ω−^1 m−^1 , calculate:
(a) The inter-collision time at 300 K, and
(b) The drift velocity in a field of 200 Vm−^1.5.18 Given that the inter-collision time in copper is 2. 3 × 10 −^14 s, calculate its
thermal conductivity at 300 K. Assume the Wiedemann-Franz constant is
CWF= 2. 31 × 10 −^8 WΩK−^2.