- #1

jazznaz

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## Homework Statement

When a uniform electric field is applied to diamond, the induced dipole moment per unit volume (ie. polarisation) is [tex]P_{0} = 1.5 \times 10^{-7} \text{C m}^{-2}[/tex]. Given that the density of diamond is [tex]3500 \text{kg m}^{-3}[/tex], 1 kmole of diamond has a mass of 12kg. Avogadro's constant is [tex]6 \times 10^{26} \text{kmol}^{-1}[/tex], the atomic number of carbon is 6.

How many carbon atoms are there per cubic metre?

What is the average induced dipole moment?

Estimate the average separation of the +ve and -ve charges.

If [tex]P = P_{0} \sin \omega t[/tex] as a result of the alternating electric field, evaluate the peak value of the resultant polarisation current density, J, at a frequency of [tex]10^{12} \text{Hz}[/tex]

## The Attempt at a Solution

First part,

[tex]\rho = \frac{m}{v} = 3500[/tex]

[tex]m = 3500[/tex]

[tex]m/0.012 = 291666.67 \text{moles}[/tex]

[tex]291666.67 \times 6 \times 10^{23} = 1.8 \times 10^{29} \text{atoms m}^{-3}[/tex]

Second part,

Average induced dipole moment is total polarisation divided by number of atoms,

[tex]= \frac{1.5 \times 10^{7}}{1.8 \times 10^{29}}= 8.3 \times 10^{-23} \text{C m}[/tex]

Third part,

Dipole moment is [tex]\vec{p} = q\vec{L}[/tex]

So [tex]L = \frac{p}{q} = \frac{8.3 \times 10^{-23}}{6 \times 1.6 \times 10^{-19}} = 8.6 \times 10^{-5} \text{m}[/tex]

Which strikes me as far too small to be an estimate of the atomic radius of a carbon atom in the diamond lattice. I can't see where I've gone wrong, maybe I've missed a subtlety in the question?

Any comments would be great, thanks...