If the density of polarization of dimond is \(1.32 \times 10^{-10} \mathrm{C} / \mathrm{m}^{2}\), calculate the shift of the centre of negative cloud of 6 electrons on each atom from the nucleus. The density of the carbon is \(3500 \mathrm{~kg} / \mathrm{m}^{3}\) and the atomic weight is 12 . Ans. \(7.8 \times 10^{-17} \mathrm{~m}\)

To find the number of atoms per unit volume, use the density and the atomic weight. The number of atoms per unit volume, \(n\), can be calculated using the formula: \[ n = \frac{\text{Density}}{\text{Atomic Weight} \times \text{Avogadro's Number}} \] Given: Density \( = 3500 \text{ kg/m}^3 \), Atomic Weight \( = 12 \text{ g/mol} = 12 \times 10^{-3} \text{ kg/mol} \), Avogadro's Number \( = 6.022 \times 10^{23} \text{ atoms/mol} \) Substitute these values into the formula: \[ n = \frac{3500}{12 \times 10^{-3} \times 6.022 \times 10^{23}} \text{ atoms/m}^3 \] Simplify: \[ n \thickapprox 2.917 \times 10^{29} \text{ atoms/m}^3 \]

The given problem provides the density of polarization, \( P = 1.32 \times 10^{-10} \text{ C/m}^2 \). The polarization density is related to the electric dipole moment per atom, \( p \), and the number of atoms per unit volume, \( n \), as follows: \[ P = n \times q \times d \] Here, \( q \) is the charge of 6 electrons, and \( d \) is the shift of the negative cloud. The charge of a single electron is \( 1.6 \times 10^{-19} \text{ C} \); hence, the charge of 6 electrons is: \[ q = 6 \times 1.6 \times 10^{-19} \text{ C} \] Substitute the values to solve for \( d \): \[ 1.32 \times 10^{-10} = (2.917 \times 10^{29}) \times (6 \times 1.6 \times 10^{-19}) \times d \] Then, \[ d = \frac{1.32 \times 10^{-10}}{2.917 \times 10^{29} \times 9.6 \times 10^{-19}} \] Simplify: \[ d \thickapprox 7.8 \times 10^{-17} \text{ m} \]