Question

An electric field is applied on a copper plate such that the electrons are displaced by \(1.1 \times 10^{-18}\,\text{m}\) relative to the nucleus. The electronic polarization (in \(\mu C \, m^{-2}\)) is (rounded off to two decimal places). Given: Atomic number of copper = 29, Copper has FCC crystal structure with lattice parameter = 0.362 nm, Electron charge = \(1.6 \times 10^{-19}\, \text{C}\).

Hint:

Electronic polarization = (charge displaced × displacement × number of electrons) / volume of unit cell.

Answer 1

Correct Answer: 0.35

Step 1: Volume of unit cell.
\[ a = 0.362 \, \text{nm} = 0.362 \times 10^{-9} \, \text{m} \] \[ V_{cell} = a^3 = (0.362 \times 10^{-9})^3 \approx 4.75 \times 10^{-29} \, \text{m}^3 \]

Step 2: Number of atoms per FCC cell.
FCC has 4 atoms per unit cell. Each atom contributes \(Z = 29\) electrons. \[ N_e = 4 \times 29 = 116 \, \text{electrons per unit cell} \]

Step 3: Dipole moment per cell.
Displacement of electrons: \(\delta = 1.1 \times 10^{-18}\, \text{m}\). \[ p_{cell} = N_e \, e \, \delta = 116 \times (1.6 \times 10^{-19})(1.1 \times 10^{-18}) \] \[ p_{cell} = 2.04 \times 10^{-35} \, \text{C·m} \]

Step 4: Polarization.
\[ P = \frac{p_{cell}}{V_{cell}} = \frac{2.04 \times 10^{-35}}{4.75 \times 10^{-29}} \approx 4.29 \times 10^{-7} \, \text{C/m}^2 \] Convert to \(\mu C/m^2\): \[ P = 0.429 \, \mu C/m^2 \] Recheck with rounding: actual result ~ \(\boxed{0.43 \, \mu C/m^2}\).