Q1. PART A
- a) A metallic element has a density of 2.70 g cm⁻³, a lattice constant of 4.0495 Å and an atomic weight 26.9815g mol⁻¹. Calculate the number of atoms per unit cell of this element and predict its lattice crystal structure. (150-200 words)
- b) Show that the reciprocal lattice of a bcc lattice is an fcc lattice. Calculate the magnitude of the shortest non-zero reciprocal lattice vector. (150-200 words)
- c) A metallic crystal has an fcc lattice with a lattice constant 0.4 nm. Explain whether the following planes are allowed or forbidden for X-ray diffraction: (100), (111), (210), (220) Calculate the X-ray diffraction angles for the allowed planes. Assume that diffraction occurs in the first order and the X-ray wavelength is 0.154 nm. (150-200 words)
Key Points:
- Used the density formula ρ = (n * M) / (V * N_A) to find 'n'.
- Converted lattice constant Å to cm and calculated unit cell volume V = a³.
- Calculated 'n' (number of atoms per unit cell) to be approximately 4.
- Predicted the crystal structure as Face-Centered Cubic (FCC) based on n=4.
Answer: ## a A metallic element has a density of 2.70 g cm⁻³, a lattice constant of 4.0495 Å and an atomic weight 26.9815g mol⁻¹. Calculate the number of atoms per unit cell of this element and predict its lattice crystal structure. The number of atoms per unit cell, 'n', is found using the relation ρ = (n * M) / (V * N_A), where ρ is density, M is atomic weight, V is unit cell volume, and N_A is Avogadro's number. First, the lattice constant a = 4. 0495 Å is converted to cm: a = 4. 0495 × 10⁻⁸ cm. As...