What is the minimum number of atoms that could be contained in the unit cell of an element with a body-centered cubic lattice? (a) \(1,(\mathbf{b}) 2,(\mathbf{c}) 3,(\mathbf{d}) 4,(\mathbf{e}) 5\)

A body-centered cubic lattice is a three-dimensional arrangement of atoms where there is an atom at each corner of the cubic unit cell, and an additional atom at the center of the cube.

In a BCC unit cell, there are a total of 9 atoms present. However, only some portion of each corner atom is present inside the unit cell, while the central atom is entirely inside the unit cell. To determine the number of corner atoms that contribute to the unit cell, we take into account only the fraction of each corner atom present in the unit cell. There are 8 corner atoms, and each corner atom contributes only 1/8th of its volume to the unit cell because it is shared between 8 neighboring unit cells. Therefore, the contribution of the 8 corner atoms to the unit cell is: \(8 \times \dfrac{1}{8} = 1\) And there is 1 additional atom at the center which contributes fully to the unit cell. So, the total contribution of atoms to the unit cell is: \(1 + 1 = 2\)

The minimum number of atoms contained in the unit cell of an element with a body-centered cubic lattice is 2. Therefore, the correct answer is (b) 2.