A solid has a '$$bcc$$' structure. If the distance of nearest approach between two atoms is $$1.73\ \text{Å}$$, the edge length of the cell is

For a body-centred cubic ($$bcc$$) lattice, the atoms touch each other along the body diagonal of the cube.

Let $$a$$ be the edge length of the cube and $$r$$ be the atomic radius.

Length of the body diagonal = $$\sqrt{3}\,a \quad -(1)$$

Along this diagonal we have the sequence: corner atom (radius $$r$$) → body-centre atom (radius $$r$$) → opposite corner atom (radius $$r$$).
Hence the total length occupied by the three touching atoms is $$4r$$.

Equating the two expressions for the body diagonal:
$$\sqrt{3}\,a = 4r \quad -(2)$$

The nearest-neighbour distance in $$bcc$$ is the distance between the body-centre atom and any corner atom, i.e. half of the body diagonal:
nearest distance $$= \dfrac{\sqrt{3}\,a}{2} = 2r \quad -(3)$$

The problem gives this nearest distance as $$1.73\ \text{Å}$$.
Since $$1\ \text{Å} = 100\ \text{pm}$$, we have
$$2r = 1.73\ \text{Å} = 1.73 \times 100\ \text{pm} = 173\ \text{pm}$$

Therefore,
$$r = \dfrac{173}{2}\ \text{pm} = 86.5\ \text{pm}$$

Substituting $$r$$ into equation (2):
$$a = \dfrac{4r}{\sqrt{3}} = \dfrac{4 \times 86.5}{\sqrt{3}}\ \text{pm}$$

Using $$\sqrt{3} \approx 1.732$$:
$$a \approx \dfrac{346}{1.732}\ \text{pm} \approx 199.8\ \text{pm} \approx 200\ \text{pm}$$

Hence the edge length of the cubic unit cell is approximately $$200\ \text{pm}$$.

Option C which is: 200 pm