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Molybdenum (Mo) has a body-centered cubic unit cell with Mo atoms in the corner of the cubic unit cell and one atom in the body center.

The body-centered cubic unit cell consists of 8 atoms in the corner, thus one atom contributing 1/8 to the unit cell and one atom in the body-center

Thus, each unit cell consists of $8脳\left(\frac{1}{8}\right)+1=2$, i.e., two atoms in the unit cell.

As in BCC, in total two atoms contribute for the unit cell. The mass of the unit cell would be the twice as the mass of single atom of Mo.

Now, calculating the volume of the atom and thus the radius of the atom,

$$\begin{gathered} \mathrm{Volume}=\frac{\mathrm{Mass}}{\mathrm{Density}} \\ \frac{4}{3}{\mathrm{蟺谤}}^3=\frac{95.95}{\left(10.28脳6.023脳{10}^{23}\right)} \\ {\mathrm{r}}^3=0.37脳{10}^{-23}{\mathrm{cm}}^3 \\ \mathrm{r}=1.54脳{10}^{-10}\mathrm{m} \end{gathered}$$

The atomic radius of molybdenum is 1.54 呛

Now, considering the body-centered cubic unit cell

Here, the diagonal of the lattice (edge length is denoted by a) will be equal to twice the diameter of the atom, thus

$$\begin{gathered} \sqrt{3}\mathrm{a}=2脳2\mathrm{r} \\ \mathrm{a}=\frac{\left(4脳1.54脳{10}^{-10}\right)}{\sqrt{3}\mathrm{m}} \\ \mathrm{a}=3.55脳{10}^{-10}\mathrm{m} \end{gathered}$$