Answer :

Given:

Atomic radius, r = 143. 1 pm

Atomic mass, M = 93 u

Crystal structure = Body centred cubic (BCC)

Atomic scattering factor for BCC, Z = 2

__Formula used:__

**(1)** Atomic radius, . . . . (1)

Where ,

r is atomic radius (pm)

a is length of side of cube(pm)

**(2)** Density of the cell, . . . . . (2)

Where,

Z is atomic scattering factor.

M is atomic mass (u)

A is length of side of cube (pm)

Avagadro no. = 6. 022 × 10^{23} atoms

Putting the value of “r” in equation (1), we get,

Solving for a, we get,

a = 330. 4 pm

Putting the value of ” z”, “m” and “a” in equation (2)

= 8. 58 g/cm^{3}

__Conclusion:__

Density of the crystal is 8. 58 g/cm^{3}

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