# Niobium crystalli

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 ,

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 × 1023 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/cm3

Conclusion:

Density of the crystal is 8. 58 g/cm3

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