Q. 133.5( 2 Votes )

Niobium crystalli

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 × 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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