An element having bcc geometry has atomic mass 50 and edge length is 290 pm. The density of unit cell is

Asked by Manoj Kumar | 22nd Jan, 2011, 07:50: PM

Expert Answer:

Dear Student
 
Length of the edge , a = 290 pm =290 x 10-10 cm 
Volume of unit cell = ( 290 x 10-10 cm )3 = 24.39 x 10-24 cm3

Since it is bcc arrangement, 
Number of atoms in the unit cell, Z = 2 
Atomic mass of the element = 50 

Mass of the atom = atomic mass/ Avogadro number = M/No = 50/6.02 x 1023

Mass of the unit cell = Z x M/No = 2 x 50/6.02 x 1023 = 100/6.23 x 1023
Therefore , density = mass of unit cell / volume of unit cell 
= 100/6.023 x 1023 x 24.39 x 10-24 = 6.81 g cm-3
 
We hope that clarifies your query.
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Answered by  | 23rd Jan, 2011, 04:59: PM

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