CBSE Class 12-science Answered
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 | 22 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
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.
Regards
Team
Topperlearning
Answered by | 23 Jan, 2011, 04:59: PM
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