Question
Assume a hypothetical cubic crystal lattice, named JEE-centered cubic (jcc) with the following characteristics:
I. An atom is present at all the corners of the cube
II. An atom is present at the center of two pairs of opposite faces
III. An atom is present at the center of all the edges of the cube
IV. One atom is present at its body-center
An element having the jcc lattice structure has a cell edge of 120 pm. The density of the element is 6.8 g/cm3. How many atoms are present in 408 g of the element?
I. An atom is present at all the corners of the cube
II. An atom is present at the center of two pairs of opposite faces
III. An atom is present at the center of all the edges of the cube
IV. One atom is present at its body-center
An element having the jcc lattice structure has a cell edge of 120 pm. The density of the element is 6.8 g/cm3. How many atoms are present in 408 g of the element?
Answer: Option C
:
C
Volume of unit cell =(120pm)3=(120×10−12)3m3=(123×10−33)m3
Volume of 408 g of the element =massdensity=4086.8=60cm3=6×10−5m3
So, number of unit cells present in 408 g of the elements =6×10−5123×10−33=3.472×1025 unit cells
Since each jcc unit cell consist of 7 atoms,
therefore the total number of atoms presents in 408 g of the given element
=7×3.472×1025=2.43×1026atoms
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:
C
Volume of unit cell =(120pm)3=(120×10−12)3m3=(123×10−33)m3
Volume of 408 g of the element =massdensity=4086.8=60cm3=6×10−5m3
So, number of unit cells present in 408 g of the elements =6×10−5123×10−33=3.472×1025 unit cells
Since each jcc unit cell consist of 7 atoms,
therefore the total number of atoms presents in 408 g of the given element
=7×3.472×1025=2.43×1026atoms
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