in diamond carbon atoms occupy fcc lattice points as well as alternate tetrahedral voids if edge length of unit cell is 356 picometre then radius of carbon atom is

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Solution

Dear student,

In a diamond structure, there are atoms at each corner and alternate tetrahedral voids.
Consider the body diagonal of unit cell (cube with side a).
It corresponds to 8R, and length of body diagonal is $\sqrt{3}$ × a.
Thus R= $\sqrt{3}$ ×a / 8 = 356 × 1.732 / 8=77.07 pm

Regards,

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in diamond carbon atoms occupy fcc lattice points as well as alternate tetrahedral voids if edge length of unit cell is 356 picometre then radius of carbon atom is

Dear student, In a diamond structure, there are atoms at each corner and alternate tetrahedral voids. Consider the body diagonal of unit cell (cube with side a). It corresponds to 8R, and length of body diagonal is 3× a. Thus R=3×a / 8 = 356 × 1.732 / 8=77.07 pm Regards,

Dear student,

In a diamond structure, there are atoms at each corner and alternate tetrahedral voids.
Consider the body diagonal of unit cell (cube with side a).
It corresponds to 8R, and length of body diagonal is $\sqrt{3}$ × a.
Thus R= $\sqrt{3}$ ×a / 8 = 356 × 1.732 / 8=77.07 pm

Regards,