In diamond- carbon atom occupy FCC lattice points as well as alternate tetrahedral voids- If edge length -160-of the unit cell is - 356- pm- then the radius -in pm- of carbon atom is- 154-14- 77-07- 251-70- 89-20-

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Density

In diamond, c...

Question

In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is $356$ pm, then the radius (in pm) of carbon atom is:

77.07

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

154.14

No worries! We‘ve got your back. Try BYJU‘S free classes today!

251.70

No worries! We‘ve got your back. Try BYJU‘S free classes today!

89.20

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $77.07$
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 $\sqrt{3} \times a$ .
Thus $R = \frac{\sqrt{3} \times a}{8} = \frac{356 \times 1.732}{8} = 77.07$ pm

Suggest Corrections

Similar questions

In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is 356 pm, then radius of carbon atom is

Q. In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is

356 p m

, then the radius of carbon atom is :

Q. In the diamond cubic structure, carbon atoms form an FCC while the other carbon atoms occupy alternate tetrahedral voids, the atoms being connected by covalent bonds. Then the packing efficiency is approximately

Q. If in diamond, there is unit cell of carbon atoms as fcc and if carbon atom is

s p^{3}

hybridized what fractions of void occupied by carbon atom?

Q. Calculate the radius of tetrahedral void present in the unit cell of an atom X which crystallizes in fcc unit cell having an edge length

a = 50 \sqrt{2} p m

Join BYJU'S Learning Program

In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is 356 pm, then the radius (in pm) of carbon atom is:

The correct option is B 77.07In 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×a8=356×1.7328=77.07 pm

In diamond, carbon atom occupy FCC lattice points as well as alternate tetrahedral voids. If edge length of the unit cell is $356$ pm, then the radius (in pm) of carbon atom is: